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Vol. XXI

No. 2

February 2013

Corporate Office: Plot 99, Sector 44 Institutional area, Gurgaon -122 003 (HR), Tel : 0124-4951200

Regd. Office 406, Taj Apartment, Near Safdarjung Hospital, Ring Road, New Delhi - 110029. e-mail : [email protected] website : www.mtg.in

Managing Editor : Mahabir Singh Editor : Anil Ahlawat (BE, MBA) Hony. Advisor : Dr. S. Malhotra Director, Delhi Public School, Faridabad (HR)

Contents Full Length JEE Advanced Practice Paper : 2013 6



Examiner’s Mind

17

Thought Provoking Problems

20

(SHM and Waves) Interview

24

NEET/PMTs Practice Paper : 2013

25

NCERT Xtract Questions for NEET

29

CBSE Board Sample Paper 2013



Class XII

36

JEE Main Practice Paper : 2013

42

Brain Map

48

Exam prep : Chapterwise MCQs

51 59

Essential Formulae for Competitive Exams 77

India Progress in Science

M

any Indian scientists are working in NASA and many reputed universities and laboratories in USA, England and Europe including CERN. One works there, gets a lot of experience and settles there because of better facilities for working. In Darmstadt, Germany, GSI Helmholtz Centre for Heavy Ion Research, is being upgraded, to become Facility for Antiproton and Ion Research (FAIR). FAIR will feature a double-ring synchrotron with a circumference of 1100 metres connected to the existing facility at GSI. We are very much involved in this work, not for just working there but as a partner. We have a variable Energy Cyclotron Centre Salt-Lake, Calcutta. Our scientists have ample experience in this field. Our scientists are making advanced components for FAIR as India is also a partner in this venture. This facility is more powerful than even the one at cern, in Geneva – The large Hadron Collider which is trying to detect the Higgs Boson or ‘God particle’. Although the energy of the particles attained in FAIR, is much lower than that of the Geneva facility, the intensity attained will be higher than that of any other facility. To create heavy particles by collision, one needs very high energy. This is the reason why there is a race for making more and more powerful cyclotrons. Energy is a very important factor. If energy is less, one cannot do the experiments that are done by very powerful accelerators. With the available energy, if one can make a very intense beam, the measurements made will be more reliable in the given energy range. The new detector can also detect collisions of the order of 10 million per second whereas the other ones can detect 10,000 collisions per second. If the collisions are more, more the new particles produced and less the error of determination. Both energy and intensity are important. These facilities can also be used for irradiation and other uses ranging from agriculture to atomic physics. We wish all the scientists working with FAIR a very bright future. The work done by every scientist in every cormer of the world is for the benefit of the whole humanity. Anil Ahlawat, Editor

for Practise NEET 2013 MBBS Practice Paper Class XI Owned, Printed and Published by Mahabir Singh from 406, Taj Apartment, New Delhi - 29 and printed by Personal Graphics and Advertisers (P) Ltd., Okhla Industrial Area, Phase-II, New Delhi. Readers are adviced to make appropriate thorough enquiries before acting upon any advertisements published in this magazine. Focus/ Infocus features are marketing incentives MTG does not vouch or subscribe to the claims and representations made by advertisers. All disputes are subject to Delhi jurisdiction only. Editor : Anil Ahlawat Copyright© MTG Learning Media (P) Ltd. All rights reserved. Reproduction in any form is prohibited.

rial

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Individual Subscription Rates Combined Subscription Rates 1 yr. 2 yrs. 3 yrs. 1 yr. 2 yrs. 3 yrs. Mathematics Today 300 500 675 PCM 800 1200 1700 Chemistry Today 300 500 675 PCB 800 1200 1700 Physics For You 300 500 675 PCMB 900 1500 2100 Biology Today 300 500 675 Send D.D/M.O in favour of MTG Learning Media (P) Ltd. Payments should be made directly to : MTG Learning Media (P) Ltd, Plot No. 99, Sector 44 Institutional Area, Gurgaon - 122003 (Haryana) We have not appointed any subscription agent.

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SECTION - I Straight Objective Type This section contains 6 multiple choice questions numbered 1 to 6. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. 1. Two cylinders of same cross-section and length L but made of two material of densities d1 and d2 are cemented together to form a cylinder of length 2L. The combination floats in a liquid of density d with a length L/2 above the surface of liquid. If d1 < d2 then 3 d (b) > d1 (a) d1 < d 4 2 d (c) > d1 (d) d < d1 4 2. A plank of mass m is placed over smooth inclined plane and a sphere is placed on plank as shown. There is sufficient friction between sphere and plank to prevent slipping. If system is released from rest, the frictional force  on sphere is (a) up the plane (b) down the plane (c) zero (d) horizontal 3



A transverse wave y = 0.05sin(20px – 50pt) m, is propagating along +ve x-axis on a string (x is in cm). A light insect starts crawling on the string with velocity 5 cm s–1 at t = 0 along +ve x-axis from a point where x = 5 cm. The difference in phase between its position at t = 5 s and its position at t = 0 is (a) 15p (b) 250p (c) –24p (d) –5p

4. Two point charges q and 2q are placed (a, 0) and (0, a). A point charge q1 is placed at point P on the quarter circle of radius a as shown in figure. If electric field at origin is zero, then  a 2  (a) point P is  , a 2q  3 3   a 2a  (b) point P is  ,  (c) q1 = –3q  5 5  (d) none of these

P q1

O

a

5. Consider a Young’s double slit set up as shown. The slits have equal width. Take O as origin. If λD λD and y2 = + average intensity between y1 = − 4d 4d equals n times the intensity of maxima, then n is

1 2  1 + π  2   2 (c) 1 +  π   (a)

 2 (b) 2  1 +  π  1 2 (d)  1 −  2 π

6. Optic axis of a thin equiconvex lens is the x-axis, the co-ordinate of a point object and its image are (– 40 cm, 1 cm) and (50 cm, –2 cm) respectively. Lens is located at x = (a) + 20 cm (b) – 30 cm (c) – 10 cm (d) 0 cm

By Momentum : JABALPUR : (0761) 2400022, NAGPUR : (0712) 3221105, GWALIOR : (0751) 3205976.

6

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WorldMags.net SECTION - II

Multiple Correct Answer Type This section contains 4 multiple choice questions numbered 7 to 10. Each question has 4 choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct. 1 7. The least count of a stop watch is s. Two 10 persons A and B use this watch to measure the time period of an oscillating pendulum. Person A takes the time period of 50 oscillations, while B takes the reading of 100 oscillations. Neglecting all other sources of error we can say that (a) accuracy in measurement by A is equal to that of B. (b) accuracy in measurement by A is less than that of B. (c) absolute error in measurement by A is greater than that of B. (d) absolute error in measurement by A is equal to that of B. 8. A light rope passes over a light frictionless pulley attached to the ceiling. An object with a large mass is tied to one end and an object with a smaller mass is tied to the other end.

Both masses are released from rest. Which of the following statement(s) is/are false for the system consisting of the two moving masses while string remains taut? (a) The center of mass remains at rest. (b) The net external force is zero. (c) The velocity of the center of mass is constant. (d) The acceleration of the center of mass is g downward.

9. An artificial satellite is moving in a circular orbit around the earth. If the universal gravitational constant start decreasing at time t = 0 at a constant rate with respect to time t. Then satellite has its (a) path gradually spiraling out away from the centre of the earth. (b) path gradually spiraling in towards the centre of the earth. (c) angular momentum about the centre of the earth remains constant. (d) potential energy increases. 10. In the network shown in figure, points A, B and C are at potential 70 V, 0 V and 10 V respectively. Choose the correct alternatives. 8

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70 V A

0V B

10  D

 20 30  10 V C

(a) Point D is at a potential of 40 V. (b) The currents in the sections AD, DB, and DC are in the ratio 3 : 2 : 1. (c) The currents in the sections AD, DB and DC are in the ratio 1 : 2 : 3. (d) The network draws a total power of 200 W.

SECTION - III Linked Comprehension Type This section contains 2 paragraphs P11-13 and P14-16. Based upon each paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. P11-13 : Paragraph for Question Nos. 11 to 13 You are on a summer tour to a remote hill station. You do not have facility of an electric heater and you want some ice. Unfortunately, the air temperature drops to only 6°C during night. Being a physicist you know that a clear, moonless night sky acts like a black body radiator at a temperature of TS = – 23°C, and decide to make ice by letting water radiate energy to such a sky. You take a cylindrical container, thermally insulated from ground and pour 4.5 g of water in it. The cross-section of container is 10 cm2. Assume that absorptivity and emissivity of water surface are same and neglect the presence of atmosphere. Also assume the average temperature of water to be 2°C. Stefan’s constant s = 5.0 × 10–8 W m–2 K–4 Emissivity of water e = 1.0 Specific heat of water, s = 4190 J kg–1 K–1 Latent heat of fusion of ice L = 3.33 × 105 J kg–1 Take 254 = 4 × 105 and 114 – 104 = 5000 11. What is the total loss in heat for the above sample of water to freeze ? (a) 1612 J (b) 1512 J (c) 1132 J (d) 1499 J 12. What is the approximate rate of energy loss by the water sample ? (a) 100 mJ s–1 (b) 8 mJ s–1 (c) 20 mJ s–1 (d) None of these 13. Is it possible to freeze the water sample during one night ? (a) Yes (b) No (c) Cannot be predicted (d) None of these

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P14-16 : Paragraph for Question Nos. 14 to 16 M is kept over another block of A block of mass 2 mass M and the heavier block is kept over smooth

horizontal surface as shown in figure. The heavier block is long enough such that the smaller block does not fall from it during the subsequent motion and the coefficient of friction between the blocks is M m. Suddenly block of mass is given an impulse J 2 in horizontal direction.

14. Which of the following is correct? (a) The friction between the blocks will be impulsive. (b) The friction between the blocks will be non-impulsive. (c) N o r m a l b e t we e n t h e b l o c k s w i l l b e impulsive. (d) None of these. 15. The work done by impulsive force will be (a)

J2 M

(b)

2J2 M

1 J2 (d) none of these 2M 16. The amount of heat evolved during the subsequent motion of the blocks is J2 J2 (b) 2 M (a) M 2J2 (c) (d) none of these 3M Section - iV Integer Answer Type (c)

This section contains 7 questions numbered 17 to 23 The answer to each of the questions is a single digit integer, ranging from 0 to 9. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following.

X 0 1 2 3 4 5 6 7 8 9

Y 0 1 2 3 4 5 6 7 8 9

Z 0 1 2 3 4 5 6 7 8 9

W 0 1 2 3 4 5 6 7 8 9

17. A barrier AB of length 12 m is hinged at A. At the lower end a horizontal spring keeps the barrier closed. The height of water is 6 m and the width of the barrier is 5 m. Water level is 4 m below the hinge A. If spring constant is 6 × 105 N m–1, find elongation of spring (in metre). 10 Physics for you |

feBruary ‘13

4m Water

A

6m

5m B

18. A wire loop ABCD is divided into two parts. AB = AD = 2 m, BC = CD = BD = 1 m. The wire loop is lying on x-y plane. The resistance per unit length of the wire is 0.2 W m–1. There exist time  ^ dependent magnetic field B = ( 2.5t) k tesla. Find the approximate value of current (in A) through part AB. Take 3 = 1.68 and

45 = 7 .

B A

C D

19. Electrons in H-like atom (Z = 3) make transitions from 4 th excited state to 3 rd excited state and from 3rd excited state to 2nd excited state. The resulting radiation are incident on a metal plate to eject photoelectrons. The stopping potential for photoelectrons ejected by the shorter wavelength is 4.20 V. What is stopping potential (in volts) for the photoelectrons ejected by longer wavelength ? 20. In a circus act, a 4 kg dog is trained to jump from B cart to A cart and then immediately back to the B cart. The carts each have a mass of 20 kg and they are initially at rest. In both cases the dog jumps at 6 m s–1 relative to the cart. If the cart moves along the same line with negligible friction, If the final magnitude of velocity of cart B with respect to the floor is x/36 then find the value of x.

21. Glycerine is filled in 25 mm wide space between two large plane horizontal surfaces. A thin plate of area 0.75 m2 at a distance of 10 mm from one of the surfaces is in horizontal position between the plates inside the glycerine. It is dragged horizontally at a constant speed of 0.5 m s –1 . Take coefficient of viscosity h = 0 . 5 N s m –2. I f t h e f o r c e r e q u i r e d to drag the plate at constant speed is 125X / 4 newton, find the value of X.

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Fixed surface

10 mm 15 mm

22. A cylinder of height h, diameter h/2 and mass M and with a homogeneous mass distribution is placed on a horizontal table. One end of a string running over a pulley is fastened to the top of the cylinder, a body of mass m is hung from the other end and the system is released. Friction is negligible everywhere. At what minimum ratio m/M will the cylinder tilt ? h/2



the rate of P, that may be varied. The rate of water in flow from tap is, F = 0.021 litres per minute. The heat generated is sufficient so that the water in the container is boiling and getting converted into steam at a steady rate. What is the minimum power P (in kW) that must be generated as heat in the steady state in resistor R so that the amount of liquid water in the container neither increases nor decreases with time? (Neglect other losses of heat, such as conduction from the container to the air and heat capacity of container) For water, specific heat s = 4.2 kJ kg–1 K–1, latent heat of vaporisation Lvap = 2.3 MJ kg–1, density r = 1000 kg m–3. Mark your answer in nearest integer.

h M

T

m

23. Water temperature 20°C flows from a tap T into a heated container C. The container has a heating element (a resistor R) which is generating heat at

R

chemistry

SECTION - I Straight Objective Type This section contains 6 multiple choice questions numbered 24 to 29. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. 24. The passage of a constant current through a solution of dilute H 2SO 4 with ‘Pt’ electrodes liberated 336 cm3 of a mixture of H2 and O2 at S.T.P. The quantity of electricity that was passed is (a) 96500 C (b) 965 C (c) 1930 C (d) 100 faraday 25. Amongst [Co(ox)3]3–, [CoF6]3– and [Co(NH3)6]3+ (a) [Co(ox)3]3– and [CoF6]3– are paramagnetic and [Co(NH3)6]3+ is diamagnetic. (b) [Co(ox)3]3– and [Co(NH3)6]3+ are paramagnetic and [CoF6]3– is diamagnetic. (c) [Co(ox)3]3– and [Co(NH3)6]3+ are diamagnetic and [CoF6]3– is paramagnetic. (d) [Co(NH3)6]3+ and [CoF6]3– are paramagnetic and [Co(ox)3]3– is diamagnetic. Ph

26. Ph OHC

Ph Ph

Ph

(a)

HOH2C

CHO

The final product is

Conc. H2SO4

HO

Ph

(c)

Ph (b) COOH

Ph

Ph

Ph

Ph

(d)

O

Ph

Ph

Ph

Ph CH2

H2C

Ph

O

Ph



Ph Ph O

O

27. Two 1 st order reactions have half-lives in the ratio 3 : 2. Then the ratio of time intervals t1 : t2, will be? Where t1 is the time period for 25% completion of the first reaction and t2 is the time required for 75% completion of the second reaction. [log 2 = 0.3, log 3 = 0.477] (a) 0.199 : 1 (b) 0.420 : 1 (c) 0.273 : 1 (d) 0.311 : 1 O

O

28. Identify the anomers. H H

Conc. NaOH,

Ph

(a)

HO H

OH

HO

OH H O OH

H

HO

and

HO H

H H H O OH

H CH2 – OH

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CH2 – OH



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(b)

H H

H Me CH2OH C

OH and

H

OH

(d)

H

D

OH H

HO

CH2OH

OH

HO

D

H OH

OH and H H OH

H H

29. Which amongst the following metal carbonyls are inner orbital complexes with diamagnetic property? (II) Fe(CO)5 ; (I ) Ni(CO)4; (IV) Cr(CO)6 (III) V(CO)6

(b) HCDO (c) HCDO (d) D2CO

Select the correct answer from the codes given below: (a) I and II only (b) II , III and IV only (c) II and IV only (d) I , II and IV only

Multiple Correct Answer Type This section contains 4 multiple choice questions numbered 30 to 33. Each question has 4 choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct. 30. Which are correct against property mentioned? (a) CH 3 COCl > (CH 3 CO) 2 O > CH 3 COOEt > CH3CONH2 (Rate of hydrolysis)

CH3

CH3

COOH >

COOH CH3

(Rate of esterification) OH

(c)

CH CH3

ON2

>

OH

>

OH



(Rate of esterification) 12 Physics for you |

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NaOH NaOEt NaOD

DCOONa + CH2DOH DCOOEt + DCH2ONa DCOONa + CD3OD

33. The van der Waals equation of state for a non-ideal PV V a gas can be rearranged to give = − RT V − b VRT for 1 mol of gas.

SECTION - II

(b) CH 3 –CH 2 –COOH > CH3

Ph – CH2 – COOH (Rate of decarboxylation)

32. Which of the following products is/are correctly mentioned in the following reactions? (a) HCHO NaOD HCOONa + CH3OD

OH Et

Et

CH3 – C – COOH > CH3 – C – CH2– COOH >

31. The correct statement(s) regarding defects in solids is (are) (a) Schottky defect is usually favoured by small difference in the sizes of cation and anion. (b) Schottky defect lowers the density of solids. (c) C o m p o u n d s h a v i n g F - c e n t r e s a r e diamagnetic. (d) Frenkel defect is a dislocation defect.

O

H

CH2OH



H

Me O

(d)

— —

OH and HO HO OH

H

O

O

H

CH2OH C

(c)

Me

HO

OH

— —

Me H

The constants ‘a’ and ‘b’ are 1.0 PV positive numbers. When RT applied to H2 at 80 K, the equation gives the curve 0 40 60 80 as shown in the figure. P, atm Which one of the following statements is(are) correct? (a) At 40 atm the two terms V/(V – b) and a/VRT are equal. (b) At 80 atm the two terms V/(V – b) and a/VRT are equal. (c) At a pressure greater than 80 atm, the term V/(V – b) is greater than a/VRT. (d) At 60 atm the term V/(V – b) is smaller than a 1+ VRT

SECTION - III Linked Comprehension Type This section contains 2 paragraphs P34-36 and P37-39. Based upon each paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct.

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P34-36 : Paragraph for Question Nos. 34 to 36 A fuel cell is a cell that is continuously supplied with an oxidant and a reductant so that it can deliver a current indefinitely. Fuel cells offer the possibility of achieving high thermodynamic efficiency in the conversion of Gibbs energy into mechanical work. Internal combustion engines at best convert only the fraction (T2 – T1)/T2 of the heat of combustion into mechanical work. While the thermodynamic efficiency of the fuel cell ∆G , where DG is the Gibbs energy is given by, η = ∆H change for the cell reaction and DH is the enthalpy change of the cell reaction. A hydrogen-oxygen fuel cell may have an acidic or alkaline electrolyte. 2.303RT = 0.06 F The above fuel cell is used to produce constant current supply under constant temperature and 30 atm constant total pressure condition in a cylinder. If 10 mol of H2 and 5 mol of O2 were taken initially. Rate of combustion of O2 is 10 millimoles per minute. The half cell reactions are Pt|H2(g)|H+(aq.)||H2O(l)|O2(g)| Pt;

1 O + 2H +( aq ) + 2 e − → H 2O( l ) ; E° = 1.246 V 2 2( g ) 2H+(aq) + 2e– H2(g); E° = 0 To maximize the power per unit mass of an electrochemical cell, the electronic and electrolytic resistances of the cell must be minimized. Since fused salts have lower electrolytic resistances than aqueous solutions, high-temperature electrochemical cells are of special interest for practical applications 34. Calculate e.m.f. of the given cell at t = 0. (log 2 = 0.3). (a) 1.255 V (b) 1.35 V (c) 1.3 V (d) 1.246 V 35. The above fuel cell is used completely as an electrolytic cell with Cu voltameter of resistance 26.94 using Pt electrodes. Initially Cu voltameter contains 1 litre solution of 0.05 M CuSO4.[H+] in solution after electrolysis (Assuming no change on volume of solution). (a) 0.015 M (b) 0.03 M (c) 0.025 M (d) 0.01 M 36.

If (Cu2+) = 0.01S m2 mole–1, (H+) = 0.035 S m2

mole–1 and

(SO42–) = 0.016 S m2 mole–1, specific conductivity of resulting solution left in sopper voltameter after above electrolysis is

(a) 2.57 S m–1 (c) 1.525 S m–1

(b) 1.75 S m–1 (d) 2.25 S m–1

P37-39 : Paragraph for Question Nos. 37 to 39 A black coloured (A) on reaction with dil. H2SO4 gives a gas (B) which on passing in a solution of an acid (C) gives a white turbidity (D). Gas (B) when passed in an acidified solution of a compound (E) gives a black precipitate (F) which is soluble in hot concentrated (C). After boiling this solution when excess of ammonia solution is added, a blue coloured compound (G) is formed. To this solution of (E), on addition of acetic acid and aqueous potassium ferrocyanide, a chocolate brown precipitate (H) is formed. On addition of an aqueous solution of BaCl2 to an aqueous solution of (E) white precipitate insoluble in HNO3 is obtained. 37. Black coloured compound (A) is (a) PbS (b) CuS (c) Ag2S (d) All of these. 38. The gas (B) on passing through an acid (C) gives a white turbidity (D) because (a) gas (B) acts as an oxidising agent (b) gas (B) acts as an reducing agent (c) acid (C) acts as an oxidising agent (d) both (b) and (c) 39. To which of the following property, the compound (E) will respond? (a) It gives white precipitate with (CH3COO)2 Pb solution soluble in ammonium acetate. (b) It gives dirty white precipitate with KI. (c) Its hydrated salt effloresces. (d) All of these. Section - iV Integer Answer Type This section contains 7 questions numbered 40 to 46. The answer to each of the questions is a single digit integer, ranging from 0 to 9. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following.

X 0 1 2 3 4 5 6 7 8 9

Y 0 1 2 3 4 5 6 7 8 9

Z 0 1 2 3 4 5 6 7 8 9

W 0 1 2 3 4 5 6 7 8 9

40. 5.6 g of an oxide of a hypothetical metal required 2.4 g coke for its complete reduction to metal along with the production of CO gas. Find the equivalent weight of metal in case of given metal oxide.

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41. How many of the given statements are correct? I : Molar entropy of a substance follows the order (Sm)Solid < (Sm)liquid < (Sm)gas II: Entropy change for the reaction H2(g) → 2H(g) is +ve. III: Molar entropy of a non-crystalline solid will be zero at absolute zero. IV: If the path of an irreversible process is reversed, then both system and surroundings shall be restored to their orginal states. V : Refractive index and molarity are intensive properties. 42. Sulphide ions reacts with sodium nitropruside giving a coloured solution. In the reaction, the change in oxidation state of iron is

44. What is the free energy change (DG) in kcal, when 144 g of water at 100°C and 1 atm pressure in converted into steam at 100°C and 4 atm pressure? (Take R = 2 cal/K mole, ln2 = 0.7).

45. Number of ‘H’ that can takes part in tautomerism in given compound are O O O



N

N

H

O

46. How many of the following compounds will give white precipitate with aqueous AgNO3 ?

43. Two solids, A and B are present in two different container having same volume and same temperature following equilibrium are established :

In container (1) A(s) equilibrium

D(g) + C(g) PT = 2 atm at



In container (2) B(s) equilibrium

E(g) + F(g) PT = 4 atm at



If excess of A and B are added to a third container having double the volume and at same temperature then, the total pressure of this container at equilibrium will be

Cl

Cl

Cl

Cl

Cl Ph Ph

O

|

|

|

|

Ph � C � C � Ph

Cl

OH Cl

Cl

mathematics

SECTION - I Straight Objective Type This section contains 6 multiple choice questions numbered 47 to 52. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. sin x + 4 π  be 47. Let g :  , π  → A defined by g( x) = sin x − 2 2  invertible function, then the set A is equal to (a) [–5, –2] (b) [2, 5] (c) [–5, 2] (d) [–3, –2] 1 9  48. If f ( x + 1) =  f ( x) + , x ∈ N and f(x) > 0 for f ( x)  2 x ∈ N all then lim f ( x) =

x →∞

(a) 3

(b) 4 10

(c) 5

(d) 6

−1

49. The value of ∫ [tan x] dx (where [⋅] denotes greatest 0

integer) is equal to  π (a)  10 −  4  (c) 10 14 Physics for you |

50. If C 0, C 1, C 2, ....., C n are binomial coefficients then 2 n   2 2 2 lim  Cn −   Cn−1 +   Cn− 2 + .... + ( −1)n   C0  = n→∞   3 3 3        

(a) 0

(b) 1

(d) None of these

feBruary ‘13

(d) 2

51. I f t h e c o m m o n t a n g e n t s t o t h e c i r c l e s x2 + y2 – 2x – 4y + 1 = 0 and x2 + y2 – 14x – 4y + 52 = 0 intersect at A and B on the line joining their centres then the equation of the ellipse with A, 1 will be B as foci and eccentricity 2 2 2 (a) ( x − 9 ) ( y − 2 ) + =1 48 64

( x − 9) + ( y − 2)

2

( x − 2) + ( y − 9) (c)

2

2

(b)

64

2

=1

64

( x − 9) + ( y − 2) 2

(d)

=1

48

48

(b) (10 – tan1)

(c) –1

24

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32

2

=1

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52. Two lines whose equations are

x − 3 y − 2 z −1 = = 2 3 λ

x−2 y−3 z−2 = = lie in the same plane, then 3 2 3 –1 the value of sin sinl is equal to (a) 3 (b) p – 3 (c) 4 (d) p – 4

and

SECTION - II Multiple Correct Answer Type This section contains 4 multiple choice questions numbered 53 to 56. Each question has 4 choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.   53. Let unit vectors a and b are perpendicular & unit   vector c is inclined at an angle ‘q’ to both a and       b . If c = αa + βb + γ( a × b ) then (a) a = b (b) 1 – 2a2 = g2 1 + cos 2θ (c) α 2 = (d) a2 – b2 = g2 2 54. The equation of the curve passing through (3, 4) & satisfying the differential equation 2

 dy  dy y   + ( x − y ) − x = 0 can be dx dx   (a) x – y + 1 = 0 (b) x2 + y2 = 25 (c) x2 + y2 – 5x – 10 = 0 (d) x + y – 7 = 0 55. The image of point having abscissa = a on y = x – 1 w.r.t. the line mirror 3x + y = 6a is the point on x = y2 + 1 with ordinate = a. Then the value of a is 1 (a) –1 (b) 3 (c) 2 (d) none of these

` 150

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` 165

56. If in a triangle ABC, sin4A + sin4B + sin4C + 8cosA = 0, then triangle may be (a) Right angled (b) isosceles (c) Equilateral (d) Right angled isosceles

SECTION - III Linked Comprehension Type This section contains 2 paragraphs P57-59 and P60-62. Based upon each paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct.

P57-59 : Paragraph for Question Nos. 57 to 59

The diagram shows the graph of the derivative of a function y = f(x) for 0 ≤ x ≤ 5 with f(0) = 0 .

x

57. Tangent line to y = f(x) at x = 0 makes an angle

of sec–1 (a) 2

k with the x-axis then k = (b) 5 (c) 10 (d) 17

58. f is increasing in the interval (a) [1, 3] (b) [0, 4] (c) [0, 1] (d) none of these 59.

1

∫0 f ( x)dx =

21 (a) − 20 (c) –1

` 150

21 20 (d) none of these (b)

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P60-62 : Paragraph for Question Nos. 60 to 62 A bag contains 6 balls of 3 different colour namely White, Green and Red, atleast one ball of each different colour. Assume all possible probability distributions are equally likely. 60. The probability that the bag contains 2 balls of each colour, is (a) 1/3 (b) 1/5 (c) 1/10 (d) 1/4

61. Three balls are picked up at random from the bag and found to be one of each different colour. The probability that the bag contained 4 Red ball is (a) 1/14 (b) 2/14 (c) 3/14 (d) 4/14 62. Three balls are picked at random from the bag and found to be one of each different colour. The probability that the bag contained equal number of White and Green balls, is (a) 4/14 (b) 3/14 (c) 2/14 (d) 5/14 Section - iV Integer Answer Type This section contains 7 questions numbered 63 to 69. The answer to each of the questions is a single digit integer, ranging from 0 to 9. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following.

X 0 1 2 3 4 5 6 7 8 9

Y 0 1 2 3 4 5 6 7 8 9

Z 0 1 2 3 4 5 6 7 8 9

b 2. Given 0 < a < 1 and a 2 < a 1 < b2 < b 1 then 4 [|b1| – |a1|] + [|b2| – |a2|] is equal to (where [⋅] denotes greatest integer)

66. The product of maximum & minimum distance z −1 z−2 between two curves = 2 and = 2 is z−2 z −1 67. The sum of infinite terms of a decreasing G.P. is equal to the greatest value of the function f(x) = x3 + 3x – 9 in the interval [–2, 3] and the difference between the first two terms is f ′(0). If the common ratio of the G.P. is k/3 then k = 68. If the area bounded by the curve f ( x) = tan x + cot x − tan x − cot x between the π lines x = 0 , x = and the x-axis is logK then K is 2 69. The sum of the factors of 7! , which are odd & are of the form 3t + 1, where ‘t’ is a whole number is

W 0 1 2 3 4 5 6 7 8 9

answer keys 1. (a) 2. (c) 5. (a) 6. (c) 9. (a,c,d) 12. (a) 13. (a) 16. (c) 17. (1) 20. (5) 21. (1) 24. (c) 25. (c) 28. (d) 29. (c) 32. (a,b,c,d) 35. (a) 36. (b) 39. (d) 40. (6) 43. (6) 44. (8)

 −1 1 1    63. If A  −1 3 2  then det [adj {adj {adj A }}] is  1 2 1   equal to 64. If (1 + ax + bx2)4 = a0 + a1x + a2x2 + .... + a8x8 where a, b, a0, a1, ..... a8 ∈ R such that a0 + a1 + a2 ≠ 0 and a0 a1 a2 a1

a2

a2 a0

47. 51. 55. 59. 63. 67.

a0 = 0 then 32(a – b) is a1

65. If roots of equation x2 – x(1 + 4a) + a + 3a2 = 0 are a1, b1 & that of x2 – ax – 2a2 = 0 are a2 and

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(a) (b) (c) (a) (1) (2)

PHYSICS 3. (b) 7. (b,d) 10. (a,b,d) 14. (b) 18. (3) 22. (1) CHEMISTRY 26. (c) 30. (a,b) 33. (c,d) 37. (d) 41. (3) 45. (4)

4. (b) 8. (a,b,c,d) 11. (a) 15. (a) 19. (1) 23. (1) 27. (d) 31. (a,b,d) 34. (c) 38. (d) 42. (0) 46. (4)

MATHEMATICS 48. (a) 49. (b) 50. (a) 54. (a,b) 52. (d) 53. (a,b,c) 56. (a,d) 57. (d) 58. (b) 60. (c) 61. (a) 62. (b) 64. (3) 65. (1) 66. (3) 68. (4) 69. (8)



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seeking Unity in diversity and diversity in unity Passage- 1 Differentiation and Integration – they are the right and left hands of science. We shall study a few cases of motion and cases similar to various types of motion in different fields. Water flow and current flow are similar. Resistance in narrow tubes will be more. although the same potential energy is applied. As the quantity of water flowing is the same, water will be faster in the narrow tube than in the wider one because the same quantity has to flow. The same current will be flowing in resistances in series. A

I R1

D

B R3

R2 +

I



– I

C

In the given circuit same current is flowing in all the three resistances in series. – IR1 – IR2 – IR3 + e = 0 Take "up the current"-positive, down the flow, negative. The potential from the positive of the battery to its negative plate is negative and negative to positive is positive. I3 m From any point, back to I 2 the same point, the total potential difference should I4 be zero. I I1

5

m The total current entering a junction will be equal to the current leaving the junction. I1 + I2 – I3 –I4 – I5 = 0 These are Kirchhoff's Laws.

2. If the given bridge is balanced and no current is flowing through the galvanometer, what is the value of S? B P

4

A R

(a) 1 A

2 2

2

12 V

(b) 2 A

(c) 3 A

8

G



Q C S

D

(a) 4 W (b) 2 W (c) 8 W (d) None of these 3. When a straight conductor is held perpendicular to the plane of the paper and the current is outwards, what is the direction of the associated magnetic lines by Oersted's law? (a) It is parallel to the conductor and outwards. (b) It is parallel to the conductor and inwards. (c) It is circular, perpendicular to the direction of current, in the plane of the paper, anticlockwise. (d) Perpendicular to the plane of the paper, clockwise. 4. What is the essential difference in the direction of action of the magnetic field on a charge and the electric field? 5. Three capacitors are connected in series as shown in figure. C1

1. The total current flowing in the circuit is 2



2

C2

C3

+ – V

2

(d) 4 A

(a) The same charge is flowing in all the three capacitors (b) Total equivalent capacitance is C = C1 + C2 + C3 (c) The equivalent capacitance C is given by 1 1 1 1 = + + C C1 C2 C3 (d) None of these

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6. The relation current, resistance and potential difference is given by V V = IR ⇒ I = R The corresponding relation for capacitance is 2 (a) Q = V (b) V = QC C (c) Q = VC (d) none of these 7. The total resistance when R1, R2, R3 are in parallel is

1 1 1 = + + R R1 R2 When capacitors are in capacitance, C is given by 1 1 1 1 (a) (b) = + + C C1 C2 C3

1 R3 parallel, the total C = C1 + C2 + C3

(c) none of these



Therefore, total resistance is, 1 W + 1 W + 2 W = 4 W. 12 V The total current in the circuit = =3A 4Ω 2. (a) : This is a Wheatstone bridge where the resistances are so balanced that no current flows through the branch BD, as the galvanometer shows no deflection. P R QR 2 × 8 Here, = ⇒ S= = =4Ω Q S P 4 3. (c) : When current is outwards, ^r to the plane of the paper, the direction of the magnetic field is anticlockwise, perpendicular to the direction of the current. 4. The magnetic field always acts perpendicular to the direction of the current. The electric field always acts parallel to the motion of the charge.

Passage-II In a CR circuit, when a capacitor is getting charged, the voltage increases exponentially and when the current is observed, initially the current is high; till the capacitor gets charged fully, the current decreases exponentially till it reaches zero exponentially. Charging and discharging a capacitor has its analogy in nuclear physics in the studies of radioactivity, production and decay of radioactive materials. The oscillations of mass and spring, L, C, R circuit are all analogous to problems in radioactivity. Students are advised to prepare Charts for comparision. 8. The functions of spring and mass in electrical oscillations are given by (a) R and L (b) L and C (c) L and R (d) none of these

5. (a, c): At any instant, the charges accumulated on all the capacitors will be equal (a) but the sign are different.

9. What is common in a leaking water tank, radioactive decay and discharge of a fully charged capacitor?





SOLUTIONS 1. (c) :





2

2 P

2

Q

2

R S

2

12 V In P and Q, 2 W and 2 W are in parallel. 1 1 1 1 1 1 i.e., = + ⇒ = + ⇒ RPQ = 1 Ω R R1 R2 R 2 2 RQR is also 1 W. They are in series with S = 2 W.



i.e.,



The total capacitance is C, where C is given by, 1 1 1 1 = + + C C1 C2 C3



6. (c) : Q = VC 7. (b) : +– V

C1

C2

C3



The total charge, Q = Q1 + Q2 + Q3 The potential applied across all the three capacitors in parallel is the same. Q = VC = VC1 + VC2 + VC3 ⇒ C = C1 + C2 + C3 8. (b) : L and C d 2Q

1 1 Q where = ω2 LC LC dt 2π The period of oscillation, T = ω 1 Frequency of oscillation = 2π LC 9. They all obey Y the exponential decay law. Let quantity of water, number of time decaying nuclei X and the charge in the capacitor be Y and time denoted by X. The decay diagram is as shown. Seek unifined approach to solve diverse problems. vvv 2

=−

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SHM and Waves By : Prof. Rajinder Singh Randhawa* R is drilled through a cylinder 4 of radius R to form a body as shown in figure. Assuming that the body is in pure rolling, find the time period T for small oscillations.

1. A hole of radius

2. A cylinder of mass M and radius R is kept on a rough horizontal platform at one extreme end of the platform at t = 0. Axis of the cylinder is parallel to z-axis. The platform is oscillating in the xy plane and its displacement from the origin is represented as x = 2cos(4pt) m. There is no slipping between the cylinder and the platform. Find the acceleration of the centre of mass of the cylinder as a function of time. 3. A particle simultaneously participates in two mutually perpendicular oscillations : x = cospt m and πt y = cos . Find the trajectory of the particle. 2 4. When a tuning fork of frequency 262 Hz is struck, it loses half of its energy after 4 s. (a) What is the decay time t ? (b) What is the Q-factor for this tuning fork? (c) What is the fractional energy loss per cycle?

5. A metallic rod of length 1 m is rigidly clamped at its midpiont. Longitudinal stationary waves are set up in such a way that there are two nodes on either side of the midpoint. The amplitude of an antinode is 2 × 10–6 m. Write the equation of motion at a point 2 cm from the midpoint and those of constituent waves in the rod. (Y = 2 × 1011 N m–2 and r = 8 × 103 kg m–3) 6. The first overtone of an organ pipe beats with the first overtone of a closed organ pipe with a beat frequency of 2.2 Hz. The fundamental frequency of the closed organ pipe is 110 Hz. Find the length of the pipes. 7. Two identical wires are stretched by the same tension of 100 N and each emits a note of frequency 200 cycles per second. The tension in one wire is increased by 1 N. Calculate the number of beats heard per second when the wires are plucked. 8. By what factor does the sound intensity increase if a sound level increases by 3 dB. SOLUTIONS 1.



Let s be the mass density of the cylinder and x be the position of centre of mass. Then

Randhawa Institute of Physics, S.C.O. 208, First Fl., Sector-36D & S.C.O. 38, Second Fl., Sector-20C, Chandigarh

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 R2  R R2 0 = x σ π  R2 −  + σ π  16 2 16 R or x = − 30 Let M be the mass of the cylinder without cavity. Moment of inertia about contact point 3 I1 = MR 2 2 And moment of inertia of an imaginary cylinder occupying the cavity is 2

I2 =

 M   3R  1M R    ×   +    2  16   4  16 2 

2

So, the moment of inertia of the system about the contact point, 3 1 9 695 I = I1 − I 2 =  − − MR 2 = MR 2 512  2 512 64  When the system is turned through a small angle clockwise, the restoring torque on it is  15  R  τ = −  Mg sin θ     16   30   15  R  d 2θ 695 MR 2 = −  Mg sin θ      30   16 512 dt 2 For small q, sinq ≈ q, d 2θ  512 g  +  θ = 0 dt 2  32 × 695 R  Comparing it with the standard equation of SHM, we get, 512 g 2π ω= = 32 × 695R T or T = 2 π

32 × 695R R = 41.4 512 g g

2. From given equation x = 2 cos(4pt) m, angular frequency, w = 4p rad s–1. \ Acceleration of the platform when displacement is x, a = – w2x = – 32p2cos(4pt) m s–2 Fictitious force acting on the cylinder = Ma = M 32p2 cos(4pt) From Newton's second law, 32 p2 M cos(4pt) – f = MaCM ...(i) Also, taking torque about C.M. of the cylinder, 1 fR = MR 2α ...(ii) 2 For pure rolling; aCM = R a ...(iii) Solving above equations, we get, 32 2 f = π M cos( 4 πt) N 3

And acceleration of C.M. of the cylinder relative 64 2 π cos( 4 πt) m s −2 to platform = 3 Acceleration of C.M. of the cylinder relative to  64  ground =  − 32  π 2 cos( 4 πt)  3  32 2 = − π cos( 4 πt) m s −2 3 3. The given equations are x = cospt ... (i) π t y = cos 2 1 + cos πt or 2 y 2 − 1 = cos πt .... (ii) 2 Put (i) in (ii), we get, 2y2 – 1 = x or 2y2 = x + 1 represents the equation of a parabola. E 4. Using E = E0e–t/t and E = 0 . The Q value can be 2 calculated from decay time and the frequency. (a) The energy at time, t = 4 s is equal to half the original energy; E0 = E0 e −4 / τ or e 4 / τ = 2 2 By taking loge both sides, we get 4 4 = ln 2 , τ = = 5.77 s τ ln 2 (b) Q - factor = w0t = (2pu)t = 2p × 262 × 5.77 = 9.5 × 103. (c) The fractional energy loss in a period is given by 1 1 | ∆E | T −4 E = τ = υτ = 262 × 5.77 = 6.6 × 10 ∴ y=

5. When a longitudinal wave is set in the rod, its free end is antinode and clamped point is node. Two λ and a node consecutive nodes are separated by 2 λ and an antinode by . 4 λ λ ∴ 4   + 2   = L = 1 m ⇒ λ = 0.4 m 2 4 Y Velocity of a transverse wave in a rod is, v = ρ 2 × 1011 N m −2 −1 ∴ v= = 5000 m s 8 × 10 3 kg m −3 v Also v = ul, υ = λ ∴ υ=

5000 m s −1 = 12500 Hz 0.4 m

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Let the incident wave on the rod be y1 = A sin(wt – kx). The reflected wave is y2 = A sin( ωt + kx + φ). From principle of superposition, y = y1 + y2 = A sin(wt – kx) + Asin (wt + kx + f)

These two produce beats frequency of 2.2 Hz when sounded together, the expressions for beat frequency are

 (ωt − kx ) + (ωt + kx + φ)  = 2 A sin    2

3 × 110 −

 (ωt − kx ) − (ωt + kx + φ)  cos    2



\ lo = 1.0067 m or lo = 0.9934 m

υ=

The free end of the rod is an antinode, here  φ amplitude is maximum, cos  kx +  = 1.  2 φ At x = 0, = 0 , hence f = 0 2 Amax = 2A = 2 × 10–6 m.

For a point 2 cm from the midpoint, x = 0.5 ± 0.02 \ y = 2 × 10–6 cos5p (0.5 ± 0.02) sin(25000pt). The equation of component waves can be obtained by applying 2cosA sinB = sin(A + B) – sin(A – B). \ y = 10–6[sin(5px + 25000pt) – sin(5px – 25000pt)]

dυ 1  T 2µ  =   dT 4 LT  T µ 2 

( 2n − 1)v , where n = 1, 2, 3,.... 4lc

For open organ pipe, υ = n v , where n = 1, 2, 3,... o 2 lo Fundamental frequency of closed organ pipe, 330 v = 110 Hz or lc = = 0.75 m 4lc 4 × 110 3v First overtone of closed organ pipe = 4lc First overtone of open organ pipe = 2222physics FOR YOU |

february ‘13

2v 2lo

.... (i)

=

1 T  4 LT  µ 

1/ 2

(Using (i))

υ  ∆T    2 T  200 1 ∆υ = × = 1 cycle per second 2 100 8. Intensity level b is given by Hence, ∆υ =

β = 10(dB)log10

The two terms in the expression for y represent the two component waves.

For closed pipe, υc =

1/ 2

dυ υ = dT 2T

= 10–6 sin(25000pt + 5px) + 10–6 sin(25000pt – 5px)

6. The frequency in nth mode of vibration of one end closed and an open organ pipe are given by;

1 T 2L µ

Differentiate equation (i) w.r.t. T, we get −1/ 2  −1/ 2 dυ 1  1 T  1 1 T  1 =    = ⋅ dT 2 L  2  µ  µ  4 L  µ  µ

The equation of the resultant standing wave is  2π  y = 2 × 10 −6 cos  x  sin 2 π υ t  λ  = 2 × 10–6cos(5px)sin(25000 pt).



330 330 = 2.2 or − 3 × 110 = 2.2 lo lo

7. The frequency of the fundamental note emitted by each wire before the tension change occurs is

  φ φ = 2 A cos  kx +  sin  ωt +  .   2 2



3v 2 v 2 v 3v − = 2.2 or − = 2.2 4lc 2lo 2lo 4lc



or

I I0

I = 10β/(10 ) I0

.... (i)

Let I′ be the intensity when the sound level is (b + 3) dB. Hence, I′ β + 3 = 10 log I0 I′ = 10(β + 3)/ 10 .... (ii) I 0 Divide (ii) by (i), we get

I ′ 10(β + 3)/ 10 = I 10β / 10

I′ = 10( 0.3) = 2 I \ An increase in sound level of 3 dB increases the intensity by a factor of 2.

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Role of Coaching Institutes in One’s Success

G

one are the days when students used to believe in self study and pass their exams with flying colors. Today, in this highly competitive world where children outdo each other in a bid to score 90 percent, professional training and coaching institutes have become a part and parcel of the modern age education. The purpose of competitive exams is to filter students with the right aptitude for a particular professional field. Simply mugging up the study material may not prove useful. It is important to have a strategy in place. Thus a mapped out plan is the need of the hour and coaching institutes play a very vital role in jotting out a good plan for the students. The coaching institutes play a very significant role in the enhancement of student’s preparation. They enhance student’s opportunities by providing a high standard of academic learning, environment and discipline. The institutes help students in developing skills in various academic fields. With the growing span of education in India, the students have become alert and conscious about their careers, they tend to seek the best quality education and for that they seek the best coaching institute in the town. Good coaching institute helps students to identify their weak areas and helps to strengthen them with well built strategies. Aspirants working hard to crack JEE Main, JEE Advanced, CAT, MAT, TOFEL, IAS etc opt for these coaching institutes for proper guidance and grooming. The growing demands of high level education have led to a fiercely competitive environment for students seeking admissions in prestigious institutes. The genesis of growth in competition lies in the education imparted in the schools. The school board exams have diverse levels of difficulty and it usually doesn’t match the competency level of the competitive exams. Schools basically focus on imparting syllabus education while entrance exams focus on application based education. The inability of traditional school set up in meeting the needs of students in preparing for

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competitive exams and choosing the right career has ushered in the concept of coaching centers. Earlier students were unaware of the various new career options available due to lack of guidance and they used to rely on their parents, siblings, friends and other such sources for information. With the emergence of media penetration and online support systems, people have become more aware. Coaching institutes have generated greater awareness of the various options available for the students. The experienced and qualified faculty at a good coaching institute helps the students to understand and analyze the concepts and test patterns and guide them to work accordingly. They clarify all the related doubts of the students and provide them with relevant suggestions notes and handouts which help students utilize their time optimally. The pattern of competitive entrance exams such as JEE, CAT etc undergo frequent change and at times this change happens on yearly basis. The coaching centers understand the changes and lead the students in the desired manner. The Mock tests taken in these coaching classes are very helpful in giving fairly good exposure to the students and encourage them to put in focused effort and to work hard. These tests give ample practice as they teach the students to respond promptly to tricky questions and manage speed and time. Coaching centers are very important but one needs to keep a few things in mind before joining a coaching institute. The students should inquire about the faculties and the reputation of the coaching institute before joining. Also, the timing should be suitable and it should not be too far from your place as then it will be tiresome to travel and you will have no time to study at home. At the end of the day, irrespective of whether you go to a coaching class or not, it is your effort that will fetch you the kind of results you are longing for.

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– Contributed by FIITJEE

nn

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1. A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter d/2 in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively 3f I I and (a) f and (b) 4 2 4 f I 3I and (c) f and (d) 2 2 4 2. The radius of the rear wheel of bicycle is twice that of the front wheel. When the bicycle is moving, the angular speed of the rear wheel compared to that of the front is (a) greater (b) smaller (c) same (d) exact double 3. When a particle oscillates in simple harmonic motion, both its potential energy and kinetic energy vary sinusoidally with time. If u be the frequency of the motion of the particle, the frequency associated with the kinetic energy is υ (a) 4u (b) 2u (c) u (d) 2 4. A wooden block is dropped from the top of a cliff 100 m high and simultaneously a bullet of mass 10 g is fired from the foot of the cliff upwards with a velocity of 100 m s–1. The bullet and wooden block will meet each other after a time (a) 10 s (b) 0.5 s (c) 1 s (d) 7 s 5. A common emitter amplifier has a voltage gain of 50, an input impedance of 100 W and an output impedance of 200 W. The power gain of the amplifier is (a) 500 (b) 1000 (c) 1250 (d) 50 6. A satellite is launched into a circular orbit of radius R around earth while a second satellite is launched into an orbit of radius 1.02R. The percentage difference in the time period is (a) 0.7 (b) 1.0 (c) 1.5 (d) 3.0

7. Keeping the banking angle same, to increase the maximum speed with which a vehicle can travel on a curved road by 10 percent. The radius of curvature of the road has to be changed from 20 m to (a) 6 m (b) 18 m (c) 24.2 m (d) 30.5 m 8. A 220 V input is supplied to a transformer. The output circuit draws a current of 2.0 A at 440 V. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is (a) 3.6 A (b) 2.8 A (c) 2.5 A (d) 5.0 A 9. A block of mass m is attached with massless spring of spring constant k. The block is placed over a fixed rough inclined surface for which the 3 coefficient of friction is µ = . The block of mass m 4 is initially at rest. The block of mass M is released from rest with spring in unstretched state. The minimum value of M required to move the block up the plane is (Neglect mass of string and pulley and friction in pulley) (Take sin 37° = 3 ) 5

3 m 5 6 (c) m 5 (a)

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4 m 5 3 (d) m 2 (b)

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10. Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. If the frequency of incident light is halved and the intensity is doubled, the photocurrent becomes (a) one fourth (b) doubled (c) halved (d) zero

11. The Poisson’s ratio of a material is 0.4. If a force is applied to a wire of this material, there is a decrease of cross-sectional area by 2%. The percentage increase in its length is (a) 3% (b) 2.5% (c) 1% (d) 0.5% 12. The centre of mass of three particles of masses 1 kg, 2 kg and 3 kg is at (2, 2, 2). The position of the fourth mass of 4 kg to be placed in the system so that the new centre of mass is at (0, 0, 0) is (a) (–3, –3, –3) (b) (–3, 3, –3) (c) (2, 3 –3) (d) (2, –2, 3) 13. A body moves with uniform acceleration, then which of the following graphs is correct? v

(a)

(b)

S

(c)



(d)

t

t

14. A charged bead is capable of sliding freely through a string held vertically in tension. An electric field is applied parallel to the string, so that the bead stays at rest at middle of the string. If the electric field is switched off momentarily and switched on (a) the bead moves downwards and stops as soon the field is switched on. (b) the bead moves downwards when the field is off and moves upwards when the field is switched on. (c) the bead moves downwards with constant acceleration till it reaches the bottom of the string. (d) the bead moves downwards with constant velocity till it reaches the bottom of the string. 26 physics for you |

february ‘13

17. The equations of motion of a projectile are given by x = 36t m and 2y = 96t – 9.8t2 m. The angle of projection is 3 4 (a) sin–1   (b) sin–1   5 5 3 4 (c) sin–1   (d) sin–1   4 3 18. A body is moving forward and backward. Change in frequency observed by the body of a source is 2%. What is velocity of the body? (Speed of sound is 300 m s–1) (a) 6 m s–1 (b) 2 m s–1 –1 (c) 2.5 m s (d) 3 m s–1 19. The refractive index of a material of a plano concave lens is 5/3, the radius of curvature is 0.3 m. The focal length of the lens in air is (a) – 0.45 m (b) – 0.6 m (c) – 0.75 m (d) – 1.0 m

t v

15. When a battery connected across a resistor of 16 W, the voltage across the resistor is 12 V. When the same battery is connected across a resistor of 10 W, voltage across it is 11 V. The internal resistance of the battery is 10 25 20 30 (a) Ω (b) Ω (d) Ω (c) Ω 7 7 7 7 16. A satellite of mass ms revolving in a circular orbit of radius rs round the earth of mass M has a total energy E. Then its angular momentum will be (a) (2Emsrs2)1/2 (b) (2Emsrs2) 1/2 (c) (2Emsrs) (d) (2Emsrs)

20. The average force that is necessary to stop a hammer with 25 N s–1 momentum in 0.05 s is (a) 500 N (b) 125 N (c) 50 N (d) 25 N 21. The dimensional formula for latent heat is (a) [MLT–2] (b) [ML2T–2] (c) [M0L2T–2] (d) [MLT–1] 22. A charge Q situated at a certain distance from a short electric dipole in the end on opposite experiences a force F. If the distance of the charge from the dipole is doubled, force acting on the charge will be F F (a) 2F (b) (c) 8 F (d) 2 8 23. The magnetic induction and the intensity of magnetic field inside an iron core of an electromagnet are 1 Wb m–2 and 150 A m–1 respectively. The relative permeability of iron is (m0 = 4p × 10–7 H m–1) (a)

106 4π

(b)

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106 6π

(c)

10 5 10 5 (d) 4π 6π

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24. The equivalent resistance between the terminals A and D in the following circuit is 10 

5

A

10  C

5

30. If the length of stretched string is shortened by 40% and the tension is increased by 44 %, then the ratio of the final and initial fundamental frequencies is (a) 2 : 1 (b) 3 : 2 (c) 3 : 4 (d) 1 : 3

10 

10 

5

B

(a) 10 W (c) 5 W

5

D

(b) 20 W (d) 30 W

25. A rain drop of radius 0.3 mm falling vertically downwards in air has a terminal velocity of 1 m s–1. The viscosity of air is 18 × 10–5 poise. The viscous force on the drop is (a) 101.73 × 10–4 dyne (b) 101.73 × 10–5 dyne (c) 16.95 × 10–5 dyne (d) 16.95 × 10–4 dyne 26. A stone is projected vertically up to reach maximum height h. The ratio of its kinetic energy 4 to its potential energy, at a height h, will be 5 (a) 5 : 4 (b) 4 : 5 (c) 1 : 4 (d) 4 : 1 27. Refer to the arrangement of logic gates. For A = 0, B = 0 and A = 1, B = 0, the values of output Y are, respectively A B



Y

(a) 0 and 1 (c) 1 and 1

(b) 1 and 0 (d) 0 and 0

28. Four massless springs whose spring constants are 2k, 2k, k and 2k respectively are attached to a mass M kept on a frictionless plane as shown in figure. If the mass M is displaced in the horizontal direction, then the frequency of the system is

2k

2k

k M

2k

(a)

1 k 2π 4M

(b)

1 4k 2π M

(c)

1 k 2π 7 M

(d)

1 7k 2π M

29. Oscillating frequency of a cyclotron is 10 MHz. If the radius of its dees is 0.5 m, the kinetic energy of a proton, which is accelerated by the cyclotron is (a) 10.2 MeV (b) 20.1 MeV (c) 5.1 MeV (d) 1.5 MeV

31. Sodium lamps are used in foggy conditions because (a) yellow light is scattered less by the fog particles. (b) yellow light is scattered more by the fog particles. (c) yellow light is unaffected during its passage through the fog . (d) wavelength of yellow light is the mean of the visible part of the spectrum. 32. In Young’s double slit experiment, an interference pattern is obtained on a screen by a light of wavelength 6000 Å coming from the coherent sources S1 and S2. At certain point P on the screen third dark fringe is formed. Then the path difference S1P – S2P in microns is (a) 0.75 (b) 1.5 (c) 3.0 (d) 4.5 33. Photoelectric effect is an example of (a) elastic collision (b) inelastic collision (c) two dimensional collision (d) oblique collision 34. The angle of minimum deviation in an equilateral prism of refractive index 1.414 is (a) 60° (b) 30° (c) 90° (d) 45° 35. A black body of mass 34.38 g and surface area 19.2 cm2 is at an initial temperature of 400 K. It is allowed to cool inside an evacuated enclosure kept at constant temperature 300 K. The rate of cooling is 0.04°C per second. The specific heat of the body in J kg–1 K–1 is (Stefan’s constant s = 5.73 × 10–8 W m–2 K–4) (a) 2800 (b) 2100 (c) 1400 (d) 1200 36. The activity of a radioactive sample is measured as N0 counts per minute at t = 0 and N0/e counts per minute at t = 5 minutes. The time (in minutes) at which the activity reduces to half its value is 2 5 (a) ln (b) 5 ln 2 (c) 5 log10 2

WorldMags.net

(d) 5 ln 2 physics for you | february ‘13

27

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37. Two concentric coils of 10 turns each are placed in the same plane. Their radii are 20 cm and 40 cm and carry 0.2 A and 0.3 A current respectively in opposite directions. The magnetic field (in tesla) at the centre is 5 3 (a) µ0 (b) µ0 4 4 7 µ 4 0

(d) 9 µ0 4 38. When 1 kg of ice at 0°C melts to water at 0°C, the resulting change in its entropy, taking latent heat of ice to be 80 cal g–1 is (a) 273 cal K–1 (b) 8 × 104 cal K–1 (c) 80 cal K–1 (d) 293 cal K–1 (c)

39. A coil has 1,000 turns and 500 cm2 as its area. The plane of the coil is placed at right angles to a magnetic field of 2 × 10–5 Wb m–2. The coil is rotated through 180° in 0.2 s. The average emf induced in the coil is (a) 5 mV (b) 10 mV (c) 15 mV (d) 20 mV 14 40. Number of neutrons in 12 6 C and 6 C are (a) 8 and 6 (b) 6 and 8 (c) 6 and 6 (d) 8 and 8

(a)

C 4

(b)

3C 4

4C (d) 3C 3 44. A particle crossing the origin of co-ordinates at time t = 0, moves in the xy-plane with a constant acceleration a in the y-direction. If its equation of motion is y = bx2 (b is a constant), its velocity component in the x-direction is (c)

2b a a b a (b) 2b (c) b (d) a 45. A galvanometer, having a resistance of 50 W, gives a full scale deflection for a current of 0.05 A. The length of a resistance wire of area of cross-section 2.97 × 10–2 cm2 that can be used to convert the galvanometer into an ammeter which can read a maximum of 5 A current is (Specific resistance of the wire = 5 × 10–7 W m) (a) 9 m (b) 6 m (c) 3 m (d) 1.5 m (a)

46. In the figure, the velocity v3 will be

41. Pressure P, volume V and temperature T for a 2 certain gas are related by P = AT − BT , where V A and B are constants. The work done by the gas as its temperature changes from T1 to T2 while pressure remains constant is (a) A − B (T2 − T1 ) 2 (b) A(T2 – T1) – B(T22 – T12)

(c)

(

)

(

A 2 B T − T12 − T23 − T13 3 T 2

)

B (T2 – T1)3 3 30 30 42. When 15 P decays to become 14 Si the particle released is (a) electron (b) a-particle (c) neutron (d) positron (d) A(T2 – T1)2 –

43. A network of six identical capacitors, each of value C is made as shown in the figure. Equivalent capacitance between points A and B is A



B

28 physics for you |

february ‘13

(a) zero (c) 1 m s–1

(b) 4 m s–1 (d) 3 m s–1

47. A travelling wave is represented by the equation 1 y= sin( 60t + 2 x), where x and y are in metres 10 and t is in seconds. This represents a wave (1) travelling with a velocity of 30 m s–1 (2) of frequency 30 Hz π (3) of wavelength p m (4) of amplitude 10 cm (5) moving in the positive x direction Pick out the correct statements from the above. (a) 1, 2, 4 (b) 3, 4, 5 (c) 1, 2, 3, 4 (d) All 48. If 4 moles of an ideal monoatomic gas at temperature 400 K is mixed with 2 moles of another ideal monoatomic gas at temperature 700 K, the temperature of the mixture is (a) 550°C (b) 500°C (c) 550 K (d) 500 K

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Cont. on Page No. 70

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1. A ball is travelling with uniform translatory motion. This means that (a) it is at rest. (b) the path can be a straight line or circular and the ball travels with uniform speed. (c) all parts of the ball have the same velocity (magnitude and direction) and the velocity is constant. (d) the centre of the ball moves with constant velocity and the ball spins about its centre uniformly. 2. Which one of the following statements isnot true  about Newton’s second law of motion F = ma ? (a) The second law of motion is consistent with the first law. (b) The second law of motion is a vector law. (c) The second law of motion is applicable to a single point particle. (d) The second law of motion is not a local law. 3. A body of mass 10 kg is acted upon by two perpendicular forces, 6 N and 8 N. The resultant acceleration of the body is 3 (a) 1 m s–2 at an angle of tan–1   w.r.t. 8 N force. 4 3 (b) 0.2 m s–2 at an angle of tan–1   w.r.t. 8 N force. 4

(c) Newton’s second law only. (d) both Newton’s second and third law. 5. Which one of the following is not a contact force? (a) Viscous force (b) Magnetic force (c) Friction (d) Buoyant force 6. A body of mass 2 kg travels according to the law x(t) = pt + qt2 + rt3 where p = 3 m s–1, q = 4 m s–2 and r = 5 m s–3.

The force acting on the body at t = 2 s is (a) 136 N (b) 134 N (c) 158 N (d) 68 N

7. A person of mass 50 kg stands on a weighing scale on a lift. If the lift is descending with a downward acceleration of 9 m s–2, what would be the reading of the weighing scale?

(Take g = 10 m s–2) (a) 50 kg (b) 5 kg

(c) 95 kg (d) 100 kg

8. Two billiard balls A and B, each of mass 50 g and moving in opposite directions with speed of 5 m s–1 each, collide and rebound with the same speed. The impulse imparted to each ball is (a) 0.25 kg m s–1 (c) 0.1 kg m s–1

(b) 0.5 kg m s–1 (d) 0.125 kg m s–1

5 (c) 1 m s–2 at an angle of tan–1   w.r.t. 8 N force. 4

9. A body with mass 5 kg is acted upon by a force  ^ ^ F = ( −3 i + 4 j ) N. If its initial velocity at t = 0 is  ^ ^ u = (6 i − 12 j ) m s–1, the time at which it will just

5 (d) 0.2 m s–2 at an angle of tan–1   w.r.t 8 N force. 4



4. Conservation of momentum in a collision between particles can be understood from (a) conservation of energy. (b) Newton’s first law only.

have a velocity along the y-axis is (a) never (b) 10 s (c) 2 s (d) 15 s

10. In figure, the coefficient of friction between the floor and the block B is 0.1. The coefficient of friction

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Physics for you | february ‘13

29



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between the blocks B and A is 0.2. The mass of A is m/2 and of B is m. What is the maximum horizontal force F can be applied to the block B so that two blocks move together? (a) 0.15mg (b) 0.05mg (c) 0.1mg (d) 0.45mg 11. A stream of water flowing horizontally with a speed of 15 m s–1 gushes out of a tube of crosssectional area 10–2 m2, and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound? (a) 1.25 × 103 N (b) 2.25 × 103 N 3 (c) 3.25 × 10 N (d) 4.25 × 103 N

12. Which one of the following statements is not true? (a) The same force for the same time causes the same change in momentum for different bodies. (b) The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts. (c) A greater opposing force is needed to stop a heavy body than a light body in the same time, if they are moving with the same speed. (d) The greater the change in the momentum in a given time, the lesser is the force that needs to be applied. 13. A block of 6 kg is suspended by a rope of length 2 m from the ceiling. A force of 50 N in the horizontal direction is applied at the midpoint P of the rope, as shown in the figure. What is the angle the rope makes with the vertical in equilibrium? Neglect the mass of the rope. (Take g = 10 m s–2) (a) 30° (b) 40° (c) 60° (d) 45° 14. A person in an elevator accelerating upwards with an acceleration of 2 m s–2, tosses a coin vertically upwards with a speed of 20 m s–1. After how much time will the coin fall back into his hand? (Take g = 10 m s–2) (a) 1.67 s (b) 2 s (c) 3.33 s (d) 5 s 15. Two masses of 5 kg and 3 kg are suspended with help of massless inextensible strings as shown in figure. The whole system is going upwards with an acceleration of 2 m s–2. The tensions T1 and T2 are (Take g = 10 m s–2) 30 Physics for you |

february ‘13

T1

5 kg

T2 3 kg

(a) 96 N, 36 N (c) 96 N, 96 N

(b) 36 N, 96 N (d) 36 N, 36 N

16. A block of mass M is held against a rough vertical wall by pressing it with a finger. If the coefficient of friction between the block and the wall is m and the acceleration due to gravity is g, what is the minimum force required to be applied by the finger to hold the block against the wall? Mg (a) mMg (b) Mg (c) (d) 2mMg µ 17. There are four forces acting at a point P produced by strings as shown in figure, which is at rest. The forces F1 and F2 are 1 3 (a) N, N 2 2 (b) (c)

3 2 1

N,

1 2

2N 45° 45° P

1N

90°



F2



F1

N

1

(d)

3

N

N,

3

N 2 2 2 2 18. A metre scale is moving with uniform velocity. This implies N,

(a) the force acting on the scale is zero, but a torque about the centre of mass can act on the scale. (b) the force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero. (c) the total force acting on it need not be zero but the torque on it is zero. (d) neither the force nor the torque need to be zero. 19. A 100 kg gun fires a ball of 1 kg horizontally from a cliff of height 500 m. It falls on the ground at a distance of 400 m from the bottom of the cliff. Find the recoil velocity of the gun. (Take g = 10 m s–2) (a) 0.2 m s–1 (b) 0.4 m s–1 –1 (c) 0.6 m s (d) 0.8 m s–1 20. A large force is acting on a body for a short time. The impulse imparted is equal to the change in (a) acceleration (b) momentum (c) energy (d) velocity 21. Figure shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 m s–2. What is the net force on the man? (Mass of the man = 65 kg)

WorldMags.net

WorldMags.net (a) T

(c) T +



(a) 35 N (b) 45 N (c) 55 N

(d) 65 N

22. Figure shows (x, t), (y, t) diagram of a particle moving in 2-dimensions.



If the particle has a mass of 500 g, the force acting on the particle is (a) 1 N along y-axis (b) 1 N along x-axis (c) 0.5 N along x-axis (d) 0.5 N y-axis

23. Block A of weight 100 N rests on a frictionless inclined plane of slope angle 30° as shown in the figure. A flexible cord attached to A passes over a frictionless pulley and is connected to block B of weight W. Find the weight W for which the system is in equilibrium. (a) 25 N (b) 50 N (c) 75 N (d) 100 N 24. The position-time graph of a body of mass 2 kg is as given in figure. What is the impulse on the body at t = 4 s? 2 (a) kg m s–1 3 2 (b) − kg m s–1 3 3 (c) kg m s–1 2

(b) T −

mv 2 l



mv 2 l

(d) 0

Where T is the tension in the string.

26. Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. The force on the 7th coin (counted from the bottom) due to all the coins on its top is (a) 3mg, vertically downwards (b) 3mg, vertically upwards (c) 7mg, vertically downwards (d) 7mg, vertically upwards 27. A car of mass m starts from rest and acquires a  ^ velocity along east v = v i (v > 0) in two seconds. Assuming the car moves with uniform acceleration, the force exerted on the car is mv (a) eastward and is exerted by the car engine. 2 mv (b) eastward and is due to the friction on 2 the tyres exerted by the road. mv (c) more than eastward exerted due to the 2 engine and overcomes the friction of the road. (d) mv exerted by the engine. 2 28. A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are Lowest Point Highest Point (a) mg – T1 mg + T2 (b) mg + T1 mg – T2 2  2   (c) mg + T1 –  mv1  mg – T2 +  mv2   R   R   2  mv 2  mg + T2 + mv2 1     R  R T1 and v1 denote the tension and speed at the lowest point. T2 and v 2 denote corresponding values at the highest point. (d) mg – T1 –



(d) −

3 kg m s–1 2

25. One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is

29. A circular racetrack of radius 300 m is banked at an angle of 15°. If the coefficient of friction between the wheels of a race car and the road is 0.2, what is the maximum permissible speed to avoid slipping?

(Take tan15° = 0.27) (a) 18.2 m s–1 (c) 38.2 m s–1

WorldMags.net

(b) 28.2 m s–1 (d) 48.2 m s–1

Physics for you | february ‘13

31

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30. A hockey player is moving northward and suddenly turns westward with the same speed to avoid an opponent. The force that acts on the player is (a) frictional force along westward. (b) muscle force along southward. (c) frictional force along south-west. (d) muscle force along south-west.

solutions

3. (a) : Here, m = 10 kg The resultant force acting on the body is F = (8 N)2 + (6 N)2 = 10 N

6N

F

1. (c) 2. (d) : The second law  of motion is a local law which means that force F at a point in space (location of the particle) at a certain instant of time is related  to a at that point at that instant. Acceleration here and now is determined by the force here and now, not by any history of the motion of the particle.

 8N

Let the resultant force F makes an q w.r.t. 8 N force. 6 N 3 From figure, tan θ = = 8 N 4 The resultant acceleration of the body is F 10 N a= = = 1 m s–2 m 10 kg The resultant acceleration is along the direction of the resultant force. Hence, the resultant acceleration of the body is 3 1 m s–2 at an angle of tan–1   w.r.t. 8 N force. 4 4. (d) : Newton’s second and third laws of motion lead to conservation of linear momentum. 5. (b) : Among the given forces magnetic force is a non-contact force. 6. (a) : Here, x(t) = pt + qt2 + rt3 where p = 3 m s–1, q = 4 m s–2 and r = 5 m s–3 m = 2 kg dx d (pt + qt2 + rt3) = p + 2qt + 3rt2 Velocity, v = = dt dt dv Acceleration, a = = 2q + 6rt dt At t = 2 s a = 2(4 m s–2) + 6(5 m s–3)(2 s) = 8 m s–2 + 60 m s–2 = 68 m s–2 The force acting on the body of mass 2 kg is F = ma = (2 kg)(68 m s–2) = 136 N 32 Physics for you |

february ‘13

7. (b) : The reading on the scale is a measure of the force on the floor by the person. By the Newton’s third law this is equal and opposite to the normal force N on the person by the floor. \ When the lift is descending downward with a acceleration of a m s–2, then 50 × 10 – N = 50 × 9 or N = 50 × 10 – 50 × 9 = 50 N \ The reading of weighing machine is 5 kg. 8. (b) :

Initial momentum of ball A = (0.05 kg)(5 m s–1) = 0.25 kg m s–1 As the speed is reversed on collision, Final momentum of the ball A = (0.05 kg)(–5 m s–1) = – 0.25 kg m s–1 Impulse imparted to the ball A = Change in momentum of ball A = Final momentum – Initial momentum = – 0.25 kg m s–1 – 0.25 kg m s–1 = – 0.5 kg m s–1 Similarly, Initial momentum of ball B = (0.05 kg)(–5 m s–1) = – 0.25 kg m s–1 Final momentum of ball B = (0.05 kg)(5 m s–1) = + 0.25 kg m s–1 Impulse imparted to ball B = (0.25 kg m s–1) – (– 0.25 kg m s–1) = 0.5 kg m s–1 Impulse imparted to each ball is 0.5 kg m s–1 in magnitude. The two impulses are opposite in direction.  ^ ^ 9. (b) : Here, m = 5 kg, F = −3 i + 4 j N ^ ^  u = 6 i − 12 j m s −1 The acceleration of the body is  ^ ^  F ( −3 i + 4 j ) N 3^ 4^ a= = = − i + j m s −2 m 5 kg 5 5 Velocity of the body along x-axis at any time t is ^ 3^ v x = u x + a xt = 6 i − i t 5 The body will have a velocity along y-axis, if its velocity along x-axis will be zero. i.e. vx = 0 ⇒ t = 10 s m 10. (d) : Here, mA = , mB = m 2 mA = 0.2, mB = 0.1 Let both the blocks are moving with common acceleration a. Then,

WorldMags.net

WorldMags.net

14. (c) : Here, v = 20 m s–1, a = 2 m s–2, g = 10 m s–2 The coin will fall back into the person’s hand after t s.

µA mA g = µA g = 0.2 g mA and F – mB(mB + mA)g = (mB + mA)a F = (mB + mA)a + mB(mB + mA)g a=

  m m =  m +  (0.2 g ) + (0.1)  m +  g   2 2 3  3  0.9 =  2 m  (0.2 g) +  2 m  (0.1g ) = mg = 0.45 mg 2 11. (b) : Here, v = 15 m s–1 Area of cross section, A = 10–2 m2 Density of water, r =

103

kg

\ t =

2 × 20 m s −1 2v 40 10 s= s = 3.33 s = = a + g ( 2 + 10) m s −2 12 3

15. (a) : The free body diagram of 3 kg block is as shown in the Fig. (a). The equation of motion of 3 kg block is T2 – 3g = 3a T2 = 3(a + g)

m–3

Mass of water hitting the wall per second

= 3(2 + 10) = 36 N

=r×A×v = 103 kg m–3 × 10–2 m2 × 15 m s–1 = 150 kg s–1

The free body diagram of 5 kg is as shown in the Fig. (b).

Force exerted on the wall = Momentum loss of water per second = 150 kg

s–1

× 15 m

s–1

= 2250 N = 2.25 ×

The equation of motion of 5 kg block is T1 – T2 – 5g = 5a

103

N

12. (d) : The greater the change in the momentum in a given time, the greater is the force that needs to be applied. 13. (b) :





5 θ = tan −1   = 40° 6

T1 = 5(a + g) + T2



= 5(2 + 10) + 36



= 96 N

(Using (i))

16. (c) : If F is the force of the finger on the book, F = N, the normal reaction of the wall on the book. The minimum upward frictional force needed to ensure that the book does not fall is Mg. The frictional force = mN. Mg Thus, minimum value of F = . µ 17. (a) :

The free body diagram of 6 kg block is as shown in Fig. (b). In equilibrium T2 = 6g = 6 × 10 = 60 N The free body diagram of the point P is as shown in Fig. (c). In equilibrium T1sinq = 50 N ...(i) T1cosq = T2 = 60 N ...(ii) Dividing (i) by (ii), we get 50 5 tan θ = = 60 6

…(i)

2N

y

2 cos45° 1 cos45°1 N 45° 45°

2 sin45° P 90° F1

1 sin45°

x

F2

Applying equilibrium conditions, SFx = 0 ⇒ F1 + 1sin45° – 2sin45° = 0 or F1 = 2sin45° – 1sin45° 2 1 1 = N − = 2 2 2 and SFy = 0 ⇒ 1cos45° + 2sin45° – F2 = 0 1 2 3 + = F2 = N 2 2 2 18. (b)

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Physics for you | february ‘13

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19. (b) : Here,

23. (b) :

Mass of the gun, M = 100 kg



T

Mass of the ball, m = 1 kg A

g = 10 m s–2



30°

\ 400 = u × 10 where u is the velocity of the ball. u = 40 m s–1

According to law of conservation of linear momentum, we get 0 = Mv + mu v=−

(1 kg )( 40 m s −1 ) mu =− = − 0.4 m s −1 100 kg M

–ve sign shows that the direction of recoil of the gun is opposite to that of the ball. 20. (b) : If a large force F acts for a short time dt the impulse imparted I is dp I = Fdt = dt dt I = dp = change in momentum 21. (d) : Here, mass of the man M = 65 kg As the man is standing stationary w.r.t. the belt, acceleration of man = acceleration of belt. \ Acceleration of man, a = 1 m s–2 \ Net force on the man

F = Ma = (65 kg)(1 m s–2) = 65 N

22. (a) : Since the graph between x and t is a straight line and passing through the origin. \ x = t Since the graph between y and t is a parabola. \ y = t2 dv \ v = dx = 1 and a = x = 0 x x dt dt and vy =

30

2 × 500 m

= 10 s 10 m s −2 Horizontal distance covered = ut



gs in m

2h = g

°

Time taken by the ball to reach the ground is t=

dy = 2t and ay = 2 m s–2 dt

The force acting on the particle is F = may = (0.5 kg)(2 m s–2) = 1 N along y-axis 34 Physics for you |

february ‘13

T

N

Height of the cliff, h = 500 m

mg 30°

mgcos30°

B W

As the system is in equilibrium, \ T = W and T = mgsin30° From (i) and (ii), we get 1 W = mgsin30° = (100 N) = 50 N 2

…(i) …(ii)

24. (d) : At t = 4 s, the body has constant velocity 3 u = m s–1. 4 After t = 4 s, the body is at rest i.e., v = 0 \ Impulse = m(v – u) 3 3 = 2 kg (0 – m s–1) = − kg m s–1 2 4 25. (a) : The net force on the particle is T and is directed towards the centre of the circle. The tension T provides the necessary centripetal force to the particle moving in the circle. 26. (a) : The force on 7th coin is due to weight of the three coins lying above it. \ F = 3mg This force acts vertically downwards. 27. (b) 28. (a) : At the lowest point, mg acts downwards and T1 upwards so that net force = mg – T1. At the highest point, both mg and T2 act downwards so that net force = mg + T2. Hence, option (a) is correct. 29. (c) : Here R = 300 m, q = 15°, g = 9.8 m s–2, m = 0.2 The maximum permissible speed is given by vmax =

Rg(µ + tan θ) = 1 − µ tan θ

300 × 9.8 × (0.2 + 0.27 ) 1 − 0.2 × 0.27

= 38.2 m s–1

30. (c)

nn

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WorldMags.net It is a matter of great satisfaction and encouragement to realise that our efforts in “Excel in Physics, Excel in Chemistry and Excel in Biology” have received overwhelming response. Nearly all the questions asked in the CBSE Board Examination 2012 were available in the books; fully solved. We feel pleased to present the revised edition of the books. The books give comprehensive account of the subject according to the current syllabus and pattern of the CBSE Board Examination. This will impart the students a clear and vivid understanding of the subject matter.

Available at leading bookshops throughout India.

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Physics for you | february ‘13

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1. Why are alloys used for making standard resistance coils? [1] 2. How does focal length of a lens change when red light is replaced by blue light? [1] 3. Name the series of hydrogen spectrum lying in the infrared region. [1] 4. Would sky waves be suitable for transmission of TV signals of 60 MHz frequency? [1] 5. W hat is the colour code for a resistor of resistance 3.5 kW with 5% tolerance? [1] 6. W hich physical quantity has the unit Wb m–2? Is it a scalar or a vector quantity? [1] 7. A solenoid is connected to a battery so that a steady current flows through it. If an iron core is inserted into the solenoid, will the current increase or decrease? Explain. [1] 8. Why is ground wave transmission of signals restricted to a frequency of 1500 kHz? [1]  – + 9. A uniform field E exists + A B – – between two charged + – plates as shown in figure. ++ – – What would be work done + – + – in moving a charge q along + – – the closed rectangular path + + – ABCDA? – + C – D + – + – + [2] 10. What is an ideal diode? Draw the output waveform across the load resistor R, if the input waveform is as shown in the figure. +6V R

0





–6V



[2]

11. Find the ratio of intensities of two points P and Q on a screen in Young’s double slit experiment when waves from sources S1 and S2 have phase 36 physics for you | February’13

π respectively. [2] 2 12. Two conductors are made of the same material and have the same length. Conductor A is solid wire of diameter 1 mm. Conductor B is a hollow tube of outer diameter 2 mm and inner diameter 1 mm. Find the ratio of resistance RA to RB. [2] difference of (i) 0° and (ii)

13. The current sensitivity of a moving coil galvanometer increases by 20% when its resistance is increased by a factor 2. Calculate by what factor the voltage sensitivity changes. [2] 14. A coil of 0.01 H inductance and 1 W resistance is connected to 200 V, 50 Hz ac supply. Find the impedance of the circuit and time lag between maximum alternating voltage and current. [2] 15. With the help of an example, explain, how the neutron to proton ratio changes during alpha decay of a nucleus. [2] 16. Two long parallel wires are hanging freely. If they are connected to a battery (i) in series, (ii) in parallel, what would be the effect on their positions? [2] 17. Obtain equivalent capacitance of the following network. For a 300 V supply, determine the charge and voltage across each capacitor. [3]

18. Which of the following waves can be polarised? (i) X-rays (ii) Sound waves. Give reasons. Two polaroids are used to study polarisation. One of them (the polariser) is kept fixed and the other (the analyser) is initially kept with its axis parallel to the poalriser. The analyser is then rotated through angles of 45°, 90° and 180° in turn. How would the intensity of light coming out of analyser be affected for these angles of rotation,

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as compared to the initial intensity and why? [3]

19. Illustrate the basic elements required for transmitting and receiving an audio signal with the help of a block diagram. [3]

20. Two cells of voltage R 10 V and 2 V and internal resistances 10 W and 5 W 2V 5Ω respectively, are connected in parallel with the positive end of 10 V 10 V 10 Ω battery connected to negative pole of 2 V battery as shown in figure. Find the effective voltage and effective resistance of the combination. [3] 21. A source contains two phosphorus radionuclides 32 33 15 P (T1/2 = 14.3 days) and 15 P (T1/2 = 25.3 days). 33 . How Initially 10% of the decays come from 15 P long one must wait until 90% do so? [3] 22. The two plates of a parallel plate capacitor are 4 mm apart. A slab of dielectric constant 3 and thickness 3 mm is introduced between the plates with its faces parallel to them. The distance between the plates is so adjusted that the rd 2 capacitance of the capacitor becomes   of its 3 original value. What is the new distance between the plates? [3]

OR Define the term electric potential due to a point charge. Find the electric potential at the centre of a square of side 2 m , having charges 100 mC, –50 mC, 20 mC and – 60 mC at the four corners of the square. [3]

23. How would you establish an instantaneous displacement current of 2.0 A in the space between the two parallel plates of 1 mF capacitor? [3] 24. The figure shows a rectangular current carrying loop, placed 2 cm away from a 25 cm long straight, current carrying conductor. What is the direction and magnitude of the net force acting on the loop?

15 A 25 A

2 cm 10 cm

[3]

25. Two lenses of powers + 15 D and – 5 D are in contact with each other forming a combination lens. (a) What is the focal length of this combination? (b) An object of size 3 cm is placed at 30 cm from this combination of lenses. Calculate the position and size of the image formed. [3] 26. In an experiment of photoelectric effect, Neeta plotted graphs for different observation between photoelectric current and collector plate potential but her friend Megha has to help her in plotting the correct graph. Neeta thanked Megha for timely help. (a) What value was displayed by Megha and Neeta. (b) Draw the correct graph between I and V. [4] 27. What is induced emf? Write Faraday’s law of electromagnetic induction. Express it mathematically. A conducting rod of length l, with one pivoted, is rotated with a uniform angular speed w in a vertical plane, normal to a uniform magnetic field B. Deduce an expression for the emf induced in this rod. In India, domestic power supply is at 220 V, 50 Hz, while in USA it is 110 V, 50 Hz. Give one advantage and one disadvantage of 220 V supply over 110 V supply. [5]

OR Explain the phenomenon of self induction. Define coefficient of self inductance. What are its units? Calculate self inductance of a long solenoid. [5]

28. (a) With the help of a circuit diagram explain the working of transistor as oscillator. (b) Draw a circuit diagram for a two inputs OR gate and explain its working with the help of input, output waveforms. [5] OR Define the terms potential barrier and depletion region for a p-n junction. Explain with the help of a circuit diagram, the use of a p-n diode as a full wave rectifier. Draw the input and output wave forms. [5] 29. Define magnifying power of an optical telescope. Draw a ray diagram for an astronomical refracting telescope in normal adjustment showing the paths through the instrument of three rays from a distant object. Derive an expression for its magnifying power. Write the singificance of diameter of the objective lens on the optical performance of a telescope. [5]

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physics for you | February’13

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OR State Huygens principle and prove laws of reflection and refraction on the basis of Huygens principle. [5] SOLUTIONS

1. Alloys are used for making standard resistance coils because they have low value of temperature coefficient (less temperature sensitivity) of resistance and high resistivity. 2. According to lens maker’s formula,



3. Paschen series, Brackett series and Pfund series. 4. No, signals of frequency greater than 30 MHz will not be reflected by the ionosphere, but will penetrate through the ionosphere. 5. Given, resistance = 3.5 kW ± 5% = 35 × 102 W ± 5% \ Colour code of given resistor is orange, green, red and gold. 6. Magnetic field induction has the unit Wb m–2. It is a vector quantity. 7. The current will decrease. As the iron core is inserted in the solenoid, the magnetic field increases and the flux increases. Lenz’s law implies that induced emf should resist this increase, which can be achieved by a decrease in current. 8. In ground wave propagation, the loss of energy due to interaction with matter increases with the increase in frequency of wave. Therefore, the waves of frequency above 1500 kHz get heavily damaged in ground wave propagation. Hence the ground wave propagation is restricted to a frequency of 1500 kHz. 9. Work done in moving a charge along the closed rectangular path would be zero, because field in the entire space is uniform and electrostatic forces are conservative forces. 10. A p-n junction diode which offers zero resistance when forward biased and infinite resistance when reverse biased is called an ideal diode. The output waveform across R is as shown in the figure below.

0

38 physics for you | February’13





As mb > mr \ fb < fr i.e. focal length of lens decreases.







 1 1 1 = (µ − 1)  −  f  R1 R2 

+6V

11. In Young’s double slit experiment, the resultant intensity at any point on the screen is I = I1 + I 2 + 2 I1I 2 cos φ where f is the phase difference between the waves at that point. Here, we consider I1 = I2 = I0 Therefore, When f = 0°, the intensity at point P is ∴ I P = I0 + I0 + 2 I0 I0 cos 0° = 4 I0 π When φ = , the resultant intensity at point Q is 2 π I = I Q 0 + I 0 + 2 I 0 I 0 cos = 2 I 0 2 IP 2 ∴ = IQ 1

12. Resistance of conductor A, ρl RA = π(0.5 × 10 −3 )2 Resistance of conductor B, ρl RB = π[(1 × 10 −3 )2 − (0.5 × 10 −3 )2 ] \

RA (1 × 10 −3 )2 − (0.5 × 10 −3 )2 0.75 3 = = = RB 0.25 1 (0.5 × 10 −3 )2

20 120 I = I , R′ = 2R 100 s 100 s Is Then, initial voltage sensitivity, Vs = R New voltage sensitivity,

13. Given, Is′ = I s +

I s′  120  1 3 I × = = V R′  100 s  2 R 5 s \ % decrease in voltage sensitivity 3 Vs − Vs Vs − Vs ′ 5 × 100 × 100 = = 40% = Vs Vs

Vs ′ =

14. Here, R = 1 W, u = 50 Hz, L = 0.01 H The inductive reactance is XL = wL = 2puL = 2 × 3.14 × 50 × 0.01 = 3.14 W The impedance of the circuit is 2 2 Z = R + X L = (1)2 + ( 3.14)2 = 10.86  3.3 Ω The phase difference between current and voltage is XL = 3.14 tan f = R 72 × π rad f = tan–1(3.14)  72°  180 φ 72 × π 1 Time lag, Dt = = = s ω 180 × 2 π × 50 250

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15. Consider the a-decay of nucleus A ZX

 →

A−4 Z−2Y

Potential difference across C4 is q 2 × 10 −8 V4 = 4 = = 200 V C4 100 × 10 −12 From (i), Vp = 300 – V4 = 300 – 200 = 100 V

+ 42 He

238 92 U

4  → 234 90 Th + 2 He Before decay, neutron to proton ratio, n/p 238 − 92 = 1.58 = 92 After decay, neutron to proton ratio, n/p



=

Thus ratio increases. 16. (i) (ii)

Charge on C1 is q1 = C1V1 = 100 × 10–12 × 100 = 10–8 C Potential difference across C2 and C3 in series

234 − 90 = 1.60 90

C1 = 100 pF

C2 = 200 pF

Charge on C3 is q3 = C3 V3 = 200 × 10–12 × 50 = 10–8 C 18. (i) Phenomenon of polarisation is shown by transverse waves only. X-rays are transverse in nature, and hence they can be polarised. According to Malus law I = I0 cos2q where I = intensity of light coming from analyser I0 = initial intensity I When q = 45°, then I = I0 cos245° = 0 2 When q = 90°, then I = I0 cos290°= 0

19.

+ 300 V –

Microphone

C4 = 100 pF



Modulator

Here, C2 and C3 are in series. 1 1 1 2 1 = + = = \ Cs 200 200 200 100 Cs = 100 pF



Cp = Cs + C1 = 100 + 100 = 200 pF

Again, Cp and C4 are in series, therefore the equivalent capacitance of the network is 1 1 1 1 1 3 = + + = = \ C C p C4 200 100 200 200 pF = 66.7 × 10–12 F 3 As Cp and C4 are in series,

\

C=

\

Vp + V4 = 300



...(i)

200 × 10 −12 × 300 3 = 2 × 10–8 C

Charge on C4 is q4 = CV =

Transmitting antenna

Receiving antenna Loudspeaker

Transmitter

Detector

Oscillator

Now Cs and C1 are in parallel. \

When q = 180°, then I = I0 cos2180° = I0



C3 = 200 pF

= 100 V

Charge on C2 is q2 = C2 V2 = 200 × 10–12 × 50 = 10–8 C

When a battery is connected in series to two long parallel wires, the currents in the two wires will be in opposite directions. Due to which a force of repulsion will be acting between them and they move further apart. When a battery is connected in parallel to two long parallel wires, the currents in the two wires will be in same direction. Due to which a force of attraction will be acting between them and they come closer to each other.

17.



Potential difference across C1 is V1 = Vp = 100 V

A brief description of the various elements is as given below. (i) A microphone converts sound waves into electrical waves i.e. audio signal. (ii) An oscillator generates carrier waves. (iii) There is a mixing of carrier waves and audio signal in a modulator. (iv) The modulated waves are fed to transmitter, these waves are then radiated through transmitting antenna. (v) T h e r e c e i v i n g a n t e n n a r e c e i ve s t h e transmitting signal. (vi) The detector demodulates (separate out) the audio signal from the modulated waves. (vii) The loudspeaker converts the audio signal back into sound waves.

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physics for you | February’13

39

20.

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R

C I

2V

B A



F

I1 10 V 10 Ω

 1

Veff

or

Reff

Applying Kirchhoff’s junction rule at junction B, we get ...(i) I1 = I + I2 Applying Kirchhoff’s loop rule for closed loop ABCDEFA gives 10 = IR + 10I1 ...(ii) Applying Kirchhoff’s loop rule for closed loop BCDEB gives 2 = 5I2 – IR = 5(I1 – I) – IR (Using (i)) ...(iii) or 4 = 10I1 – 10I – 2IR Equation (i) – (iii) gives 6 = 3IR + 10I  10  or 2 = I  R +   3 or 2 = (R + Reff)I Comparing with Veff = (R + Reff)I, we get Veff = 2 V and Reff =

10 Ω 3

21. Initially, the source has 90% of 10% of

33 15 P

32 15 P

nuclides and

nuclides. Finally, say after t days, the

source has 10% of nuclides.

32 15 P

nuclides and 90% of

33 15 P

32 15 P nuclides = 9x; 33 Initial no. of 15 P nuclides = x 32 Final no. of 15 P nuclides = y; 33 Final no. of 15 P nuclides = 9y

\ Initial no. of

n

As or For

 1 N  1 = =   2 N 0  2  N = N0 ( 2) 32 P 15

t / T1/ 2

= ( 2)

1 

−  1 t 9 = ( 2)  14.3 25.3  or 81 = 211t / 14.3 × 25.3 9 11t ∴ log10 81 = log10 2 14.3 × 25.3

I

E

I2

5Ω

Dividing Eq. (ii) by Eq. (i), we get

R

D

1.9085 =

11t × 0.3010 14.3 × 25.3

1.9085 × 14.3 × 25.3 = 208.5 days 11 × 0.3010 22. Here, distance between parallel plates d = 4 mm = 0.004 m, K = 3, thickness, t = 3 mm = 0.003 m Let the new distance between the plates be d1. ε0A ε0 A and C1 = ∴ C= d  1 d1 − t  1 −   K 2 Since C1 = C (Given) 3 ε0 A 2 ε0 A ∴ =  1 3 d d1 − t  1 −   K 1 2 = d 3   1 d1 − t  1 −   K 1 2 =  1  3 × 0.004 d1 − 0.003  1 −   3 1 1 = 2 0.006 d1 − 0.003 × 3 1 1 = d1 − 0.002 0.006 d1 – 0.002 = 0.006 d1 = 0.006 + 0.002 = 0.008 m = 8 mm ∴



− t / T1/ 2

t=

OR Electric potential at a point is the amount of work done to bring a unit positive charge from infinity to that point against the electrostatic forces.

− t / T1/ 2

isotope,

N0 = 9x; N = y; T1/2 = 14.3 days, \ For

y = 9x(2)–t/14.3 33 P 15

...(i)

isotope,

\

9y = x(2)–t/25.3

40 physics for you | February’13

2m



N0 = x; N = 9y; T1/2 = 25.3 days ...(ii)



AC = BD =

( 2) + ( 2)

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2

2

=2m

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\ AO = OC = BO = OD = 1 m



Potential at the centre of the square O is V=



Focal length, f =

q q q  1  qA + B + C + D   4 πε0  AO BO OC OD 

(b)

 100 × 10 −6 50 × 10 −6 20 × 10 −6 60 × 10 −6   V = 9 × 109  − + − 1 1 1 1  

\

= 9 × 109 × 10 × 10–6 = 9 × 104 V



23. Here, ID = 2.0 A, C = 1 mF = 10–6 F dφ d dE ID = ε0 E = ε0 (EA) = ε0 A dt dt dt



24.

B

15 A

C

(b)

P

2 cm



10 cm

D

Q

Force between wires PQ and CD µ 0 I1I 2 l F1= 2 πr 2 × 10 −7 × 15 × 25 × 0.25 = = 93.75 × 10–5 2 × 10 −2 = 9.375 × 10–4 N (Repulsive) Force between wires PQ and AB 2 × 10 −7 × 15 × 25 × 0.25 0.12 = 1.56 × 10–4 N (Attractive) Net force on the rectangular loop F = F1 – F2 = (9.375 – 1.56) × 10–4 = 7.815 × 10–4 N (Repulsive i.e., towards left).

1 3−1 2 = = v 30 30



hI =

I3 > I2 > I1

I3 I2 I1

Stopping potential 0

Collector plate potential

27. Refer point 4.1(3, 4, 5, 7 (d)) page no. 224, 225 (MTG Excel in Physics). Advantage: At 220 V supply power loss due to heating effect is lesser. Disadvantage: At 220 V peak value of current is more. Thus, it is more dangerous.

25 cm

A

or

v = 15 cm v hI m= = u ho

–V0 Retarding potential

25 A



1 1 1 1 1 = + = + v f u 10 −30

Photocurrent

d V  A dV dV =C   = ε0 dt  d  d dt dt  E = V and C = ε0 A     d d  dV ID 2.0 or = = = 2 × 106 V s −1 dt C 10 −6 Thus a displacement current of 2.0 A can be set up by changing the potential difference across the parallel plates of capacitor at the rate of 2 × 106 V s–1.

1 1 1 = − f v u

15 v = × 3 = – 1.5 cm u −30 26. (a) The values displayed by them is sharing and caring. or

= ε0 A



1 1 = × 100 = 10 cm P 10

F2 =

25. (a) Here, P1 = 15 D, P2 = –5 D Power of the combination, P = P1 + P2 = 15 D – 5 D = 10 D



OR Refer point 4.2(1) page no. 226 (MTG Excel in Physics).

28. (a) Refer point 9.4(10) page no. 541, (MTG Excel in Physics).



(b) Refer point 9.5(1) page no. 542, (MTG Excel in Physics). OR Refer point 9.3[(2), 6(ii)] page 532, 534 (MTG Excel in Physics).

29. Refer point 6.9(3), page 347 (MTG Excel in Physics).

OR Refer point 6.10(6) and point 6.11 page no. 403, 404 (MTG Excel in Physics). mmm

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1. The random error in the arithmetic mean of 100 observations is x, then the random error in the arithmetic mean of 400 observations would be 1 (a) 4x (b) x 4 1 (c) 2x (d) x 2 2. After one second the velocity of a projectile makes an angle of 45° with the horizontal. After another one second it is travelling horizontally. The magnitude of its initial velocity and angle of projection are (Take g = 10 m s–2) (a) 14.62 m s–1, 60° (b) 14.62 m s–1, tan–1(2) (c) 22.36 m s–1, tan–1(2) (d) 22.36 m s–1, 60° 3. A block of mass 1 kg is 1 kg placed over a plank of 2 kg F = 30 N mass 2 kg. The length of the plank is 2 m. 2m Coefficient of friction between the block and the plank is 0.5 and the ground over which plank is placed is smooth. A constant force F = 30 N is applied on the plank in horizontal direction. The time after which the block will separate from the plank is (Take g = 10 m s–2) (a) 0.73 s (b) 1.2 s (c) 0.62 s (d) 1.6 s 4. A boy of mass 30 kg starts running from rest along a circular path of radius 6 m with constant tangential acceleration of magnitude 2 m s–2. After 2 s from start he feels that his shoes started slipping on ground. The friction between his shoes and ground is (Take g = 10 m s–2) 1 1 1 1 (a) (b) (c) (d) 5 3 4 2 42 physics for you |

february‘13

5. A light rod of length 200 cm is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross-section 0.1 cm2 and the other of brass of cross-section 0.2 cm2. Along the rod at which distance a weight may be hung to produce equal stresses in both the wires? (a) 4 m from steel wire 3 4 m from brass wire (b) 3 (c) 1 m from steel wire 1 m from brass wire 4 6. Two cylinders A and B, fitted with pistons, contain equal amounts of an ideal diatomic gas at 300 K. The piston of A is free to move, while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30 K then the rise in temperature of the gas in B is (a) 30 K (b) 18 K (c) 50 K (d) 42 K (d)

7. A spherical conductor A of radius r is placed concentrically inside a conducting shell B of radius R(R > r). A charge Q is given to A, and then A is joined to B by a metal wire. The charge flowing from A to B will be  r   R  (a) Q  (b) Q    R + r   R + r (c) Q

(d) zero

8. A 1 mF capacitor is connected in the circuit shown below. The emf of the cell is 3 V and internal resistance is 0.5 W. The resistors R1 and R2 have

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14. A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m s–2. He reaches the ground with a speed of 3 m s–1. At what height did he bail out? (a) 293 m (b) 111 m (c) 91 m (d) 182 m

4

15. Two blocks A and B of masses 2m and m,

values 4 W and 1 W respectively. The charge on the capacitor in steady state must be 3 V 0.5  1 F R2



1

(a) 1 mC (c) 1.33 mC

(b) 2 mC (d) zero

9. A long cylindrical wire kept along z-axis carries a  ^ current of density J = J0r k , where J0 is a constant and r is the radial distance from the axis of the cylinder. The magnetic field inside the conductor at a distance d from the axis of the cylinder is µ 0 J0 d (a) µ 0 J0 (b) 2 µ 0 J0 d 3 µ 0 J0 d 2 (c) (d) 4 3 10. An object of mass 0.2 kg executes simple harmonic oscillations along the x-axis with a frequency 25 Hz. At the position x = 0.04 m, the object has π kinetic energy 0.5 J and potential energy 0.4 J. The amplitude of oscillation is (Potential energy is zero at mean position) (a) 6 cm (b) 4 cm (c) 8 cm (d) 2 cm 11. Velocity of sound in an open organ pipe is 330 m s–1. The frequency of wave is 1.1 kHz and the length of tube is 30 cm. To which harmonic does this frequency correspond? (a) 2nd (b) 3rd (c) 4th (d) 5th 12. A point object is placed at a distance of 25 cm from a convex lens of focal length 20 cm. If a glass slab of thickness t and refractive index 1.5 is inserted between the lens and the object the image is formed at infinity. The thickness t is (a) 10 cm (b) 5 cm (c) 20 cm (d) 15 cm 13. An ideal massless spring S can be compressed 2 m by a force of 200 N. This spring is placed at the bottom of the frictionless inclined plane which makes an angle q = 30°­ with the horizontal. A 20 kg mass is released from rest at the top of the inclined plane and is brought to rest momentarily after compressing the spring 4 m. Through what distance does the mass slide before coming to rest? (a) 2.2 m (b) 4 m (c) 8.17 m (d) 1.9 m

respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as A 2m shown in the figure. The magnitudes of acceleration A and B immediately after B m the string is cut, are respectively g g ,g (a) g , (b) 2 2 g g ,  (c) g, g (d) 2 2 16. If the work done in blowing a soap bubble of volume V is W, then the work done in blowing a soap bubble of volume 2V will be (a) W (b) 2W (c) 2W (d) W(4)1/3 17. A hydrogen atom in an excited state emits a photon which has the longest wavelength of the Paschen series. Further emissions from the atom cannot include the (a) longest wavelength of the Lyman series (b) second longest wavelength of the Lyman series (c) longest wavelength of the Balmer series (d) second longest wavelength of the Balmer series 18. A horizontal rod rotates about a vertical axis through one end. A ring, which can slide along the rod without friction, is initially close to the axis and then slides to the other end of the rod. In this process, which of the following quantities will be conserved? [L = angular momentum, KT = total kinetic energy, KR = rotational kinetic energy] (a) L only (b) L and KT only (c) L and KR only (d) KT only 19. A photon of energy 10.2 eV corresponds to light of wavelength l0. Due to an electron transition from n = 2 to n = 1 in a hydrogen atom, light of wavelength l is emitted. If we take into account the recoil of the atom when the photon is emitted, (a) l = l0 (b) l < l0 (c) l > l0 (d) the data is not sufficient to reach a conclusion

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20. An electron of mass me initially at rest moves through a certain distance in a uniform electric field in time t1. A proton of mass mp also initially at rest takes time t2 to move through an equal distance in this uniform electric field. Neglecting t the effect of gravity, the ratio 2 is nearly equal to t1  mp  (b)    me 

(a) 1 m  e  (c)   mp 

1/ 2



1/ 2

(d) 1836

21. A spherical ball A of mass 4 kg, moving along a straight line strikes another spherical ball B of mass 1 kg at rest. After the collision, A and B move with velocities v1 m s–1 and v2 m s–1 respectively making angles of 30° and 60° with respect to the v original direction of motion of A. The ratio 1 v2 will be 4 1 3 (a) (b) (c) (d) 3 4 3 3

P2

5 V, 2 



10 

2 V, 1 

P1

(a) 0.27 A, P1 to P2 (c) 0.03 A, P1 to P2

(b) 0.27 A, P2 to P1 (d) 0.03 A, P2 to P1

26. A galvanometer of 50 W resistance has 25 divisions. A current of 4 × 10 – 4 A gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25 V, it should be connected with a resistance of (a) 2500 W as a shunt (b) 2450 W as a shunt (c) 2550 W in series (d) 2450 W in series 27. What is the charge induced in coil of 100 turns of resistance 100 W, if magnetic flux changes from 2 T m2 to – 2 T m2? (a) 4 C (b) 2 C (c) 2.8 C (d) 0.4 C 28. If an electron revolves around a proton, then its time period T is (R = radius of orbit) (a) ∝ R2 (b) ∝ R3/2 (c) ∝ R3 (d) ∝ R

22. A bullet of mass 20 g moving with 600 m s–1 collides with a block of mass 4 kg hanging with the string. What is the velocity of bullet when it comes out of block, if block rises to height 0.2 m after collision? (Take g = 10 m s–2) (a) 200 m s–1 (b) 150 m s–1 –1 (c) 400 m s (d) 300 m s–1

29. A uniform circular disc of mass 12 kg is held by two identical springs as shown in the figure. When the disc is pressed down slightly and released, it executes SHM with a time period of 2 s. The spring constant of each spring is (a) 236 N m–1 (b) 118.3 N m–1 (c) 59.15 N m–1 (d) 108.3 N m–1

23. A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest its time period is T. With what acceleration should the lift be accelerated upwards in order to reduce its period T to ? (g is the acceleration due to gravity) 2 (a) 4g (b) g (c) 2g (d) 3g

30. Two thin uniform circular rings each of radius 10 cm and mass 0.1 kg are arranged such that they have common centre and their planes are perpendicular to each other. The moment of inertia of this system about an axis passing through common centre and perpendicular to the plane of either of the rings in kg m2 is (a) 15 × 10–3 (b) 5 × 10 –3 (c) 15 × 10 –4 (d) 18 × 10–4

24. Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres of A and B is (a) 1 : 2

(b) 2 : 1

(c) 1 : 4

(d) 4 : 1

25. A 5 V battery with internal resistance 2 W and 2 V battery with internal resistance 1 W are connected to a 10 W resistor as shown in figure. The current in 10 W resistor is 44 physics for you |

february‘13

SOLUTIONS 1. (b) : Since error is measured for 400 observations instead of 100 observations. So error will reduce 1 by factor. 4 Hence, the random error in the arithmetic mean x of 400 observations would be . 4 2. (c) : Total time of flight is T = 4 s and if u is its initial speed and q the angle of projection. Then

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2u sin θ = 4 g or usinq = 2g ...(i) After 1 s velocity vector makes an angle of 45° with horizontal i.e., vx = vy ucosq = usinq – gt ucosq = usinq – g ( t = 1 s) ucosq = 2g – g (Using (i)) or ucosq = g ...(ii) Squaring and adding (i) and (ii), we get



T=

u2sin2q + u2cos2q = (2g)2 + (g)2

u2 = 5g2 = 5(10)2 m2 s–2 \ u = 22.36 m s–1 Dividing (i) by (ii), we get, u sin θ 2 g = =2 u cos θ g tanq = 2 or q = tan–1 (2) 3. (a) : Maximum frictional force between the block and the plank is f max = mmg = (0.5)(1)(10) = 5 N The free body diagrams of the block and the plank are as shown in the figure (a) and (b) respectively. 5N



(a)

5N

30 N (b)

5 = 5 m s −2 1 30 − 5 25 Acceleration of plank, a2 = m s −2 = 2 2 \ Relative acceleration of plank a = a2 – a1 25 a= − 5 = 7.5 m s −2 2 2S 2×2 ∴ t= = = 0.73 s a 7.5 Acceleration of block, a1 =

4. (b) : After 2 s speed of boy will be v = 2 × 2 = 4 m s–1 At this moment centripetal force on boy is mv 2 30 × 16 Fc = = N = 80 N R 6 Tangential force on boy is Ft = ma = 30 × 2 N = 60 N Total force acting on boy is Fc2

Ft2

2

2

F= + = (80) + (60) = 100 N At the time of slipping, F = mmg 1 or 100 = m × 30 × 10 or µ = 3

5. (a) : A

B

2m T1

T2

x W

As stresses are equal, T1 = T2 A1 A2 T1 A1 0.1 i.e. ...(i) = = or T2 = 2T1 T2 A2 0.2 Now for translatory equilibrium of the rod, T1 + T2 = W ...(ii) From (i) and (ii), we get 2W W T1 = ; T2 = 3 3 Now if x is the distance of weight W from steel wire, then for rotational equilibrium of rod, W 2W T1x = T2 (2 – x) or x= ( 2 − x) 3 3 4 ∴ x= m 3 6. (d) : In cylinder A, heat is supplied at constant pressure while in cylinder B, heat is supplied at constant volume. \ (DQ)A = nCPDTA and (DQ)B = nCVDTB Given: (DQ)A = (DQ)B nCPDTA = nCVDTB C or ∆TB = P ∆TA CV C 7  For diatomic gas, P = CV 5 7 ∴ ∆TB = × 30 K = 42 K 5 7. (c) : When charge of amount q has flown from A to B, the charge on A is (Q – q). The potentials of A and B are 1 Q−q 1 q VA = + πε πε 4 r 4 0 0 R 1 Q−q 1 q VB = + 4 πε 0 R 4 πε 0 R 1 1 1 (Q − q )  −  > 0 ∴ VA − VB =  r R 4 πε 0 \ VA > VB for all values of q. Charge will flow A to B till q = Q. 8. (b) : In steady state current in the branch containing the capacitor is zero and hence emf e is shared between r and R2 in the ratio of their resistances.

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Potential difference across R2 is εR2 ( 3 V)(1 Ω) = =2V VR = 2 R2 + r 1 Ω + 0.5 Ω Potential difference across capacitor of 1 mF is same across R2. \ Potential difference across 1 mF capacitor is VC = 2 V \ Charge on capacitor, Q = CVC = (1 mF)(2 V) = 2 mC  ^ 9. (c) : Current density, J = J0r k Current within a distance d, d 2π   d 2π ^ ^ I = ∫ ∫ J ⋅ dS = ∫ ∫ ( J0rrdφdr )( k ⋅ k ) r = 0 φ= 0

0 0

 ^ [∴ dS = rdφdr k ] z



d3 3 From Ampere’s Law,   d3 ∫ B ⋅ dl = µ0 2πJ0 3 C = 2 πJ0



y

10. (a) : ω = 2 πυ =

k m \ k = (2pu)2m Total energy of oscillation is E = (0.5 J + 0.4 J) = 0.9 J 1  E = KA2 2 where A is the amplitude of oscillation 1 ∴ 0.9 = kA2 2 A= =

1.8 1.8 = k ( 2 πυ)2 m 1 1.8 = 2 πυ 0.2

1  25  2π    π

...(i)

(Using (i)) 1.8 3 = m = 6 cm 0.2 50

11. (a) : Fundamental frequency of an open organ pipe is v 330 m s −1 = = 550 Hz 2l 2 × 0.3 m The given frequency u = 1.1 kHz = 1100 Hz = 2u0 υ0 =

46 physics for you |

february‘13

12. (d) : Image will be formed at infinity if object is placed at focus of the lens i.e., at 20 cm from the lens. Hence,  1 shift = 25 – 20 =  1 −  t µ   1  or 5 =  1 − t  1.5  or

 5 × 1.5  t= = 15 cm  0.5 

13. (c) : As the spring is compressed by 2 m with the application of a force of 200 N, hence its spring constant k is given by g 0k

Here loop C is a circle of radius d, B is the magnetic field at d. d3 B( 2 πd ) = µ 0 2 πJ0 3 µ 0 J0 d 2 B= 3

or

Therefore, the given frequency corresponds to 2nd harmonic.

F 200 N k= = = 100 N m −1 2 2m

2

h  = 30°

Suppose l be the distance along the inclined plane which the mass travels before it comes to rest. Applying the conservation of energy, 1 2 kx = mgh = mgl sin θ 2 1 1 1 or × 100 × 4 2 = 20 × 9.8 × l × 2 2 800 ∴ l= = 8.17 m 98 14. (a) : Initial velocity of parachutist after bailing out u2 = 2gh = 2 × 9.8 × 50 = 980 ...(i) When it reaches the ground 32 = u2 – 2 × 2 × h1 or 32 = 980 – 2 × 2 × h1 (Using (i)) 980 − 9 971 or h1 = or h1 = 242. 75 m = 4 4 \ Total height = 242.75 m + 50 m  293 m 15. (b) : Just before the string is cut, force on the spring pulling up = kx = 3mg. After string is cut, free body diagram of block A gives 3mg 2maA = 3mg – 2mg A mg g or aA = = 2mg 2m 2 Free body diagram of block B gives maB = mg or aB = g

B

mg 16. (d) : W = TD A = T(2 × 4pR ) 4 3 and V = πR 3 When volume is doubled new radius becomes R′ = (2)1/3R \ W′ = T × 2 × 4pR′2 = T × 2 × 4p(2)2/3R2 = T × 2 × 4p(4)1/3R2 = (4)1/3W

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2

17. (d) : n = 4 n=3 n=2

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Applying the law of conservation of linear momentum along a direction perpendicular to the direction of motion (i.e. along y-axis), we get 0 + 0 = 4v1sin30° – v2sin60° 4v1sin30° = v2sin60°

A B

C

D

n=1 The photon emitted comes from the transition A. Further transitions possible are B (longest l of the Balmer series), C (longest l of the Lyman series) and D (second longest l of the Lyman series). 18. (b) : As no external torque acts on the system, its angular momentum is conserved. Since there is no loss of energy due to friction, the total energy is conserved. However, the ring acquires some translational kinetic energy as it slides outwards, and hence rotational kinetic energy is not conserved. 19. (c) : The total energy available from the transition = 10.2 eV = energy of emitted photon + kinetic energy of recoiling atom. \ Energy of emitted photon < 10.2 eV \ l > l0 20. (b) : The acceleration of the electron is eE ...(i) ae = m e Starting from rest, the distance travelled by the electron in time t1 in a uniform electric field E is 1 2d d = ae t12 or t1 = ...(ii) 2 ae The acceleration of the proton is eE ap = ...(iii) mp Starting from rest, the same distance d travelled by the proton in time t2 in the same electric field is 1 2d d = apt22 or t2 = ...(iv) 2 ap Divide (iv) by (iii), we get mp ae t2 (Using (i) and (iii)) ∴ = = t1 ap me 21. (a) :

v1 sin 60° 3 = = v2 4 sin 30° 4 22. (a) : Here, Mass of the bullet, m1 = 20 g = 20 × 10–3 kg = 0.02 kg Mass of the block, m2 = 4 kg According to law of conservation of linear momentum, we get m1u1 + m2u2 = m1v1 + m2v2 where v1 and v2 be the velocities of the bullet and block after the collision. 0.02 × 600 + 4 × 0 = 0.02 × v1 + 4v2 or 0.02 × 600 = 0.02v1 + 4v2 ...(i) Here, v2 = 2 gh = 2 × 10 × 0.2 = 2 m s −1 Substituting this value of v2 in Eq. (i), we get

12 = 0.02v1 + 8

or 0.02v1 = 12 – 8 4 ∴ v1 = = 200 m s −1 0.02 l ...(i) g When lift is accelerated upwards with acceleration T a, let time period becomes . Then 2 T l = 2π ...(ii) 2 g+a Dividing Eq. (i) by Eq. (ii), we get

23. (d) : Here, T = 2π



2=

g+a  a = 1 +  g g 

1/ 2

Squaring both sides, we get a 4 = 1 + or a = 3g g 24. (b) : Here, rA = 1 mm, rB = 2 mm, d = 5 cm When spheres are connected by a conducting wire, charge flows from the sphere at higher potential to the sphere at lower potential, till their potentials become equal. Now,

CA rA 1 = = CB rB 2

As potential V is same,

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BRAIN MAP Digital Electronics and Logic Gates

A

Y

B Inputs A B 0 0 0 1 1 0 1 1

Output Y 0 1 1 1

A ×= A 0

A+ A= 1 A= A

A= A

A+ B= A× B

A× B= A+ B

A+ A× B= A+ B

A× (A + B) = A× B

A+ B= A× B

Inputs A B 0 0 0 1 1 0 1 1

Output Y 0 1 1 0

Inputs A B 0 0 0 1 1 0 1 1

Y= A× B+ A× B= AÅ B

Output Y 1 1 1 0 Y= A× B

A× B= A+ B

Inputs A B 0 0 0 1 1 0 1 1

Output Y 1 0 0 1

Inputs A B 0 0 0 1 1 0 1 1

Y= A× B+ A× B= AÅ B

Output Y 1 0 0 0

Y= A+ B

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Inputs A B 0 0 0 1 1 0 1 1

Input A 0 1

Output Y 0 0 0 1

Output Y 1 0 Y= A



∴

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qA CAV CA 1 = = = qB CBV CB 2 qA

Now,

2

2 EA 4 πε 0rA2 qA  rB  1  2 = = ×  =   = 2 qB EB qB  rA  2  1 2 4 πε 0rB

25. (d) : A

P2 (I – I1)

I

I1

I

10 

5 V, 2  D

P1 (I – I ) 1

I

B

(I – I1) C

10I1 + 2 I – 5 = 0



2I + 10I1 = 5

Applying Kirchhoff’s second law for closed loop P2 BCP1P2, we get

–2 + 1(I – I1) – 10I1 = 0

or I – 11I1 = 2

...(ii)

or 2I – 22I1 = 4

...(iii)

Subtracting Eq. (iii) from Eq. (i), we get 32 I1 = 1 or I1 =

1 A ≈ 0.03 A from P2 to P1 32

26. (d) : G = 50 W Ig = Current for full scale deflection = Current per division × total no. of divisions = 4 × 10–4 × 25 = 10–2 A Given V = 25 V Hence, required resistance, V 25 R= −G = − 50 Ig 10 −2 = 2500 – 50 = 2450 W This resistance of 2450 W should be connected in series to convert the galvanometer into a voltmeter.

e

m m = 2π keff 2k

T = 2π or

T2 =

or

k=

4π 2m 2k

2 π2m T2 2

∴ k=

2 × ( 3.14 ) × 12 2 = 6 × ( 3.14 ) = 59.15 N m −1 4

30. (c) : Because both the rings have common centre and their planes are mutually perpendicular, hence, an axis which is passing through the centre of one of the rings and perpendicular to the plane of its plane, will be along the diameter of other ring. Hence, moment of the inertia of the system about the given axis is 1 3 I = ICM + Idiameter = MR 2 + MR 2 = MR 2 2 2

3 = (0.1 kg )(0.1 m )2 = 15 × 10 −4 kg m 2 2 vvv

Life’s True Beauty “I think your education is imperfect, if you do not realize my young friends, that life is not merely a question of getting food, clothes and shelter. Man does not live by bread alone. This has been realized from ancient times. I think that the finest thing in life are

27. (a) 28. (b) : Time period of revolution of electron around a proton is given by



  e    4 πε mR  0

2 π 4 πε 0mR 3

29 . (c) : Let k be spring constant of each spring. Here, two springs are connected in parallel, the effective spring constant is given by keff = k + k = 2k

...(i)

=

\ T ∝ R3/2

2 V, 1 

Applying Kirchhoff’s second law for closed loop AP2 P1DA, we get

2 πR

=

T=

Orbit circumference 2πR = Electron speed v

50 physics for you |

february‘13

not

these,

beauty,

but

aesthetic

music, sense,

colour, the

flowers,

satisfaction

derived from these. It is these finer things in life that makes life worth living.

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- C V Raman

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Atoms and nuclei 1. In the following reaction, the energy released is 1

4 He 2

+ 2e+ + Energy



4 1H →



Given :



Mass of 1H = 1.007825 u



Mass of 2 He = 4.002603 u



1

4

e+

Mass of = 0.000548 u (a) 12.33 MeV (b) 24.67 MeV (c) 25.7 MeV (d) 49.34 MeV

2. The total energy of an electron in the excited state corresponding to n = 3 state is E. What is its potential energy with proper sign? (a) –2E (b) 2E (c) –E (d) E 3. The energy levels of a certain atom for 1st, 2nd and 3rd levels are E, 4E/3 and 2E respectively. A photon of wavelength l is emitted for a transition 3 → 1. What will be the wavelength of emission for transition 2 → 1? 4λ λ (a) (b) (c) 3 λ (d) 3l 3 4 3 4. In Bohr model of hydrogen atom, the ratio of period of revolution of an electron in n = 2 and n = 1 orbit is (a) 2 : 1 (b) 4 : 1 (c) 8 : 1 (d) 16 : 1 5. A radioactive isotope A with a half life of 1.25 × 1010 years decays into B which is stable. A sample of rock from a planet is found to contain both A and B present in the ratio 1 : 15. The age of the rock is (a) 9.6 × 1010 years (b) 4.2 × 1010 years 10 (c) 5 × 10 years (d) 1.95 × 1010 years 6. A heavy nucleus at rest breaks into two fragments which fly off with velocities in the ratio 3 : 1. The ratio of radii of the fragments is

(a) 1 : 31/3 (c) 4 : 1

(b) 31/3 : 4 (d) 2 : 1

7. A hydrogen atom and a Li 2+ ion are both in the second excited state. If lH and lLi are their respective electronic angular momenta, and EH and ELi their respective energies, then (a) lH > lLi and EH > ELi (b) lH = lLi and EH < ELi (c) lH = lLi and EH > ELi (d) lH < lLi and EH < ELi 8. A radioactive nucleus emits 3a-particles and 5b-particles. The ratio of number of neutrons to that of protons will be (b) A − Z (a) A − Z − 12 Z−6 Z−1 A − Z − 11 (c) A − Z − 11 (d) Z−1 Z−6 9. A radioactive sample has half life of 5 days. To decay from 8 microcurie to 1 microcurie, the number of days taken will be (a) 40 (b) 25 (c) 15 (d) 10 10. A free neutron decays spontaneously into (a) a proton, an electron and anti-neutrino (b) a proton, an electron and a neutrino (c) a proton and electron (d) a proton, and electron, a neutrino and an anti-neutrino. 11. The radius of germanium (Ge) nuclide is measured 9

to be twice the radius of 4Be. The number of nucleons in Ge are (a) 72 (b) 73 (c) 74 (d) 75 12. The ratio of the longest and shortest wavelengths in Brackett series of hydrogen spectra is 9 4 25 17 (a) (b) (c) (d) 5 3 9 6

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13. Fusion reaction takes place at high temperature because (a) nuclei break up at high temperature (b) atoms get ionised at high temperature (c) kinetic energy is high enough to overcome the coulomb repulsion between nuclei (d) molecules break up at high temperature

14. Which of the following is more effective in inducing nuclear fission? (a) Fast neutron (b) Fast proton (c) Slow proton (d) Slow neutron 15. The electric potential between a proton and an   electron is given by V = V0 ln  r  , where r0 is a  r0  constant. Assuming Bohr’s model to be applicable, write variation of rn with n, n being the principal quantum number (a) rn ∝ n (b) rn ∝ 1 n 1 2 (c) rn ∝ n (d) rn ∝ n2 16. To generate power of 3.2 MW, the number of fissions of 235U per minute is (Energy released per fission = 200 MeV, 1 eV = 1.6 × 10–19 J) (a) 6 × 1018 (b) 6 × 1017 17 (c) 10 (d) 6 × 1016 17. A hydrogen atom is excited up to 9th level. The total number of possible spectral lines emitted by the hydrogen atom is (a) 36 (b) 35 (c) 37 (d) 38 18. The 232 90Th atom has successive alpha and beta decays to the end product 208 82Pb. The numbers of alpha and beta particles emitted in the process respectively are (a) 4, 6 (b) 4, 4 (c) 6, 2 (d) 6, 4 19. The radius of the hydrogen atom in its ground state is a0. The radius of a muonic hydrogen atom in which the electron is replaced by an identically charged muon with mass 207 times that of an electron, is am equal to a (a) 207a0 (b) 0 207 a0 (c) 207 (d) a0 207 20. A nucleus with mass number 220 initially at rest emits an a particle. If the Q value of the reaction is 5.5 MeV, the kinetic energy of the a particle is (a) 4.4 MeV (b) 5.4 MeV 52 physics for you | February ’13

(c) 5.6 MeV

(d) 6.5 MeV

21. An element X decays first by positron emission and then two a-particles are emitted in successive radioactive decay. If the product nucleus has mass number 227 and atomic number 89, the mass number and atomic number of element X are (a) (273, 93) (b) (235, 94) (c) (238, 93) (d) (237, 92) 22. The intensity of gamma radiation from a given source is I. On passing through 36 mm of lead, I it is reduced to . The thickness of lead which 8 I will reduce the intensity of will be 2 (a) 12 mm (b) 18 mm (c) 9 mm (d) 6 mm 23. F p p , F n n a n d F n p a r e t h e n u c l e a r f o r c e s between proton-proton, neutron-neutron and neutron-proton respectively. Then relation between them is (a) Fpp = Fnn ≠ Fnp (b) Fpp ≠ Fnn = Fnp (c) Fpp = Fnn = Fnp (d) Fpp ≠ Fnn ≠ Fnp 24. Which energy state of doubly ionised lithium has the same energy as that of the ground state of hydrogen ? Given Z for lithium = 3. (a) 4 (b) 3 (c) 2 (d) 1 25. When the electron in hydrogen atom is excited from the 4th stationary orbit to the 5th stationary orbit, the change in the angular momentum of the electron is (Planck’s constant, h = 6.63 × 10–34 J s) (a) 4.16 × 10–34 J s (b) 3.32 × 10–34 J s (c) 1.05 × 10–34 J s (d) 2.08 × 10–34 J s

Semiconductor electronics 26. In the following figure, the diodes which are forward biased, are +10 V

(i)

R

+5V (ii)

R

–10 V

(iii) –12 V

(iv)



R –5 V R



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+5V



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(a) only (iii) (c) (ii) and (iv)

32. Consider the following statements A and B and identify the correct choice of the given answers. (A) The width of the depletion layer in a p-n junction diode increases in forward bias. (B) In an intrinsic semiconductor the fermi energy level is exactly in the middle of the forbidden energy gap. (a) A is true and B is false (b) Both A and B are false (c) A is false and B is true (d) Both A and B are true

(b) (i) and (iii) (d) (i), (ii) and (iv)

27. Truth table for the given circuit is A

C Y



D

B



A 0 0 1 1

B 0 1 0 1



A 0 0 1 1

B 0 1 0 1

(a)

(c)

Y 1 0 1 0 Y 0 1 0 1

(b) (d)

A 0 0 1 1

B 0 1 0 1

Y 1 0 0 1

A 0 0 1 1

B 0 1 0 1

Y 0 1 0 0

33. A common emitter transistor amplifier has a current gain of 50. If the load resistance is 4 kW and input resistance is 500 W, the voltage gain of the amplifier is (a) 100 (b) 200 (c) 300 (d) 400

28. For a cubic crystal structure which one of the following relations indicating the cell characteristics is correct? (a) a ≠ b ≠ c and a = b = g = 90° (b) a = b = c and a ≠ b ≠ g = 90° (c) a = b = c and a = b = g = 90° (d) a ≠ b ≠ c and a ≠ b and g ≠ 90° 29. If an intrinsic semiconductor is heated, the ratio of free electrons to holes is (a) greater than one (b) less than one (c) equal to one (d) decreases and becomes zero 30. While a collector to emitter voltage is constant in a transistor, the collector current changes by 8.2 mA when the emitter current changes by 8.3 mA. The value of forward current ratio is (a) 82 (b) 83 (c) 8.2 (d) 8.3

34. The voltage drop across a forward biased diode is 0.7 V. In the following circuit, the voltages across the 10 W resistance in series with the diode and 20 W resistance are 10  20 



Y X

(a) (b)

Y X Y

(c) X (d) X

Z

Z Z Z

(b) 3.58 V, 4.28 V (d) 3.58 V, 9.3 V

36. In the given circuit, the current through the resistor 2 kW is 1 k + 20 V 12 V –

X

(a) 0.70 V, 4.28 V (c) 5.35 V, 2.14 V

35. To obtain p-type extrinsic semiconductor, the impurity element to be added to germanium should be of valency (a) 2 (b) 3 (c) 4 (d) 5

31. The logic circuit shown below is equivalent to

Z

10 V 10 

(a) 2 mA (b) 4 mA

2 k

(c) 6 mA

(d) 10 mA

37. In a p-n junction photodiode, the value of the photo electromotive force produced by monochromatic light is proportional to (a) the barrier voltage at p-n junction (b) the intensity of light falling on the cell (c) the frequency of light falling on the cell (d) the voltage applied at the p-n junction. 38. Two pieces, one of germanium and the other of the aluminium are cooled from T1 K to T2 K. The resistance of

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(a) aluminium increases and that of germanium decreases. (b) each of them decreases. (c) aluminium decreases and that of germanium increases. (d) each of them increases.

39. Pure Si at 500 K has equal number of electron (ne) and hole (nh) concentrations of 1.5 × 1016 m–3. Doping by indium increases nh to 4.5 × 1022 m–3. The doped semiconductor is of (a) n-type with electron concentration ne = 5 × 1022 m–3. (b) p-type with electron concentration ne = 2.5 × 1010 m–3. (c) n-type with electron concentration ne = 2.5 × 1023 m–3. (d) p-type having electron concentration ne = 5 × 109 m–3. 40. A full wave p-n diode rectifier uses a load resistor of 1500 W. No filter is used. The forward bias resistance of the diode is 10 W. The efficiency of the rectifier is (a) 81.2% (b) 40.6% (c) 80.6% (d) 40.2%

Communication systems 41. If the maximum amplitude of an amplitude modulated wave is 25 V and the minimum amplitude is 5 V, the modulation index is 1 1 3 (b) (c) (d) 2 (a) 5 3 2 3 42. A signal wave of frequency 12 kHz is modulated with a carrier wave of frequency 2.51 MHz. The upper and lower side band frequencies are respectively (a) 2512 kHz and 2508 kHz (b) 2522 kHz and 2488 kHz (c) 2502 kHz and 2498 kHz (d) 2522 kHz and 2498 kHz 43. The frequency band used in the downlink of satellite communication is (a) 9.5 to 2.5 GHz (b) 896 to 901 MHz (c) 3.7 to 4.2 GHz (d) 840 to 935 MHz 44. A basic communication system consists of (A) transmitter (B) information source (C) user of information (D) channel (E) receiver Choose the correct sequence in which these are arranged in a basic communication system. (a) ABCDE (b) BADEC (c) BDACE (d) BEADC 54 physics for you | February ’13

45. 1000 kHz carrier wave is amplitude modulated by the signal frequency 200 – 4000 Hz. The channel width of this case is (a) 8 kHz (b) 4 kHz (c) 7.6 kHz (d) 3.8 kHz 46. A n e x a m p l e o f p o i n t t o p o i n t m o d e o f communication is (a) FM radio (b) standard FM radio (c) television (d) telephony 47. If both the length of an antenna and the wavelength of the signal to be transmitted are doubled, the power radiated by the antenna (a) is doubled (b) is halved (c) remains constant (d) is quadrupled 48. Identify the incorrect statement from the following. (a) AM detection is carried out using a rectifer and an envelope detector. (b) Pulse position denotes the time of rise or fall of the pulse amplitude. (c) Modulation index m is kept ≥ 1, to avoid distortion. (d) Facsimile (FAX) scans the contents of the document to create electronic signals. 49. Through which mode of propagation, the radio waves can be sent from one place to another (a) ground wave propagation (b) sky wave propagation (c) space wave propagation (d) all of them 50. A television tower of height 140 m can broadcast its signal upto a maxiumum area of (Radius of earth = 6.4 × 106 m) (a) 1.56 × 106 km2 (b) 5.6 × 103 km2 10 2 (c) 5.6 × 10 km (d) 1.56 × 109 km2

solutions 1. (c) : The given nuclear reaction is 4

1

4 1H → 2 He + 2e+ + Energy The energy released during the process is

4

1

Q = [4m(1H) – m( 2He) – 2(me+)]c2

= [4 × 1.007825 – 4.002603) – 2 × 0.000548]u × c2 = [4.0313 – 4.002603 – 0.001096]u × c2 = (0.027601 u)c2 = (0.027601 u)(931.5 MeV/u) = 25.7 MeV

2. (b) 3. (d) : 2 E 4 3E E

3 

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

2 1

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E3 – E1 = 2E – E = E hc =E λ From figure, E2 – E1 = E2 – E3 + E3 – E1 4 = E − 2E + 2E − E 3 2 1 =− E+E= E 3 3 hc 1 hc = λ′ 3 λ l′ = 3l

....(i)

(Using (i))

4. (c) : In Bohr model of hydrogen atom, the period of revolution of electron in nth orbit is given as Tn =

4ε02 h 3n3 4

\

T2  2  3 8 =  = 1 T1  1 

me Tn ∝ n3

5. (c) : According to Rutherford and Soddy law for radioactive decay, Number of atoms remained undecayed after time t is N = N0e–lt N N e λt = 0 or λt = ln 0 N N 1 N t = ln 0 λ N T1/ 2 N0 ln t= ln 2 N 1.25 × 1010 16 ln 1 ln 2 1.25 × 1010 × 4 × ln 2 = ln 2 = 5 × 1010 years =

2

\ R1 : R2 = 1 : 31/3

8. (d) : During the emission of a-particle, the mass number and atomic number decreases by four and two respectively. During the emission of b-particle the mass number remains the same while the atomic number increases by 1.

A ZX

3α

→

( A −12 ) ( Z − 6 )Y

5β

→

( A −12 ) ( Z −1) Y ′

No. of neutrons A − 12 − ( Z − 1) = No. of protons Z−1 A − Z − 11 = Z−1 9. (c) : Here, Half life, T1/2 = 5 days Initial activity, R0 = 1 microcurie Final activity, R = 8 microcurie ∴

As

R 1 =  R0  2 

n

where n is the no. of half lives 3

n

n

1 1 1 1   =   or   =   8 2 2 2 or n = 3 t As n = or t = nT1/ 2 = ( 3)( 5 days) = 15 days T1/ 2 ∴

10. (a) : A free neutron is unstable. It decays spontaneously into a proton, an electron and anti-neutrino. – n → p + e– + u neutron

proton

electron

anti-neutrino

11. (a) : Nuclear radii, R = R0(A)1/3 (where R0 = 1.2 fm) or R ∝ (A)1/3 ∴

6. (a) : As the heavy nucleus at rest breaks, therefore according to law of conservation of momentum, we get m 1v 1 + m 2v 2 = 0 v1 m2 3 .... (i) or = = v2 m1 1 As nuclear density is same, 4 3 m1 ρ 3 πR1 R13 ∴ = = m2 ρ 4 πR 3 R 3 2 2 3 3 R1 m1 1 = = or (Using (i)) R 3 m2 3

7. (b) : In second excited state, n = 3, So, lH = lLi = 3  h   2π  while E ∝ Z2 and ZH = 1, ZLi = 3 So, ELi = 9 EH or EH < ELi

or

RBe RGe

=

(9)1/ 3 ( A)1/ 3

RBe (9)1/ 3 = 2 RBe ( A)1/ 3

\ (A)1/3 = 2 × (9)1/3 or A = 23 × 9 = 72 The number of nucleons in Ge is 72. 12. (a) : For Brackett series, 1 1 1 = R −  λ  4 2 n2  where n = 5, 6, 7, 8, ........ For longest wavelength, n = 5 1 1 1 ∴ = R −  λ Longest  4 2 52 

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1 1 9 = R −  = .... (i) R  16 25  400 For shortest wavelength, n = ∞ 1 1 1  R ∴ = R − .... (ii) = λ Shortest  4 2 ∞ 2  16

Dividing (ii) by (i), we get λ Longest R 400 25 = × = 9 λ Shortest 16 9 R

13. (c) : For fusion to take place, high temperature is needed because at high temperature, the kinetic energy becomes large enough to overcome the coulomb repulsion between nuclei. 14. (d) 15. (a) : Given : V = V0 ln  r   r  0 \ Potential energy, U = eV

or | F | =

eV0 r

eV dU =− 0 dr r

This force provides the necessary centripetal force. 2

eV eV0 mv ∴ = 0 or v = r r m By Bohr’s postulate, mvr = nh 2π nh or v = 2πmr

.... (i)



.... (ii)

From equations (i) and (ii), we get

eV0 nh = 2πmr m

or r =

nh m × 2πm eV0

1  ×n 2 π meV 0   rn ∝ n







=

.... (ii)



β+

A ZX

→

A Z −1Y

α 2 →

A− 8 Z− 5 Y′

Since the product nucleus has mass number 227 and atomic number 89. \ A – 8 = 227 or A = 235 and Z – 5 = 89 or Z = 94

22. (a) : As I = I0e–mx 1 ∴ = e − µ 36 8 1 and = e − µx 2 From Eq. (i), we get

.... (i) .... (ii)

1 − µ 36   = e 2 Using (ii), we get

16. (a) : Here, Power = 3.2 MW = 3.2 × 106 W Energy released per fission = 200 MeV = 200 × 106 eV = 200 × 106 × 1.6 × 10–19 J Number of fissions per minute

.... (i)

3

 h

or r =  ∴

h 2ε0 19. (b) : a0 = πme 2 h 2ε0 aµ = π( 207 m)e 2 Dividing (ii) by (i), we get aµ a 1 = or aµ = 0 207 a0 207

20. (b) : Here, A = 220, Q = 5.5 MeV The kinetic energy of the a particle is A−4 220 − 4 Q= KEα = × 5.5 MeV= 5.4 MeV A 220 21. (b) : Let A and Z be mass number and atomic number of element X.

r  1 r dU or U = eV0 ln   ∴ = eV0  0  dr  r  r0  r0  Force, F = −

9(9 − 1) 9 × 8 = = 36 2 2 18. (d) : Let number of a particles emitted be x and number of b particles emitted be y. Difference in mass number = 4x = 232 – 208 = 24 or x = 6 Difference in charge number = 2x – y = 90 – 82 = 8 12 – y = 8 or y = 4 Here n = 9 \ N =

3.2 × 106 × 60 6

200 × 10 × 1.6 × 10 = 6 × 1018

17. (a) : Number of spectral lines emitted is n(n − 1) N= 2 56 physics for you | February ’13

−19

36 = 12 mm 3 23. (c) : Nuclear force is charge independent. \ Fpp = Fnn = Fnp or e–3 mx = e–m36 or x =

24. (b) : The energy of nth state of a hydrogen like atom is given as CZ2

, where C is a constant. n2 For ground state of hydrogen atom, Z = 1, n = 1 En =

∴

E1 =

C(1)2 (1)2

=C

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For nth state of Li2+ ion (Z = 3) C ( 3) 2

9C = n2 n2 As En = E1 9C ∴ = C or n2 = 9 or n = 3 n2 25. (c) : According to Bohr’s quantisation condition nh Ln = 2π 4h ∴ For n = 4 , L4 = 2π 5h and for n = 5 , L5 = 2π \ Change in angular momentum when an electron is excited from n = 4 to n = 5 is 5h 4 h ∆L = L5 − L4 = − 2π 2π h 6.63 × 10 −34 J s = 1.05 × 10–34 J s = = 2π 2 × 3.14 En =

26. (b) : p-n junction diode is said to be forward biased if p side of junction is at higher potential and n side of junction is at lower potential. Therefore, option (b) is true. A

27. (c) :

0

0 0 1



B A

0

0

0

0 1 1

B



A

1

1

1

1 0 0



B A

0

0

1

1 1 0



B

1

1

0 C 0 0

0

0

Y

0 C

1

1

1

Y

D

0 0

0

Y

D

1 C 1 0

0 D

30. (a) : Here, DIE = 8.3 mA DIC = 8.2 mA As DIE = DIB + DIC \ DIB = DIE – DIC = 8.3 mA – 8.2 mA = 0.1 mA  ∆I  Forward current ratio, h fe =  C   ∆IB VCE is constant  8.2 mA  =  = 82  0.1 mA 

31. (d) :

X Y

– X – A

A

– Y

C

– C

Z

Output of first OR gate is A = (X + Y ) Inputs of second OR gate are X and A. Output of second OR gate is C = X + A = X + X + Y = X + (X ⋅ Y ) = X + (X ⋅ Y ) = X(1 + Y ) = X Final output is Z = C = X = X

33. (d) : Here, b = 50, R L = 4 kW = 4 × 103 W Ri = 500 W R 4 × 10 3 = 400 Voltage gain, Av = β L = 50 × 500 Ri 34. (b) : Let the currents through the 20 W (parallel) and 10 W (in series with the diode) be I1 and I2 respectively. C

0 C

0

29. (c)

32. (c) : In a p-n junction the width of the depletion layer decreases in forward bias.

D

0

28. (c) : For cubic crystal, a = b = c and a = b = g = 90°

1

Y

B (I1 + I2)

I2 I1

10  20 

D E (I1 + I2) F

A 10 V 10  Applying Kirchhoff’s second law for closed loop ABEFA, we get 20I1 + 10(I1 + I2) – 10 = 0 .... (i) Applying Kirchhoff’s second law for closed loop BCDEB, we get 0.7 + 10I2 – 20I1 = 0 .... (ii) Solving (i) and (ii), we get I1 = 0.214 A and I2 = 0.358 A

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Thus, voltage across the 10 W resistance in series with the diode = 0.358 A × 10 W = 3.58 V And voltage across the 20 W resistance = 0.214 A × 20 W = 4.28 V

35. (b) : To obtain p-type extrinsic semiconductor a trivalent impurity should be added to germanium.

42. (d) : Here, us = 12 kHz, uc = 2.51 MHz = 2510 kHz The upperside band frequency = uc + us = (2510 + 12) kHz = 2522 kHz The lower side band frequency = uc – us = (2510 – 12) kHz = 2498 kHz 43. (c) : The frequency band used in the downlink of satellite communication is 3.7 to 4.2 GHz.

36. (c) : In the circuit, Zener diode is used as a voltage regulating device. Therefore, the current in the 12 V circuit, I = = 6 × 10 −3 A = 6 mA 2 kΩ

44. (b) : The block diagram of a communication system is shown in the figure below.

37. (b) : In photodiode, the photoelectromotive force is produced by photo-voltaic action i.e. a potential difference is created between two points whose magnitude depends upon the intensity of incident light.

45. (a) : Bandwidth is equal to twice the frequency of modulating signals. Therefore, Band width = 2um = 2 × 4000 Hz = 8 × 103 Hz = 8 kHz

38. (c) : The temperature coefficient of resistance of aluminium is positive and of germanium is negative. Hence the resistance of aluminium decreases and that of germanium increases with decreases in temperature.

47. (c) : Power radiated by the antenna is proportional

39. (d) : As ni2 = ne × nh (1.5 × 1016 m–3)2 = ne × (4.5 × 1022 m–3) ∴

ne =

(1.5 × 1016 m −3 )2 22

−3

= 5 × 109 m −3

( 4.5 × 10 m ) As nh >> ne, so semiconductor is p-type and ne = 5 × 109 m–3. 40. (c) : Here, ri = 10 W, RL = 1500 W Efficiency of the full wave rectifier is η=

0.812 RL r f + RL

=

0.812 × 1500 = 0.806 = 80.6% 10 + 1500

Transmitter

µ=

User of Information

2

to  l  . When both the length of the antenna l and λ wavelength of the signal l are doubled, the power radiated by the antenna remains constant. 48. (c) : Modulation index m is kept ≤ 1 to avoid distortion. 49. (d) 50. (b) : Here, h = 140 m R = 6.4 × 106 m

25 V − 5 V 20 2 = = 25 V + 5 V 30 3

58 physics for you | February ’13

Receiver

46. (d) : In point-to-point communication mode, communication takes place over a link between a single transmitter and a receiver. Telephony is an example of such a mode of communication.

41. (d) : Maximum amplitude, Amax = Ac + Am ...(i) ...(ii) Minimum amplitude, Amin = Ac – Am Solving (i) and (ii), we get A + Amin , A − Amin Ac = max Am = max 2 2 A A − Amin Modulation index, µ = m = max Ac Amax + Amin

Channel

Coverage range, d = 2 Rh Area covered = pd2 = p2Rh 22 = × 2 × 6.4 × 106 × 140 7 = 5632 × 106 m2 = 5.6 × 109 m2 = 5.6 × 103 km2

Pdc ( 2 Im / π)2 RL = Pac ( I / 2 )2 (r + R ) m f L

=

Information Source

WorldMags.net

nn

WorldMags.net Questions for Practice

a a − t2 in the equation P = , b bx where P is pressure, x is distance and t is time are (a) [M2LT–3] (b) [ML0T–2] 3 –1 (c) [ML T ] (d) [MLT–3]

1. The dimensions of

2. A balloon is going upwards with velocity 12 m s–1. It releases a packet when it is at a height of 65 m from the ground. How much time the packet will take to reach the ground ? [Take g = 10 m s–2] (a) 5 s (b) 6 s (c) 7 s (d) 8 s 3. A block of mass 10 kg is placed on rough horizontal surface whose coefficient of friction is 0.5. If a horizontal force of 100 N is applied on it, then acceleration of block will be [Take g = 10 m s–2] (a) 10 m s–2 (b) 5 m s–2 –2 (c) 15 m s (d) 0.5 m s–2 4. A ball with charge – 50e is placed at the centre of a hollow spherical shell has a net charge of –50e. What is the charge on the shell’s outer surface? (a) –50e (b) Zero (c) –100e (d) +100e 5. A voltmeter of resistance 998 W is connected across a cell of emf 2 V and internal resistance 2 W. The potential difference across the voltmeter is (a) 1.99 V (b) 3.5 V (c) 5 V (d) 6 V 6. The magnetic moment produced in a substance of 1 g is 6 × 10–7 A m2. If its density is 5 g cm–3, then the intensity of magnetisation will be (a) 8.3 × 106 A m–1 (b) 3.0 A m–1 (c) 1.2 × 10–7 A m–1 (d) 3 × 10–6 A m–1 7. A planet is revolving around a star in an elliptic orbit. The ratio of the farthest distance to the closest distance of the planet from the star is 4.

The ratio of kinetic energies of the planet at the farthest to the closest position is (a) 1 : 16 (b) 16 : 1 (c) 1 : 4 (d) 4 : 1 8. A homogeneous disc with a radius 0.2 m and mass 5 kg rotates around an axis passing through its centre. The angular velocity of the rotation of the disc as a function of time is given by the formula w = 2 + 6t. The tangential force applied to the rim of the disc is (a) 1 N (b) 2 N (c) 3 N (d) 4 N 9. A body of mass 0.01 kg F(N) executes simple harmonic 80 0 0.2 motion (SHM) about x = 0 –0.2 x(m) under the influence of a force –80 as shown in the adjacent figure. The period of the SHM is (a) 1.05 s (b) 0.52 s (c) 0.25 s (d) 0.03 s 10. A transverse wave in a medium is described by the equation y = A sin 2(wt – kx). The magnitude of the maximum velocity of particles in the medium is equal to that of the wave velocity, if the value of A is 2λ λ λ λ (a) (b) (c) (d) π 2π 4π π 11. The temperature of an air bubble while rising from bottom to surface of a lake remains constant but its diameter is doubled. If the pressure on the surface is equal to h metre of mercury column and relative density of mercury is r, then the depth of lake in metre is (a) 2rh (b) 4rh (c) 8rh (d) 7rh 12. A stone of mass 2 kg is projected upwards with kinetic energy of 98 J. The height at which the kinetic energy of the body becomes half its original value is

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59



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(Take g = 10 m s–2) (a) 5 m (b) 2.5 m (c) 1.5 m (d) 0.5 m

13. A particle of mass m collides with another stationary particle of mass M. If the particle m stops just after collision, the coefficient of restitution of collision is equal to m (a) 1 (b) m (c) M − m (d) M+m M+m M 14. A body cools down from 60°C to 55°C in 30 s. Using Newton’s law of cooling, calculate the approximate time taken by same body to cool down from 55°C to 50°C. Assume that the temperature of surroundings is 45°C. (a) 40 s (b) 55 s (c) 50 s (d) 60 s 15. A conducting circular loop is placed in a uniform magnetic field, B = 0.025 T with its plane perpendicular to the loop. The radius of the loop is made to shrink at a constant rate of 1 mm s–1. The induced emf when the radius is 2 cm, is (a) 2p mV (b) p mV π (c) (d) 2 mV µV 2 16. In the series LCR circuit as shown in the figure, the voltmeter V and ammeter A readings are 400 V

400 V

9 ]

V

A



R = 50 

L

C

100 V, 50 Hz

(a) V = 100 V, I = 2 A (b) V = 100 V, I = 5 A (c) V = 400 V, I = 2 A (d) V = 300 V, I = 2 A

17. The electric field part of an electromagnetic wave in a medium is represented by Ex = 0,   N rad  −2 rad   Ey = 2.5 cos  2π × 106  x ,  t −  π × 10  C s m   Ez = 0. The wave is (a) moving along the + x direction with frequency 106 Hz and wavelength 100 m. (b) moving along + x direction with frequency 106 Hz and wavelength 200 m. (c) moving along – x direction with frequency 106 Hz and wavelength 200 m. (d) moving along + y direction with frequency 2p × 106 Hz and wavelength 200 m. 60 physics for you |

february ‘13

18. A point source of light is placed at a depth of h below the surface of water of refractive index m. A floating opaque disc is placed on the surface of water so that light from the source is not visible from the surface. The minimum diameter of the disc is 2h (a) (b) 2h(m2 – 1)1/2 2 (µ − 1)1/ 2 h (c) (d) h(m2 – 1)1/2 [2(µ 2 − 1)1/ 2 ] 19. An electron is moving in an orbit of a hydrogen atom from which there can be a maximum of six transitions. An electron is moving in an orbit of another hydrogen atom from which there can be a maximum of three transitions. The ratio of the velocity of the electron in these two orbits is 1 2 3 5 (a) (b) (c) (d) 2 1 4 4 20. One milliwatt of light of wavelength 4560 Å is incident on a cesium surface of work function 1.9 eV. Given that quantum of efficiency of photoelectric emission is 0.5%, Planck’s constant, h = 6.62 × 10–34 J s, velocity of light, c = 3 × 108 m s–1, the photoelectric current liberated is (a) 1.836 × 10–6 A (b) 1.836 × 10–7 A –5 (c) 1.836 × 10 A (d) 1.836 × 10–4 A 21. The distance between poles of horse shoe magnet is 10 cm and its pole strength is 10–4 A m. The magnetic field at a point P midway between the poles is (a) zero (b) 8 × 10–9 T –7 (c) 2 × 10 T (d) 8 × 10–7 T 22. Find the potential difference between the points E and F in the figure given below. Assume E and F are the midpoints of AB and DC respectively. (+q)



6m

(+q)

A

E

B

D +q/2

F

C +q/2

(a) (1.2 × 109q) volt (c) (1.5 × 109q) volt 23. As the switch S is closed in the circuit shown in figure, current passed through it is (a) zero (b) 1 A (c) 2 A (d) 1.6 A

WorldMags.net

4m

(b) (1.8 × 109q) volt (d) (3 × 109q) volt 10 V 4 

2 2 S

5V

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24. When a spring is stretched by a distance x, it exerts a force, given by F = (– 5x – 16x3) N. The work done, when the spring is stretched from 0.1 m to 0.2 m is (a) 8.1 × 10–2 J (b) 12.2 × 10–2 J –1 (c) 8.1 × 10 J (d) 12.2 × 10–1 J

103 kg m–3. The bubble is at a depth of 10.0 cm below the free surface. By what amount is the pressure inside the bubble greater than the atmospheric pressure? (a) 1030 N m–2 (b) 1230 N m–2 –2 (c) 1130 N m (d) 1330 N m–2

25. A satellite is orbiting around the earth with total energy E. What will happen if the satellite’s kinetic energy is made 2E? (a) Radius of the orbit is doubled (b) Radius of the orbit is halved (c) Period of revolution is doubled (d) Satellite escapes away

32. When a system is taken from state a to state b along the path acb as shown in figure, 60 J of heat flows into the system and 30 J of work is done by the system. Along the path adb, if the work done by the system is 10 J, heat flow into the system is

26. Sand drops vertically at the rate of 2 kg s–1 on to a conveyor belt moving horizontally with a velocity of 0.2 m s–1. Then the extra force required to keep the belt moving is (a) 0.4 N (b) 0.08 N (c) 0.04 N (d) 0.02 N 27. A solid sphere is rolling on a frictionless surface, shown in figure with a translational velocity v m s–1. If it is to climb the inclined surface then v should be

(a) 100 J (b) 20 J (c) 80 J (d) 40 J 33. In the circuit shown, the cell is ideal, with emf = 15 V. Each resistance R is of 3 W. The potential difference across the capacitor of capacitance 3 mF is C = 3 F

R

R

R

(a) ≥ 10 gh 7

(b) ≥ 2gh

(c) 2gh

(d)

10 gh 7 28. The average translational kinetic energy of O2 (molar mass 32) molecules at a particular temperature is 0.048 eV. The translational kinetic energy of N2 (molar mass 28) molecules in eV at the same temperature is (a) 0.0015 (b) 0.003 (c) 0.048 (d) 0.768 29. Two simple harmonic motions are represented  π by the equations y1 = 0.1 sin  100 πt +  and  3 y2 = 0.1 cospt. The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is π π π π (a) (b) − (c) (d) − 3 3 6 6 30. A pipe 30.0 cm long is open at both ends. Which harmonic mode of the pipe is resonantly excited by a 1.1 kHz source? Take the speed of sound in air as 330 m s–1. (a) First harmonic (b) Second harmonic (c) Third harmonic (d) Fourth harmonic 31. There is an air bubble of radius 1.0 mm in a liquid of surface tension 0.075 N m–1 and density

R

R 15 V

(a) zero

(b) 9 V

(c) 12 V (d) 15 V

34. A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field B. The magnitude of B (in tesla) is

(Take g = 10 m s–2) (a) 2 (b) 1.5

(c) 0.55

(d) 0.66

35. A 50 Hz ac of crest value 1 A flows through primary of a transformer. If mutual inductance between primary and secondary be 1.5 H, the crest voltage in secondary is (a) 75 V (b) 150 V (c) 471 V (d) 300 V 36. When 100 V dc is applied across a coil, a current of 1 A flows through it. When 100 V 50 Hz ac is applied to the same coil, only 0.5 A flows. The inductance of the coil is (a) 5.5 mH (b) 0.55 mH (c) 55 mH (d) 0.55 H 37. Young’s experiment is performed with light of wavelength 6000 Å wherein 16 fringes occupy a certain region on the screen. If 24 fringes occupy

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61

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the same region with another light of wavelength l, then l is (a) 6000 Å (b) 4500 Å (c) 5000 Å (d) 4000 Å

38. A plane electromagnetic wave propagating in the x-direction has wavelength of 6.0 mm. The electric field is in the y-direction and its maximum magnitude of 33 V m–1. The equation for the electric field as function of x and t is x x (a) 11 sinp(t – ) (b) 33 sinp × 1011(t – ) c c x x (c) 33 sinp (t – ) (d) 11 sinp × 1011 (t – ) c c 39. A convex lens of focal length 1.0 m and a concave lens of focal length 0.25 m are 0.75 m apart. A parallel beam of light is incident in the convex lens. The beam emerging after refraction from both lenses is (a) parallel to principal axis (b) convergent (c) divergent (d) none of these 40. The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion. The atomic numer Z of hydrogen like ion is (a) 4 (b) 1 (c) 2 (d) 3 41. A particle moving with uniform acceleration has velocity 6 m s–1 at a distance of 5 m from the initial position. After moving another 7 m, the velocity becomes 8 m s–1. The initial velocity and acceleration of the particle are (a) 2 m s–1, 4 m s–2 (b) 4 m s–1, 2 m s–2 –1 –2 (c) 4 m s , 4 m s (d) 6 m s–1, 1 m s–2 42. A particle is moving along a circular path of radius 5 m with uniform speed of 5 m s–1. What will be average acceleration when the particle completes half revolution ? (a) zero (b) 10 m s–2 10 (c) 10p m s–2 (d) m s–2 π 43. The system is pushed by a force F as shown in figure. All surfaces are smooth except between B and C. Friction coefficient between B and C is m. Minimum value of F to prevent block B from downward slipping is F

A 2m

B m

C 2m

 3  (a)   mg  2µ 

 5  (b)   mg  2µ 

5 (c)   mmg 2

(d)  3  mmg 2

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44. A 1000 kg elevator rises from rest in the basement to the fourth floor, a distance of 20 m. As it passes the fourth floor its speed is 4 m s–1. There is a constant frictional force of 500 N. The work done by the lifting mechanism is (a) 196 × 103 J (b) 204 × 103 J 3 (c) 214 × 10 J (d) 203 × 105 J 45. A body is projected vertically upwards from the surface of a planet of radius R with a velocity equal to half of the escape velocity for that planet. The maximum height attained by the body is R R R R (a) (b) (c) (d) 2 3 4 5 SOLUTIONS 2



a−t bx [a] = [T2], as t2 is subtracted from a.



From, P =

1. (b) : P =

a − t2 t2 = bx bx

 t2  [T 2 ] = [M–1L0T4] [b] =   =  Px  [ML−1T −2 ][L] a [T 2 ] ∴  = = [ML0T–2]  b  [M −1L0T 4 ]





2. (a) : Here, u = – 12 m s–1, as packet is to fall to the ground and it is released initially with the velocity of balloon, so a = + g = 10 m s–2, S = 65 m, t = ? 1 S = ut + at2 2 65 = – 12t + 5t2 or 5t2 – 12t – 65 = 0

12 ± 144 + 1300 12 ± 38 = = 5 s or − 2.6 s 10 10 Time cannot be negative. Therefore, t = 5 s. t=

3. (b) :

N m f

mg

F

Here, m = 10 kg, g = 10 m s–2, m = 0.5 F = 100 N Force of friction, f = mN = mmg = 0.5 × 10 kg × 10 m s–2 = 50 N Force that produces acceleration F′ = F – f = 100 N – 50 N = 50 N F ′ 50 N a= = = 5 m s −2 m 10 kg

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4. (d) : The net charge on the outer surface is – (– 50e – 50e) = 100e. 5. (a) : 2V 2



I

I

998  V

From Ohm’s law 2V ε I= = = 2 × 10–3 A R + r 998 Ω + 2 Ω



Potential difference across the voltmeter is V = IR = (2 × 10–3 A) × 998 W = 1.996 V

6. (b) : Intensity of magnetisation M M I= = V  Mass   Density  Given: Mass = 1 g = 10–3 kg



5 × 10 −3 kg and density = 5 g cm–3 = = 5 × 103 kg m–3 −2 3 3 ( 10 ) m



Hence, I =

6 × 10 −7 A m 2 × 5 × 10 3 kg m −3 10 −3 kg

= 3 A m −1

7. (a) : Angular momentum remains conserved during the revolution of planet. Because gravitational force is a central force.

Planet

rmax

vmin

Star vmax



According to law of conservation of angular momentum, we get



mvmaxrmin = mvminrmax where vmin is the speed of the planet when it is farthest from the star and vmax is the speed of the planet when it is closest to the star. vmin rmin ∴ = ...(i) v r



max

max



As per question rmax =4 rmin ∴

1 2 2 r  KEFarthest 2 mvmin = =  min  1 KEClosest r mv 2max  max  2 2



1 1 =  = 4 16

F=

(Using (i))

1 1 MRα = × 5 × 0.2 × 6 = 3.0 N 2 2

9. (d) : The slope of F-x gives the spring constant k. From the given graph, (80 − 0) N k = Slope of F-x graph = = 400 N m −1 (0.2 − 0) m Time period of SHM is



rmin

8. (c) : Given : w = 2 + 6t dω α= =6 dt t = FR = Ia where R is the radius of the disc. 1 2 I α 2 MR α ∴ F= = R R

T = 2π

m 0.01 kg = 2π = 0.03 s k 400 N m −1

10. (b) : The given equation of the transverse wave is y = A sin2(wt – kx) dy Velocity of the particle = = 2Awcos2(wt – kx) dt Maximum velocity = 2Aw Coefficient of t 2ω ω Velocity of the wave = = = Coefficient of x 2 k k As per question

2 Aω =

ω 1 λ λ or 2 A = = ⇒ A= k k 2π 4π

11. (d) : From Boyle’s law (T = constant) P1V1 = P2V2 4 \ (Hrwater + hrmercury) g ( pr3) 3 4 = hrmercury g ( p(2r)3) 3 or Hrwater = 8hrmercury – hrmercury ρmercury or H = 7h ρ water \ H = 7hr 12. (b) : At the time of projection kinetic energy of the stone, 1 K = mu2 2 where m is the mass of the stone and u is the velocity of the projection

2 K 2 × 98 = = 98 m 2 Using, v2 = u2 – 2gh or

u2 =

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u2 ( ∵ v = 0) ∴ h= 2g 98 h= =5m 2 × 9.8 1 Also , K = m ( 2 gh) 2 1 1 and K ′ = mv′ 2 = m × ( 2 gh′ ) 2 2 K ′ h′ ∴ = K h



...(i)

(Using (i))



T + T  T1 − T2 = K  1 2 − Ts   2  t where Ts is the temperature of the surroundings.



For 1st case



 60 + 55  60 − 55 = K − 45  2  30



Similarly, for 2nd case

As V = ∴

100 =

VR2 + (VL − VC )2 VR2 + 0 = VR or VR = 100 V

Hence, the reading of the voltmeter V is 100 V and the reading of ammeter A is 2 A.

17. (b) : Given, Ex = 0,   rad  N −2 rad   cos  2π × 106 Ey = 2.5  x ,  t −  π × 10  s m   C

Ez = 0. This shows that the wave is propagating along + x direction. Comparing the given equation with Ey = E0cos(wt – kx), we get

ω = 2π × 106 or 2πυ = 2π × 106 or υ = 106 Hz 2π and k = π × 10 −2 or k = = π × 10 −2 λ 2π = 200 m or λ = π × 10 −2 18. (a) : The figure shows incidence from water at critical angle qc for the limiting case.

Disc

...(i)

 55 + 50  55 − 50 ...(ii) = K − 45  2  t Dividing Eq. (i) by Eq.(ii), we get 5 30 = 12.5 or t = 12.5 or t = 12.5 × 30 = 50 s 5 30 7.5 7.5 7.5 t 15. (b) : Here, Magnetic field, B = 0.025 T Radius of the loop, r = 2 cm = 2 × 10–2 m Constant rate at which radius of the loop dr shrinks, = 1 × 10 −3 m s −1 dt Magnetic flux linked with the loop is f = BAcosq = B(pr2)cos0° = Bpr2 The magnitude of the induced emf is dφ d ( dr ε = = Bπr 2 ) = Bπ2r dt dt dt

64 physics for you |

16. (a) : As VL = VC \ XL = XC The circuit is resonance circuit. 100 V Current, I = =2A 50 Ω



14. (c) : According to Newton’s law of cooling

= 0.025 × p × 2 × 2 × 10–2 × 1 × 10–3 = p × 10–6 V = p mV





According to the problem K h′ K = or K′ = 2K h 2 h 5 h′ = = m = 2.5 m 2 2 13. (b) : According to law of conservation of linear momentum, we get mu mu + 0 = 0 + Mv2 ⇒ v2 = ...(i) M As per definition, mu v2 − 0 v2 ( v2 − v1 ) m (Using(i)) e=− =− = = M = u 0−u (u2 − u1 ) u M





february ‘13

r h

c

Air Water

c

Light source

1 1 so that tan θc = µ (µ 2 − 1)1/ 2 r From figure, tan θc = h where r is the radius of the disc. Therefore, diameter of the disc is 2h 2r = 2 h tan θc = (µ 2 − 1)1/ 2 19. (d) : Number of spectral lines obtained due to transition of an electron from nth orbit to lower orbits is n(n − 1) N= 2 In the first case, N = 6 n(n − 1) ∴ 6= ⇒ n=4 2 Now, sin θc =

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In the second case, N = 3 n(n − 1) ⇒ n=3 ∴ 3= 2 Velocity of an electron in hydrogen atom in nth orbit is 2 πe 2 1 vn = ;v ∝ 4 πε0nh n n

21. (b) : At a point P, midway between the poles, magnetic field due to each pole is in the same direction. µ m m  \ B = B1 + B2 = 0  21 + 22  4π  r r  As m1 = m2 and r1 = r2

−4

µ 2 m 10 × ( 2 × 10 ) \ B = 0 = 8 × 10–9 T = 4π r 2 ( 5 × 10 −2 )2 22. (a) :

(+q)



6m

(+q)

A

E

B

D +q/2

F

C +q/2

DE = CE = ( AE)2 + ( AD)2 = ( 3)2 + ( 4)2 = 5 m



Potential at E, q q / 2 q / 2 1  q VE = + + + 4 πε0  AE BE DE CE 



=

1  q q q / 2 q / 2  = 1  2q + q  + + + 4 πε0  3 3 5 5  4 πε0  3 5 

AF = BF = ( AD)2 + (DF )2 = ( 4)2 + ( 3)2 = 5 m Potential at F, q q / 2 q / 2 1  q VF = + + + 4 πε0  AF BF DF CF  1  q q q / 2 q / 2  = 1  2q + q  = + + + 4 πε0  5 3  4 πε  5 5 3 3  0

I1

4 C

2 I3 2

I2 B

5V

S



Let V be the potential at C. Applying Kirchhoff’s first law at C, we get I1 + I 2 = I 3 10 − V 5 − V V − 0 + = 4 2 2 10 – V + 10 – 2V = 2V or V = 4 V ∴

I3 =

4V = 2A 2Ω

24. (a) : Given: F = – 5x – 16x3 N xf



4m





10 V A

2

−7

q 1 1 2q − = 4 πε0  3 5  15 × 4 πε0

=

P hc / λ \ No. of photoelectrons emitted per second 0.5 1 Pλ = N= × 100 200 hc 1 Pλ \ Photoelectric current = ×e 200 hc (10 −3 ) × ( 4560 × 10 −10 ) × (1.6 × 10 −19 ) 1 = × 200 (6.62 × 10 −34 ) × ( 3 × 108 ) –6 = 1.836 × 10 A





No. of photons emitted per second, N =

1

=

1  2q q 2q q  + − − 4 πε0  3 5 5 3 

2 × 9 × 109 × q = 1.2 × 109 q volt. 15 23. (c) : The currents through various arms will be as shown in figure.

v 3 ∴ 4 = v3 4

hc 20. (a) : Energy of photon = λ Power of lamp = P



∴ VE − VF =

W=

∫

xi

Fdx =

0.2

∫ ( −5x − 16 x

0.1

3

)dx

0.2

 5  = − x2 − 4x4   0.1  2 5 5 = − (0.2)2 − 4(0.2)4 + (0.1)2 + 4(0.1)4 2 2 = –0.1 – 0.0064 + 0.025 + 0.0004 = – 0.081 = – 8.1 × 10–2 J

25. (d) : Kinetic energy of the satellite orbiting the earth is 1 E = mv 2, where v is the orbital velocity. 2 2 \ 2 E = 1 m ( 2 v ) 2 2v is the escape velocity of the satellite from the earth. \ When the kinetic energy of satellite is made 2E, satellite escapes away. 26. (a) : Force required to keep the belt moving = F dm = 0.2 m s–1 × 2 kg s–1 = 0.4 N F=v dt 27. (a) : From conservation of energy Potential energy = Translational + Rotational kinetic energy kinetic energy

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1 1 mv 2 + I ω 2 2 2 v2 1 1 2 mgh = mv 2 +   mR2 2 2 5 R2 1 2 1 2 or mgh = mv + mv 2 5



or

mgh =

mgh =



32. (d) : According to first law of thermodynamics, For the path acb,

7 10 mv 2 or v 2 ≥ gh 10 7

28. (c) : The average translational kinetic energy of a 3 molecule of a gas = kBT which is independent 2 of mass of molecule but depends upon the temperature of the gas. π 29. (d) : y1 = 0.1sin(100pt + ) 3 dy Velocity, v1 = 1 dt π = 0.1 × 100pcos(100pt + ) 3 y = 0.1cospt



c

π or = 0.1sin(pt + ) 2 π dy Velocity, v2 = 2 = 0.1 × pcos(pt + ) 2 dt \ Phase difference of the velocity of particle 1 w.r.t. particle 2 is π π = (100pt + ) – (pt + ) 2 3 π π = 99pt + – 3 2 At t = 0, phase difference =

π π π – =– 3 6 2

30. (b) : Here, l = 30 cm = 30 × 10–2 m, v = 330 m s–1 In case of an open organ pipe, fundamental frequency = frequency of 1st harmonic 330 v υ1 = = = 550 Hz 2l 2 × 30 × 10 −2 The frequncies of 2nd harmonic, 3rd harmonic, 4th harmonic,.... are 2 × 550 Hz = 1100 Hz; 3 × 550 Hz = 1650 Hz; 4 × 550 Hz = 2200 Hz and so on. Hence, the source of frequency 1.1 kHz i.e., 1100 Hz will resonantly excite second harmonic. 31. (c) : Here, r = 1.0 mm = 10–3 m, S = 0.075 N m–1, r = 103 kg m–3; h = 10 cm = 10 × 10–2 m The pressure inside the bubble which is greater than the atmospheric pressure is 2S = + hρg r 66 physics for you |

february ‘13

b

Qacb = DUacb + Wacb \ DUacb = Qacb – Wacb P = 60 J – 30 J = 30 J a d For the path adb, V Qadb = DUadb + Wadb  DUacb = DUadb, change in internal energy is path independent. \ Qadb = 30 J + 10 J = 40 J

33. (c) : When capacitor is fully charged, it draws no current. Hence potential difference across capacitor = Potential difference across C and F. 3

B I1

2



2 × 0.075

+ 10 × 10 −2 × 10 3 × 9.8 10 −3 = 150 + 980 = 1130 N m–2

=

A I H

C

3 F

I1

3

3 I2 3 

D I

E

3

F I G

15 V Refer figure. Effective resistance of the network between A and D is ( 3 + 3) × 3 R1 = =2 Ω ( 3 + 3) + 3 Total resistance of the circuit = 2 W + 3 W = 5 W 15 V Current, I = =3A 5Ω ( 3 A)( 3 Ω) ∴ I1 = = 1A 3Ω+6Ω

and

I2 =

( 3 A)(6 Ω) = 2A 3Ω+6Ω

Potential difference across A and D = 3 W × 2 A = 6 V Potential difference across D and F, VD – VF = 3 W × 3 A = 9 V Potential difference across C and D, VC – VD = 3 W × 1 A = 3 V Potential difference across C and F = VC – VF = (VC – VD) + (VD – VF) = 3 V + 9 V = 12 V \ Potential difference across capacitor = VC – VF = 12 V 34. (d) : Magnetic force on a straight wire F = BIl sin90° = BIl Weight of the wire, W = mg Since the wire remains suspended in mid-air so,

WorldMags.net



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BIl = mg

mg ( 200 × 10 −3 ) × 10 or B = = = 0.66 T Il 2 × 1.5

35. (c) : If the alternating current in the primary is I = I0 sinwt, the emf induced in the secondary is dI d εs = M P = M [ I 0 sin ωt ] dt dt εs = MωI 0cos ωt = 2 πυ MI 0 cos ωt [as w = 2pu]

\ (es)max = 2puMI0



Substituting the given data,



(es)max = 2p × 50 × 1.5 × 1 = 150p = 471 V



Note:



[as (coswt)max = 1]

dI 1 = 300 V is wrong as it (es)max = M = 1.5 × dt 1 1 × 4 50 will give the average value (over a quarter cycle) and not the required peak value.

36. (d) : R =

V 100 V = = 100 Ω I 1A

Vrms 100 V = = 200 Ω 0.5 A I rms



Z=



As R2 + X L2 = Z 2

As XL = wL = 2puL = 2p × 50 × L



\ L =

37. (d) : l1 = 6000 Å, n1 = 16 fringes \ n2 = 24 fringes As n1l1 = n2l2 λ1 n2 ⇒ = λ 2 n1

⇒

6000 24 = λ2 16



⇒

λ2 =

ω = 2 πυ =



=

1  1  0.75 +  −  −0.25 1 (1)( −0.25)

=1–4+3=–3+3=0 Since, power of the system is zero therefore, the incident parallel beam of light will remain parallel after emerging from the system.

40. (c) : The wavelength of the first line of lyman series for hydrogen atom is 1 1 1 3 4 = R  −  = R or λ = 2 2 λ R 4 3 2  1

The wavelength of the second line of Balmer series for hydrogen like ion is 1 1 1 3 2 16 = Z2 R  −  = Z R or λ ′ = 2 2 16 λ′ 4  3Z 2 R 2 As per question, l = l′ 4 16 = or or Z2 = 4 or Z = 2 3 3Z 2

41. (b) : Let u be the initial velocity and a be uniform acceleration of the particle. Using v2 – u2 = 2aS 62 – u2 = 2a × 5 ...(i) and 82 – u2 = 2a(5 + 7) = 2a × 12 ...(ii) Solving (i) and (ii), we get a = 2 m s–2 and u = 4 m s–1 42. (d) : Change in velocity when the particle completes half revolution is

6000 × 16 96000 = 4000 Å = 24 24

38. (b) : Angular frequency,

x  c

= 33 sinp × 1011 (t –





XL 100 3 = = 0.55 H 100 π 100 π

 Ey = E0 sin ω  t − 

x ) c d 1 1 39. (a) : Power of system = + − f1 f2 f1 f2



X L = Z 2 − R2 = 200 2 − 100 2 = 100 3 Ω





2 πc 2 π × 3 × 108 = λ 6 × 10 −3

= p × 1011 rad s–1 The equation for the electric field, along y-axis in the electromagnetic wave is



Dv = 5 m s–1 – (– 5 m s–1) = 10 m s–1 Time taken to complete the half revolution is πr π × 5 m t= = = πs v 5 m s −1 ∆v 10 m s −1 10 Average acceleration = = m s −2 = πs π t

43. (b) : Horizontal acceleration of the system is F F ...(i) a= = 2 m + m + 2 m 5m

Let N be the normal reaction between B and C. C Free body diagram of C gives 2 N N = 2 ma = F (Using (i)) 5 a

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Now B will not slide downward if mN ≥ mBg

or



So



5 2  mg µ  F  ≥ mg or F ≥ 5  2µ Fmin =

5 mg 2µ

44. (c) : Work done against gravitational force = mgh = 1000 × 9.8 × 20 = 196 × 103 J Work done to impart velocity to the body = 1 mv 2 2 1 = × 103 × 16 = 8 × 103 J 2 Work done against frictional force = 500 × 20 = 10 × 103 J 3 Total work done = 214 × 10 J

45. (a) : Escape velocity, ve =

Velocity of projection, v =



The total energy of the body when it is projected is Ei = KE + PE 1 GMm 1 2GM GMm = mv 2 − = m − 2 R 2 4R R 3 GMm 4 R Let the maximum height attained by the body be h. At this height, the total energy of the body is GMm E f = KE + PE = 0 − R+h =−

From the principle of conservation of energy, Ei = Ef −

2GM R



where M and R are the mass and radius of the planet respectively.

ve 1 2GM = 2 2 R



3 GMm GMm =− 4 R R+h

or 3(R + h) = 4R or 3 R + 3h = 4 R ⇒ h =

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Cont. from Page No. 28

Voltage gain = 50 Input resistance, Ri = 100 W Output resistance, Ro = 200 W 200 Ω R Resistance gain = o = =2 Ri 100 Ω (Voltage gain)2 50 × 50 Power gain = = = 1250 Resistance gain 2

3R from the centre 2 of a charged conducting spherical shell of radius R R is E. The electric field at a distance from the 2 centre of the sphere is

49. The electric field at a distance

E E (a) zero (b) E (c) (d) 2 3   50. A and B are two vectors of equal magnitude and q is the angle between them. The angle between   A or B with their resultant is θ θ (a) (b) (c) 2q (d) q 4 2

2 2 or mr rr ω r = m f r f ω f

where the subscripts r and f denote the rear and front. [rr = 2rf (Given)] m(2rf)2wr = mrf2wf ωf \ 4wr = wf or ω r = 4 Hence, the angular velocity of rear wheel will be smaller compared to front wheel. 3. (b) : The frequency of kinetic energy is twice the frequency of simple harmonic motion. 4. (c) : Let the block and bullet meet after time t, at a distance x vertically below the top of the cliff. Taking vertical downwards motion of block for time t 1 ( u = 0) ...(i) x = 0 + gt 2 2 Taking vertical upwards motion of bullet for time t 1 100 − x = 100t − gt 2 ...(ii) 2 Adding Eqs. (i) and (ii), we get 100 = 100t or t = 1 s

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SOLUTIONS 1. (c) : By covering the aperture of diameter d/2, focal length of lens is not affected. Area reduces by (1/4)th. So does the intensity. \ New focal length = f, and I 3I = New intensity = I − 4 4 2. (b) : As no external torque acts on the system, angular momentum will remain conserved. According to law of conservation of angular momentum, I1w1 = I2w2

5. (c) : Here,

6. (d) : According to Kepler’s third law, 3/ 2

T2  1.02 R  = = 1.03 T1  R  T − T1 Percentage difference = 2 × 100 T1

T ∝ R3/ 2 ∴

T  =  2 − 1 × 100  T1  = (1.03 − 1) × 100 = 0.03 × 100 = 3%



7. (c) : As vmax = Rg tan θ For the same banking angle

vmax ∝ R

∴

vmax ′ = vmax

′  R′ R′  vmax or = R R  vmax 

2

As per question 10 11 vmax = vmax + v = v = 1.1 vmax ′ 100 max 10 max 2

∴

R′  1.1 vmax  = 1.21 = R  vmax 



R′ = 1.21R = 1.21 × 20 m = 24.2 m

8. (d) : Here, Vp = 220 V, Is = 2 A, Vs = 440 V h = 80%, Ip = ? η=

Vs I s VI or I p = s s Vp I p ηVp

Substituting the given values, we get Ip =

(440 V)(2 A) =5 A  80    (220 V) 100

9. (a) : As long as the block of mass m remains stationary, the block of mass M released from 2Mg (before coming it rest rest comes down by k momentarily again). Thus the maximum extension in the spring is

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2 Mg ...(i) k For block of mass m to just move up the incline kx = mgsin 37° + mmgcos 37° ...(ii) 3 3 4 (Using (i)) 2 Mg = mg × + mg × 5 4 5 3 or M = m 5 10. (d) : When the frequency of incident light is halved, its frequency becomes 0.75u0 which is less than threshold frequency (u0). Hence, no photoelectric emission will take place. x=

πD 11. (b) :  Cross - sectional area, A = 4 where D is the diameter of the wire. ∆A 2 ∆D ∴ = A D ∆D 1 ∆A 1 = = × 2% = 1 % or D 2 A 2 Poisson’s ratio, σ = or

2

GMms rs2

=

ms vs2 rs

 GM  ∴ vs =   rs  Kinetic energy of the satellite is K=

1 1  GM  GMms m v2 = m = 2 s s 2 s  rs  2rs

…(i)

(Using (i))

Potential energy of the satellite is GMms U=− rs Total energy of the satellite is ...(i)

E = K +U =

GMms GMms GMms − =− 2rs rs 2rs

...(ii)

The angular momentum L of the satellite is given by

∆D / D ∆l / l

∆l ∆D / D 1 = = = 2.5 % l σ 0.4

(Using (i))

12. (a) : According to definition of centre of mass, we can imagine one particle of mass (1 + 2 + 3) kg at (2, 2, 2). Let the second particle of mass 4 kg be put at (x2, y2, z2). Given (xCM, yCM, zCM) = (0, 0 , 0) m x + m2 x2 Using xCM = 1 1 m1 + m2 6 × 2 + 4 × x2 5+4 x2 = –3 Similarly, we get y2 = –3, z2 = –3 0=

13. (a) 14. (a) 15. (b) : Let e be the emf and r be the internal resistance of the battery. As per question ε × 16 12 = ...(i) 16 + r ε × 10 10 + r Dividing (i) by (ii), we get 12 8(10 + r) 3 2 (10 + r) = or = 11 5(16 + r) 11 5 (16 + r) and 11 =

15(16 + r) = 22(10 + r) 240 + 15r = 220 + 22r 20 Ω 7r = 20 or r = 7 16. (a) : The velocity vs of the satellite is given by

...(ii)

L = msvsrs = ms  GM   r  s 2 1/2 = (GMms rs)

1/ 2

rs

(Using(i)) …(iii)

From Eqs. (ii) and (iii), we get L = (2Emsrs2)1/2 17. (a) : Given : x = 36t and 2y = 96t – 9.8t2 or y = 48t – 4.9t2 Let the initial velocity of projectile be u and angle of projection is q. Then, Initial horizontal component of velocity,  dx  ux = ucosq =   = 36  dt  t =0 or ucosq = 36 Initial vertical component of velocity,  dy  uy = usinq =   = 48  dt  t =0

…(i)

or

…(ii)

usinq = 48

Dividing (ii) by (i), we get \

48 4 = 36 3 4 4 sinq = or q = sin–1   5 5

tanq =

18. (d) : Here, body is acting as an observer. According

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to the given case, let u be the velocity of the body. v − u v + u and υ2 = υ  Then, υ1 = υ   v   v  \

Du = u2 – u1 = u

2u v

∆υ 2u × 100 = × 100 = 2 υ v v 300 ⇒ u= = = 3 m s–1 100 100 So,

19. (a) : According to lens maker’s formula  1 1 1  = (µ − 1)  − f  R1 R2  Here, m = \

5 , R = ∞, R2 = 0.3 3 1

26. (c) : At maximum height h, v = 0

\ Total mechanical energy, E1 = mgh + 0 4 h. 5 4 1 \ Total mechanical energy, E2 = mg h + mv′ 2 5 2 According to the law conservation of mechanical energy, Let v′ be velocity of stone at height h′ =



0.9 or f = – = – 0.45 m 2 25 N s −1 I = = 500 N 20. (a) : Fav = ∆t 0.05 s



Heat energy Mass



E2 = E1

4 1 mgh + mv′ 2 = mgh 5 2  4  2 or v′ = 2 g  h − h   5  ∴

2 ...(i) gh 5 Required ratio of kinetic energy to its potential 4 energy at h′ = h is 5 1 1 2 mgh K.E. 2 mv′ (Using (i)) = = 5 P.E. 4 4 mgh mgh 5 5 1 = 4 or v′ =

[ML2T −2 ] = [M 0L2T −2 ] [M] 1 22. (d) : In case of electric dipole, F ∝ 3 r 3 1 F ′  2r  1 F = or F ′ = = 3 F 8 8 1   r

25. (a) : Here, h = 18 × 10–5 poise, r = 0.3 mm = 0.03 cm v = 1 m s–1 = 100 cm s–1 According to Stoke law, Viscous force, F = 6phrv F = 6p × 18 × 10–5 poise × 0.03 cm × 100 cm s–1 = 101.73 × 10–4 dyne



 1 1 5 1  2 = − 1  − =– f  3   ∞ 0.3  0.9

21. (c) : Latent heat =

\ Effective resistance between A and D = 5 W + 10 W + 5 W = 20 W

=

23. (d) : Here, B = 1 Wb m–2 , H = 150 A m–1 m0 = 4p × 10–7 H m–1 As B = m0mrH

1 Wb m −2 10 5 B µr = = = 7 1 1 − − − 6π µ0 H 4 π × 10 H m × 150 A m

A 27. (b) : B

24. (b) : The equivalent circuit is as shown in figure a and b.

A B

0

0

0

1

0 0

0

0 0

1

1 0

0

1 0

0

28. (b) : 2k

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2k

0 1

0 0

k M

2k

1

Y

0

Y

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Springs on the left of the block are in series, hence their equivalent spring constant is ( 2 k )( 2 k ) =k k1 = 2k + 2k Springs on the right of the block are in parallel, hence their equivalent spring constant is k2 = k + 2k = 3k Now again both k1 and k2 are in parallel. \ keq = k1 + k2 = k + 3k = 4k Hence, frequency of the system is 1 keq 1 4k υ= = 2π M 2π M 29. (c) : The oscillator frequency should be same as proton’s cyclotron frquency. Bq Cyclotron frequency, υ = 2 πm or Bq = 2pmu ...(i) Kinetic energy of proton is

q 2 B2r 2 ( 2 πmυ)2 r 2 (Using (i)) = 2m 2m = 2p2mu2r2 Substituting the given values, we get K=

K=

2 × ( 3.14)2 × (1.67 × 10 −27 ) × (107 )2 × (0.5)2

= 5.1 MeV

1.6 × 10 −13

1 T …(i) 2L µ where the symbols have their usual meanings. When the length of a stretched string is shortened by 40% and the tension is increased by 44%, then its length and tension becomes 40 3 L′ = L − L= L 100 5 44 36 T= T 100 25

Dividing (ii) by (i), we get υ′ 2 = υ 1



5 5 × 6 × 10 −7 λ= m = 15 × 10–7 m 2 2

= 1.5 × 10–6 m = 1.5 mm

33. (a)  A + δm  sin   2  34. (b) : µ = sin( A / 2) where m is the refractive index, A is the angle of prism and dm is the angle of minimum deviation. Given : A = 60°, m = 1.414  60° + δ m  sin   2 2= sin 30° 1 2

 60° + δ m   60° + δ m  = sin   or sin 45° = sin   2 2 60° + δ m or dm = 90° – 60° = 30° 2

35. (c) : According to Stefan’s law dT Aσ 4 = [T − Ts4 ] dt ms Aσ or s= [T 4 − Ts4 ] dT m dt

Here, A = 19.2 cm2 = 19.2 × 10–4 m2, m = 34.38 g = 34.38 × 10–3 kg, s = 5.73 × 10–8 Wm–2 K–4,

dT = 0.04°C s–1 dt

T = 400 K, Ts = 300 K

1 T′ 2L′ µ 1 36 T 2 T =  3  25 µ 2 L µ 2  L 5 

\ S1P – S2P =



Then, final fundamental frequency is

=

For 3rd dark fringe n = 3

45° =

υ=

υ′ =

32. (b) : Here, l = 6000 Å = 6000 × 10–10 m = 6 × 10–7 m For nth dark fringe, λ Path difference, S1P – S2P = (2n – 1) 2

MeV

30. (a) : Initial fundamental frequency of a stretched string is

T′ = T +

31. (a) : The yellow light is scattered less by the fog particles.

∴ s= =

…(ii)

(19.2 × 10 −4 )( 5.73 × 10 −8 )[( 400)4 − ( 300)4 ] 34.38 × 10 −3 × 0.04

(19.2 × 10 −4 )( 5.73 × 10 −8 )108 [( 4)4 − ( 3)4 ] 34.38 × 10 −3 × 0.04

= 1400 K

36. (d) : From N = N0 e–lt N0 = N0e–l×5 e

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1 5 The activity reduces to half its initial value in half life period, ln 2 ln 2 T1/ 2 = = = 5 ln 2 λ 1/ 5 5l = 1 or l =

\

37. (b) : The magnetic field at the centre O due to current in the inner coil is µ 10 × 2 × 0.2  B1 = 0 4π 0.2

0.3 A

0.2A O

AT − BT 2 V AT − BT 2 or V = P

41. (b) : P =

Since P is constant AdT − 2 BTdT P or PdV = (A – 2BT) dT

\ dV =

Workdone, W = ∫ PdV =

0.2 m

0.4 m

The magnetic field at the centre O due to current in the outer coil is µ 10 × 2 × 0.3 ⊗ B2 = 0 4π 0.4

20µ0 = 4π

+ 30 30 15 P →14 Si + e

positron

43. (c) :

Net magnetic field at the centre O is B = B1 – B2 [Since B1 and B2 are in opposite directions]

A

C



Q 8 × 10 4 cal = T 273 K = 293 cal K–1

–4

2

A

C

C

C

C

C

C

2C

E B

C

2C

B

2C/3

A

−6 −6   = − 1000  −10 − 10  V 0.2

2C/3

B

The equivalent capacitance between A and B is

= 10 × 10–3 V = 10 mV 40. (b) : Number of neutrons in 126 C = 12 – 6 = 6

february ‘13

D

C

 − φB N φ B N ∆φ B Initial  = −  Final ∆t ∆t

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B

G

= 2 × 10–5 × 500 × 10–4 × cos180° = – 10–6 Wb The average induced emf in the coil is

Number of neutrons in

C G

The equivalent circuit diagrams are as shown in the figure below.

F

= 2 × 10–5 × 500 × 10–4 cos0° = 10–6 Wb Final flux after rotation fBFinal = BAcos180°



E

C

F

A

39. (b) : Here, N = 1000, A = 500 cm = 500 × 10 m B = 2 × 10–5 Wb m–2 Initial flux through the coil fBInitial = BAcosq 2

ε=−

C

C

Change in entropy, ∆S =

D C

 3 5 1 − 4  = 4 µ0 T

38. (d) : Heat required to melt 1 kg ice at 0°C into water at 0°C is Q = mice Lice = (1 kg) (80 cal g–1) = (1000 g) (80 cal g–1) = 8 × 104 cal

∫ ( A − 2 BT )dT

T1

= A(T2 – T1) – B(T22 – T12)

42. (d) :

T2

14 6C

= 14 – 6 = 8



CAB =

2C 2C 4C + = 3 3 3

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5

` 17

5

` 17

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44. (b) : y = bx2

Differentiating w.r.t to t on both sides, we get

dy dx = b2 x dt dt



(iii) Angular wave number, k = 2 rad m–1

vy = 2bxvx



Again, differentiating w.r.t. to t on both sides, we get dvy

dv dx = 2bvx + 2bx x = 2 bvx2 + 0 dt dt dt



dvx = 0, because the particle has constant dt acceleration along y-direction] [

As per question dvy



2 = a = 2bvx2 or vx =

a or vx = 2b

dt 45. (c) : Here, G = 50 W, Ig = 0.05 A , I = 5 A,

a 2b

A = 2.97 × 10 –2 cm2 = 2.97 × 10–2 × 10–4 m2 = 2.97 × 10–6 m2 r = 5 × 10 –7 W m I gG 0.05 × 50 50 = = S= Ω I − Ig 5 − 0.05 99 Now, S = \ l=

1 m = 10 cm 10 (ii) Angular frequency, w = 60 rad s–1 (i) Amplitude , A =

ρl SA or l = A ρ

50 ( 2.97 × 10 −6 ) × = 3.0 m 99 ( 5 × 10 −7 )

46. (c) :

According to steady flow, A1v1 = A2v2 + A3v3 or A3v3 = A1v1 – A2v2 1 or v3 = [ A v − A2 v2 ] A3 1 1 1 [0.2 × 4 − 0.2 × 2] = 1 m s −1 = 0.4 47. (c) : The given wave equation is 1 y= sin( 60t + 2 x) 10 Compare it with the standard wave equation y = Asin(wt + kx)

\ Velocity of the wave,

ω 60 rad s −1 = = 30 m s −1 k 2 rad m −1

v=

\ Frequency of the wave, ω 60 30 υ= = = Hz 2π 2π π

2π 2π = = πm k 2 As there is positive sign between t and x terms, the given wave is moving in the negative x direction. \ Wavelength of the wave, λ =

48. (d) : Here, n1 = 4, T1 = 400 K n2 = 2, T2 = 700 K



The temperature of the mixture is Tmixture =

n1T1 + n2T2 n1 + n2



=

4 × 400 K + 2 × 700 K 4+2



=

1600 K + 1400 K = 500 K 6

49. (a) : Electric field at a point inside the charged conducting spherical shell is zero as charge only resides on the outer surface of the conducting spherical shell.   50. (b) : Here, A = B  Let the resultant R makes an angle a  with the vector A . Then

tan α =

  B sin θ A sin θ =  A = B Given A + B cos θ A + A cos θ

θ θ 2 sin cos sin θ 2 2 = tan θ = = 1 + cos θ 2 2θ 2 cos 2 θ α= 2 mmm

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(

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)

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units & measurements

kinematics

‰‰ Physical quantity = Numerical value × unit ‰‰ Homogeneity Principle

Dimensions of [LHS] = Dimensions of [RHS] ‰‰ Mean absolute error

| ∆a1 | + | ∆a2 | +... + | ∆an | n 1 n = × ∑| ∆ai | n i =1

∆amean = ∆amean

€€ Instantaneous speed

a + a2 + .... + an ‰‰ Arithmetic mean amean = 1 amean =

n

1 ai n i∑ =1

total path length ‰‰ Speed = time taken total distance travelled €€ Average speed = total time taken s1 + s2 + s3 ...... i.e. vav = t1 + t2 + t3 ......

n

‰‰

‰‰ Relative error or fractional error

mean absolute error ∆amean = amean mean value ∆amean × 100% ‰‰ Percentage error δa = amean =

‰‰ If in a vernier callipers n VSD coincide with (n – 1)

MSD, then vernier constant or its least count is  n − 1 1 VC =  1 − (value of 1 MSD) or (value of n  n  MSD). ‰‰ Least count of screw gauge or spherometer

=

Pitch and Number of divisions on circular scale

Pitch =

Number of divisions moved on linear scale Number of rotaations given





= Linear distance moved in one rotation.

n , where n = number of events or n = number of quantities. of curvature using spherometer ‰‰ Radius ‰‰ Random error =

R=

l2 h + 6h 2

= lim

∆s

∆t → 0 ∆t

ds dt

=

displacement time taken total displacement €€ Average velocity = total time taken Velocity =

‰‰ Acceleration a =

Change in velocity time taken

€€ Average acceleration



 ∆v  aav = ∆t

€€ Instantaneous acceleration

   ∆v  a = lim   . ∆t → 0  ∆t 

‰‰ Equation of motion for a uniform accelerated

motion

€€ v = u + at

1 2 at 2 2 2 €€ v – u = 2as a €€ sn = u + ( 2n − 1) 2 where u is initial velocity, v is final velocity, a is uniform acceleration, s is distance travelled in time t, sn is distance covered in nth second. These equations are not valid if the acceleration is nonuniform. ‰‰ Equation of motion for a body under gravity €€ v = u + gt €€ s = ut +

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1 2 gt 2 2 2 €€ v – u = 2gh 1 €€ hn = u + g( 2n − 1) 2 ‰‰ Relative velocity €€ If two bodies are moving along the same line in the same direction with velocities vA and vB relative to earth, the velocity of B relative to A will be given by vBA = vB – vA. €€ If the bodies are moving towards or away from each other as directions of vA and vB are opposite velocity of B relative to A will be vBA = vB – (–vA) = vB + vA. ‰‰ Relative velocity of rain

vm vr Where, vm = velocity of man vr = velocity of rain and a is the angle with the vertical direction at which man should hold umbrella to save himself from the rain.  a ^ ‰‰ Unit vector, a =  |a| where, a^ is the unit vector drawn in the direction   of a and| a |is the magnitude of the vector. ‰‰ Dot or scalar product   a ⋅ b = ab cos θ, 0≤q≤p ‰‰ Properties of dot product     €€ a ⋅ b = b ⋅ a        €€ a ⋅ ( b + c ) = a ⋅ b + a ⋅ c         €€ m( a ⋅ b ) = ma ⋅ b = a ⋅ ( mb ) = ( a ⋅ b )m where m is a scalar tan α =

^ ^

^ ^

^ ^

^ ^

^ ^

^ ^

€€ i ⋅ i = j⋅ j = k⋅ k = 1, €€ i ⋅ j = j⋅ k = k⋅ j = 0



^

^

^



^

^

^

€€ If a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k 

 a ⋅ b = a1b1 + a2b2 + a3b3   a ⋅ a = a 2 = a12 + a22 + a33 .   b ⋅ b = b 2 = b12 + b22 + b33 .     €€ If a ⋅ b = 0 and a and b are not null vectors, then  a and b are perpendicular. ‰‰ Cross or vector product   ^ a × b = ab sin θ n .

0≤q≤p

‰‰ Properties of vector product









€€ a × b = − b × a

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









          €€ m( a × b ) = ( ma ) × b = a × ( mb ) = ( a × b )m , where m is a scalar. €€ a × ( b + c ) = a × b + a × c

€€ h = ut +

€€



^

^

^

€€ If a = a1i + a2 j + a3 k

 ^ ^ ^ and b = b1i + b2 j + b3k , then ^

^

i j   a × b = a1 a2 b1 b2



^

k a3 b3

    a and b .     €€ If a × b = 0 and a and b are not null vectors,   then a and b are parallel. ‰‰ Parallelogram law of vector addition     R = a + b , then R = a 2 + b 2 + 2 ab cos θ b sin θ and tan β = a + b cos θ      €€ If R = a − b = a + ( − b ) €€ a × b = the area of a parallelogram with sides

then R = a 2 + b 2 − 2 ab cos θ b sin(180° − θ) b sin θ = and tan β = a + b cos(180° − θ) a − b cos θ   Where, q is the angle between a and b . ‰‰ Equation of trajectory

y = x tan θ −

gx 2 2u2 cos 2 θ

Where u is initial velocity makes an angle q with the horizontal. ‰‰ Time of flight 2u sin θ T= g ‰‰ Horizontal range R=



u2 sin 2θ g

€€ Range will be maximum.



If q = 45° 2 u Rmax = g

€€ If angle of projection is changed from q to

q = (90° – q) then range

R′ =

u2 sin 2θ u2 sin[2(90° − θ)] u2 sin 2θ = =R = g g g

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‰‰ Maximum Height

€€ Centripetal acceleration

u2 sin 2 θ H= 2g €€ Height attained by projectile is maximum if q = 90° u2 Rmax H max = = 2g 2 €€ Here, range of projectile u2 sin 2 × 90° =0 g €€ When the range is maximum, (q = 45°)



at = ra Resultant acceleration a = ac2 + at2 a tan β = t . ac

‰‰ Centripetal force

u2 sin 2 45° u2 Rmax = = 2g 4g 4



F=



(Motion up the Plane) 2u sin(θ − β) €€ Time of flight T = g cos β €€ Range

‰‰ Linear momentum

  p = mv

Where, m is mass of a body moving with velocity  v.

u2 [sin( 2θ − β) − sin β]

g cos 2 β R will be maximum maximum. i.e. sin(2q – b) = 1 Rmax =

‰‰ Newton’s second law

when

sin(2q –b)

is

u2 up the plane g(1 + sin β)

‰‰ Motion down the plane €€ Time of flight T =

2u sin(θ + β) g cos β

u2  sin( 2θ + β) + sin β    g  1 − sin 2 β  R will be maximum, if sin(2q + b) = 1

€€ Range, R =

 1 + sin β  u2 down the plane  = 2  1 − sin β  g(1 − sin β) ‰‰ At the highest point of a projectile motion given angular projection, the angular momentum of projectile. Rmax =

u2 g

L = mu cos θ ×

u2 sin 2 θ 2g

‰‰ In case of angular projection, the angle between

velocity and acceleration varies from 0° < q < 180°. ‰‰ Angular acceleration

dω d 2θ = dt dt 2 ‰‰ When a body moves in a circular path with increasing angular velocity, it has two linear acceleration. α=

mv 2 . r Laws of Motion

‰‰ Projectile on an inclined plane

R=

v2 = rω 2 = vω = v( 2 πυ)2 r

€€ Tangential acceleration



R=

H=



ac =

 Force, F = rate of change of linear momentum   dp = = ma. dt  Where a is acceleration produced in the body. ‰‰ Impulse = Change in linear momentum = F × t = m(v – u) ‰‰ Equilibrium of concurrent forces :     F1 + F2 + F3 + .... + Fn = 0 ‰‰ Lami’s theorem : F F1 F = 2 = 3 sin α sin β sin γ   where, α = angle between F2 and F3   β = angle between F3 and F1   γ = angle between F1 and F2 ‰‰ Apparent weight of a man in a lift : €€ When the lift is at rest or moving with constant velocity, the apparent weight = mg. Thus apparent weight = true weight. €€ When the lift is accelerating upwards with acceleration a, then apparent weight = m(g + a). €€ Thus apparent weight is more than the true weight. €€ When the lift is accelerating downwards with acceleration a, then apparent weight = m(g – a). Thus apparent weight is less than the true weight of man.

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‰‰ In the cable supporting the lift breaks, the lift falls

freely with a = g, then apparent weight = m(g – g) = 0.

‰‰ When a person of mass m climbs up a rope

with acceleration a, the tension in the rope is T = m(g + a).

‰‰ When

the person climbs down the rope with acceleration a, the tension in the rope is T = m(g – a).



v2 rg ‰‰ Circular turning of roads €€ The velocity with which a car can take a circular path of radius r without slipping is given by

vmax = µs rg

slipping,

speed, the tension in the rope is T = mg.

‰‰ Thrust on the rocket F = − u 





fk = mk R where mk is coefficient of kinetic friction. ‰‰ Acceleration of a body down a rough inclined plane, a = g(sinq – mcosq)

where, q is the angle of inclination and m is the coefficient of friction. ‰‰ Angle of repose m = tana Where, a is angle of repose ‰‰ Work done in moving a body over a rough horizontal surface. W = mR × s = mmg × s Where, R is normal reaction and s is distance moved by body. ‰‰ Work done in moving a body up a rough inclined plane. 80 physics for you |

february ‘13

 rg(µ s + tan θ)  vmax =    1 − µ s tan θ 

1/ 2

v2 or tan θ = 0 rg v02 h €€ tan θ = = rg b2 − h2 €€ v02 = rg tanq

‰‰ Motion in a vertical circle €€ Tension at any position of angular displacement,

(q) along a vertical circle is given by

mv 2 + mg cos θ r €€ At the lowest point of vertical circle, q = 0° Tension at the lowest point is given by

‰‰ Laws of friction : €€ The magnitude of the force of static friction

between any two surfaces in contact can have the values fs ≤ ms R ...(i) where the dimensionless constant ms is called the coefficient of static friction, R is the magnitude of normal reaction force. The equality in equation €€ holds when the surfaces are on the verge of slipping i.e., fs = (fs)max = (fl) ≡ ms R. €€ The magnitude of the force of kinetic friction acting between two surfaces is

angle of bending tan θ =

€€ The maximum permissible speed to avoid

‰‰ When the person climbs up or down with uniform

 dm   dt  dm Where, is mass of burnt gases escaping per dt second and u = exhaust speed of the burnt gases. ‰‰ Velocity of rocket at any time t. m  v = u log e  0  m ‰‰ Acceleration of rocket at any instant upthrust − weight a= mass

W = (mgsinq + mR)s.

‰‰ Bending cyclist,



T=

mvL2 TL = + mg r €€ At the highest point of the vertical circle, q = 180°. Tension at the highest point is given by 2 mvH TH = − mg r €€ Minimum velocity at the highest point,

vH = gr

€€ Minimum velocity at the lowest point for

looping the loop, vL = 5 gr .

€€ When the string is horizontal, q = 90°, minimum

velocity, v = 3 gr .

€€ Height through which a body should fall for

looping the vertical loop h = 5r/2.

WORK, ENERGY AND POWER   ‰‰ W = F ⋅ S = FS cos θ   Where q is angle between F and S ‰‰ Work done by a variable force, W =

xf

∫ F( x)dx

xi

1 2 mv . 2 ‰‰ Relation between kinetic energy (K) and linear momentum (p) ‰‰ Kinetic energy : K =

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‰‰ A ball dropped from a height h and rebounding.

p2 or p = 2mK 2m Work done by a spring force   W = Fspring ⋅ ds ‰‰ Work energy theorem : W = Kf – Ki

K=



The time taken by the ball in rising to height h1

∫

ROTATIONAL MOTION ‰‰ The coordinates of centre of mass are given by

‰‰ Elastic potential energy : U = 1 kx 2

N

2

XCM =

‰‰ Gravitational potential energy : U = mgh ‰‰

W Power, P = total t

 dW  d s   ‰‰ Instantaneous power, P = = F⋅ = F⋅v dt dt









(m1 − m2 )u1 2m2u2 + m1 + m2 (m1 + m2 )

v2 =

2m1u1 (m − m1 ) u + 2 m1 + m2 m1 + m2 2

i =1 N

=

∑ mi

∑ mi xi i =1

M

N

∑ mi yi ∑ mi yi i =1 N

=

∑ mi

i =1

M

i =1

N

A

v1 =

∑

N

mi xi

i =1 N

YCM =

‰‰ Elastic collision in one dimension

B

2 h1 2h = 2e . g g

and coming back is 2

ZCM =



‰‰ Perfectly inelastic collision in one dimension

N

∑ mi zi ∑ mi zi i =1 N

=

∑ mi

i =1

M

i =1

where M = m1 + m2 + m3 ..... mN (total mass of system)

‰‰ For a continuous distribution of mass, the

coordinates of centre of mass are given by

B

v=



1 1 1 x dm; YCM = y dm ; ZCM = z dm M M M ‰‰ Velocity of centre of mass is given by XCM =

A

m1u1 + m2u2 (m1 + m2 )

‰‰ Loss in kinetic energy in elastic collision is 1 m1m2 (u − u2 )2 ∆K = 2 (m1 + m2 ) 1 ‰‰ Coefficient of restitution v − v1 e= 2 u1 − u2 ‰‰ Kinetic energy lost in inelastic collision is



1 m1m2 (u − u2 )2 (1 − e 2 ) ∆K = 2 (m1 + m2 ) 1

‰‰ A ball falls from a height h, it strikes the ground

with a velocity u = 2 gh . Let it rebound with a velocity v and rise to a height h1.

e= or

v = u

2 gh1 2 gh

h1 = e h

=

h1 h or

h1 = e 2 h.

∫

N



 vCM =

∑

i =1 N

∫

N

 mi vi

∑ mi

=

∫



∑ mi vi i =1

M

i =1

‰‰ Acceleration of centre of mass is given by N

 aCM

=

N





∑ mi ai ∑ mi ai i =1 N

∑ mi

=

i =1

M

i =1

‰‰ Angular velocity : ω =

dθ dt

dω dt ‰‰ Equations of rotational motion €€ w = w0 + at 1 2 €€ θ = ω 0 t + αt 2 €€ w 2 – w02 = 2aq ‰‰ Angular acceleration : α =

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

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

‰‰ Theorem of perpendicular axis : Iz = Ix + Iy

‰‰ Torque τ = r × F

In magnitude t = rF sinq    ‰‰ Angular momentum L = r × p In magnitude, L = rp sin q ‰‰ Relationship between torque momentum   dL i.e., τ ext = dt

where, x and y are two perpendicular to axes in the plane and z axis is perpendicular to its plane.



‰‰ Moment of inertia : I =

and

angular

‰‰ Theorem of parallel axes : I = ICM + Md2



N

∑ miri2

where, ICM is the moment of inertia of the body about an axis passing through the centre of mass and d is the perpendicular distance between two parallel axis.

i =1

S.No.

Body

Axis of rotation

Moment of inertia (I)

Radius of gyration (K)

MR2

R

(ii) about a diameter

1 MR 2 2

R

(iii) about a tangent in its own plane

3 MR 2 2

3 R 2

(iv) about a tangent perpendicular to its plane

2MR2

R 2

(i) about an axis passing through its centre and perpendicular to its plane

1 MR 2 2

R

(ii) about a diameter

1 MR 2 4

R 2

(iii) about a tangent in its own plane

5 MR 2 4

5

(iv) about a tangent perpendicular to its own plane

3 MR 2 2

(i) about an axis passing through its centre and perpendicular to its plane

1.

2.

3.

4.

Uniform circular ring of mass M and radius R

Uniform circular disc of mass M and radius R

Solid sphere of radius R and mass M

Hollow sphere of radius R and mass M

82 physics for you |

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2

2

R 2

3 R 2

(i) about its diameter

2 MR 2 5

2 R 5

(ii) about a tangential axis

7 MR 2 5

7 R 5

(i) about its diameter

2 MR 2 3

2 R 3

(ii) about a tangential axis

5 MR 2 3

5 R 3

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6.

(i) about its own axis

1 MR 2 2

Solid cylinder of (ii) about an axis passing through its centre length l, radius R and perpendicular to its own axis and mass M

 l2 R2  M +   12 4 

l2 R2 + 12 4

(iii) about the diameter of one of the faces of cylinder

 l2 R2  M +  4   3 

l2 R2 + 3 4

(i) about an axis through its centre and perpendicular to the rod

ML2 12

L

(ii) about an axis through one end and perpendicular to the rod

ML2 3

L

Thin rod of length L

‰‰

Relation between torque and moment of inertia



Torque t = Ia



where a is the angular acceleration. between angular momentum and moment of inertia, L = Iw 1 ‰‰ Kinetic energy of rotational motion, K R = I ω 2 . 2 ‰‰ Kinetic energy of a rolling body = translational kinetic energy (KT) + rotational kinetic energy (KR) ‰‰ Relation

 K2  1 1 1 Mv 2 + I ω 2 = Mv 2 1 +  2 2 2 R 2   ‰‰ When a body rolls down an inclined plane of inclination q without slipping its velocity at the =

2 gh

bottom of incline is given by v =

1+

K

2

R2

without slipping, its acceleration down the g sin θ . inclined plane is given by a = K2 1+ R2 ‰‰ When a body rolls down on an inclined plane without slipping, time taken by the body to reach

the bottom is given by t =

 K2  2l  1 +  R2   g sin θ

g=



3

GMe Re2

4 G πRe3ρ 4 = 3 = πGRe ρ 3 Re2

‰‰ The acceleration due to gravity at height h above

the surface of earth is given by









gh =

 h  = g 1 + 2 Re   ( Re + h) GMe

−2

 GMe   g = 2   Re 

For h R

E=−

GM

r2 €€ At a point on the surface of the shell i.e. r = R

GM E=− R2 €€ At a point inside the shell i.e. r < R, E = 0 ‰‰ For solid sphere gravitational field intensity change only at a point inside the sphere i.e., r < R. GMr E=− R3 GM ‰‰ Gravitational potential : V = − r ‰‰ The gravitational potential due to a spherical shell of radius R and mass M at a point distant r from the centre of the shell is given as follows: €€ At a point outside the shell i.e. r > R GM V=− r €€ At a point on the surface of the shell i.e. r = R



GM V=− R €€ At a point inside the shell i.e. r < R

GM( 3R 2 − r 2 )

V=− 2R3 ‰‰ Relation between gravitational field intensity and gravitational potential







vo =

dV dr

‰‰ Time period of a satellite :

T = 2π



( Re + h)3 2 π ( Re + h)3 = GMe Re g

‰‰ Height of satellite above the earth’s surface  T 2R2 g  e h=   4π 2 

GMm

m at height h above the surface of the earth is given by −GMe m Uh = ( Re + h)

1/ 3

− Re

‰‰ Angular momentum of a satellite





GM = [m2GMr ]1/ 2 r

L = mvo r = mr

‰‰ Kinetic energy of a satellite,

1 2 1 GMe m |U | . mv = = 2 o 2 ( Re + h) 2 −GMe m ‰‰ Potential energy of a satellite, U = . Re + h K=

‰‰ Total energy (mechanical) of a satellite





GMe m 2( Re + h)

‰‰ Escape speed : ve =

2GMe Re

PROPERTIES OF SOLIDS restoring area

‰‰

Stress =

‰‰

Longitudinal Stress =

FN A

FV A FT ‰‰ Tangential Stress = A change in length ∆L ‰‰ Longitudinal strain = = original length L ‰‰

‰‰ Gravitational potential energy : U = − r ‰‰ Gravitational potential energy of a body of mass

 GMe   As g = 2   Re 

GMe g = Re Re + h Re + h

E = K +U = −

GM V=− R ‰‰ The gravitational potential due to a solid sphere at a point inside the sphere i.e. r < R

E=−

−GMe m Re ‰‰ Orbital speed of satellite, when it is revolving around earth at a height h is given by Us =



Volumetric Stress =

‰‰ Volumetric strain =

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Stress =E Strain normal stress €€ Young’s modulus, Y = longitudinal strain

‰‰ Hooke’s law : Stress = E × Strain or





=



–ve sign shows that volume is decreasing when force is applied. tangential stress €€ Modulus of rigidity (h) = shearing strain

F/A F = = . θ Aθ ‰‰ In case of a solids and liquids bulk modulus is almost constant. ‰‰ In case of a gas, it is process dependent €€ In isothermal process, K = Ki = P €€ In adiabatic process K = Ka = gP



1 Bulk modulus (B) when pressure is applied on a substance, its volume decreases while mass remains constant. Compressibility =

Hence, its density will increases,  ρ ∆P  ∆P ρ′ = or ρ′ ≈ ρ  1 + s). In this situation as weight will be more than upthrust, the body will sink. €€ The density of body is equal to the density of liquid (i.e., rB = s). In this situation W = W′ so the body will float fully submerged anywhere in the liquid. €€ The density of body is lesser than that of liquid (i.e. rB < s). In this situation W < W′ so the body will move upwards and in equilibrium will float partially immersed in the liquid such that



2S P= . r ‰‰ When an air bubble of radius r is at depth h below the free surface of liquid of density r and surface tension S, then the excess pressure inside the bubble, 2S P= + hρg. r ‰‰ If r1 and r2 are the radii of curved liquid surface, then excess pressure inside the liquid surface is given by 1 1 P = S + .  r1 r2  ‰‰ When two soap bubbles of radii r1 and r2 coalesce to form a new soap bubble of radius r, under

isothermal conditions then r = r12 + r22 . ‰‰ When two soap bubbles of radii r1 and r2 are in

Force F ‰‰ Surface tension, S = = Length L ‰‰ Work done in forming a liquid drop of radius r, surface tension S is, W = 4pr2S. ‰‰ Work done in forming a soap bubble of radius r, surface tension S is,

given by

W = Vinsg or VrBg = Vinsg or VrB = Vins

‰‰ Equation of continuity A1v1 = A2v2



‰‰ Excess pressure inside an air bubble in a liquid is

W = 2 × 4pr2S = 8pr2S.

‰‰ Work done in increasing the radius of a liquid

contact with each other and r is the radius of the r1r2 . interface, then r = r2 − r1

‰‰ The total pressure inside an air bubble of radius r

at a depth h below the surface of liquid of density r is 2S P = P0 + hρg + r ‰‰ The rise or fall in a capillary tube is given by

drop from r1 to r2 is





W = 4 πS

(

r22

−

h=

).

r12

‰‰ Work done in increasing the radius of a soap

bubble from r1 to r2 is

(

 r   cos θ =  R

where θ is the angle of contact.

‰‰ According to Newton viscous force (F) of a liquid

)

W = 8 πS r22 − r12 ‰‰ When n number of smaller drops of a liquid, each of radius r, surface tension S are combined to form a bigger drop of radius R, then



2S cos θ 2S = rρg Rρg



between two layers is given by dv F = − ηA dx where h = coefficient of viscosity of the liquid

πPr 4 P = 8 ηl R 8 ηl R = is called liquid resistance. πr 4 ‰‰ Stoke’s law : F = 6πηrv.



R = n1/3r ‰‰ The surface area of bigger drop = 4pR2 = 4pn2/3r2. It is less than the area of n smaller drops. ‰‰ Work done in breaking a liquid drop of radius R into n equal small drops

‰‰ Poiseuille’s equation : Q =



‰‰ Terminal velocity : v = T

W = 4pR2 (n1/3 – 1) S where S is the surface tension. ‰‰ Excess pressure inside a liquid drop is given by 2S P= . r ‰‰ Excess pressure inside a soap bubble is given by 4S P= . r

2r 2 (ρ − σ ) g . 9η Kη ‰‰ Critical velocity : vc = ρr ‰‰ Reynold number : vc =

ρDvc NR η or NR = ρD η

‰‰ Bernoulli’s theorem :

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1 P + ρgh + ρv 2 = constant 2 1 1 or P1 + ρgh11 + ρv12 = P2 + ρgh2 + ρv22 2 2 ‰‰ Velocity of efflux v = 2 gh



‰‰ Time after which liquid strikes the horizontal

surface

€€ t =

2( H − h) g



where, K = thermal conductivity



and s = electrical conductivity

‰‰ Thermal resistance of the bar, RH = ‰‰ Stefan Boltzmann law : E = sT4

€€ Range = R = vt =

2 gh × 2 h( H − h) H €€ Rmax = H at h = 2 €€ If the hole is at the bottom of the tank, time t taken by the tank to emptied. where a is the area of the hole. A t= 2H/ g a THERMAL, PROPERTIES OF MATTER Relationship between different temperature scales:

‰‰

KA(T1 − T2 ) L K ‰‰ Wiedemann-Franz law : = a constant, where K = therm σT H=

TC − 0 TF − 32 TR − 0 TRa − 460 TK − 273.15 = = = = 100 180 80 212 100

‰‰ If the body is not a perfectly black body, then



E = esT4 ‰‰ The energy radiated per second by a body of area A = eAsT4 ‰‰ Newton’s law of cooling :

dQ = − k(T − TS ) dt

‰‰ Wien’s displacement law : λ T = constant . m

 R 2S  ‰‰ Temperature of sun is given by T =    R 2σ  s

increase in area ∆A β= = original area × rise in temperature A∆T ‰‰ Coefficient of volume expansion of a solid, increase in volume ∆V = original volume × rise in temperature V ∆T ‰‰ Relation between a, b and g γ=

β γ = . 2 3 The specific heat of a substance is given by 1 ∆Q s= m ∆T The molar specific heat of a substance is given by 1 ∆Q C= µ ∆T Thermal capacity, S = s × m Q The latent heat of a substance’s given by L = m Principal of calorimetry : α=

‰‰

‰‰

‰‰ ‰‰ ‰‰



Heat lost by one body = Heat gained by the other. ‰‰ When a bar of length L and uniform area of cross section A with its ends maintained at temperatures T1 and T2, the rate of flow of heat (or heat current) H is given by 88 physics for you |

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.

THERMODYNAMICS

‰‰ Coefficient of linear expansion of a solid,

increase in length ∆L α= = original length × rise in temperature L∆T ‰‰ Coefficient of area expansion of a solid,

L . KA

‰‰ The work done by a gas is W =

Vf

∫ dW = ∫ PdV Vi



where Vi and Vf are the initial and final volume of the gas. ‰‰ First law of thermodynamics : DQ = DU + DW ‰‰ Equation of isothermal process PV = constant. €€ Work done during isothermal process, P   Vf  i W = µRT ln   ; W = µRT ln  P  V f    i



€€ Equation of adiabatic process, PV g = constant

where g = CP/CV.

€€ Work done during adiabatic process,



W=

( PV i i − Pf V f ) ( γ − 1)

; W=

µR(Ti − T f )

γ −1 V ‰‰ Equation of isobaric process = constant. T €€ Work done during isobaric process, W = P(Vf – Vi) = mR(Tf – Ti). ‰‰ Efficiency of a heat engine, Q work done W Q1 − Q2 η= = = =1− 2 heat absorbed Q1 Q1 Q1 ‰‰ The coefficient of performance of a refrigerator, β=



heat extracted from the reservoir at low temperature T2 work done to transfer the heat

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WorldMags.net =

Q2 Q2 = W Q1 − Q2

€€ γ ( rigid diatomic ) =

‰‰ The efficiency of a Carnot engine is given,





Boltzmann constant

R NA NA is the Avogadro’s number. m N Here, µ = = M NA where, m is the mass of the gas containing N molecules, M is the molar mass ‰‰ Equation of a real gas :  µ2 a   P + 2  (V − µb) = µRT V   kB =



where, a and b are Van der waals constants 8a ‰‰ Critical temperature : TC = 27 Rb ‰‰ Critical pressure : PC =

a 27 b 2

‰‰ According to kinetic theory of an ideal gas

pressure exerted by an ideal gas is given by

1 P = mn v 2 3 ‰‰ Root mean square speed, 3RT = M

3kBT m 8 kBT . πm

‰‰ Most probable speed, vmp =

2 RT = M

vrms > v > vmp

3 k T 2 B ‰‰ The molar specific heats are given by 5 €€ CV ( rigid diatomic ) = R 2 7 €€ CP ( rigid diatomic ) = R 2



v = ω A2 − x 2 ‰‰ Acceleration of a particle in S.H.M. is given by

a = –w2x

‰‰ The kinetic energy of a particle in S.H.M. is given by

1 mω 2 ( A2 − x 2 ) 2 ‰‰ The potential energy of a particle in S.H.M. is 1 given by, = mω 2 A2 sin 2 (ωt + φ) 2 ‰‰ Total energy of a particle in S.H.M. is given by 1 E = mω 2 A2 2 ‰‰ Spring pendulum K=

m k ‰‰ The time period of a simple pendulum is given by

T = 2π

‰‰ If the length of a simple pendulum is comparable

with the radius of earth (Re), then time period T is given by 1 T = 2π 1 1  g +  L Re 

is accelerating downwards with an acceleration a, then its time period is given by T = 2π

2 kBT . m

‰‰ Average translational kinetic energy of a gas

molecule is E =

2π T ‰‰ Velocity of a particle in S.H.M. is given by ‰‰ Angular frequency w = 2pu =

‰‰ If a simple pendulum is suspended in a lift and lift

8 RT = πM

‰‰ Average speed, v =

2nπd 2

T = 2π L / g .

‰‰ Critical volume : VC = 3b

vrms =

1

OSCILLATIONS

‰‰ Equation of an ideal gas : PV = µRT = kBNT

‰‰

‰‰ The mean free path, λ =

T η = 1− 2 T1 KINETIC THEORY OF GASES



7 5

L g−a

‰‰ For upwards motion, T = 2π

L g+a

‰‰ For upwards or downwards with constant velocity

v, T = 2π L g ‰‰ If a simple pendulum is suspended in a lift and

lift is freely falling with acceleration g, then its L time period is given by T = 2π =∞ g−g

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‰‰ If a simple pendulum is suspended in a carriage

containing capacitance C and inductance L is given by

which is accelerating horizontally with an acceleration a, then its time period is given by T = 2π

T = 2π LC

‰‰ If a wire of length L, area of cross-section A,

L

Young’s modulus Y is stretched by suspending a mass m, then the mass can oscillate with time period

( g 2 + a2 )

‰‰ If a simple pendulum is suspended from the roof

of a trolley which is moving down an inclined plane of inclination q, then the time period is given by T = 2π



L g cos θ

‰‰ If a gas is enclosed in a cylinder of volume V fitted

with piston of cross section area A and mass M and the piston is slightly depressed and released, the piston can oscillate with a time period

‰‰ If a simple pendulum whose bob is of density r

oscillates in a non-viscous liquid of density s(s < r), then its time period is given by T = 2π

‰‰ Torsional pendulum :

L  σ  1 − ρ  g

T = 2π

‰‰ Speed, frequency and wavelength relation



I C where I is the moment of inertia of the disc about the suspension wire as axis of rotation and C is the restoring torque per unit twist. πηr 4 2L where r is the radius, L is the length and h is the modulus of rigidity of a wire respectively. ‰‰ The time period of oscillation of a liquid in U-tube is given by C=

L h T = 2π = 2π 2g g where L = total length of liquid column in a U-tube, h = height of liquid column in each limb of U-tube Also h = L/2 ‰‰ The time period of oscillation of floating cylinder m Aσg where m is the mass of a cylinder, A is the area of cross section of a cylinder, s is the density of a liquid hρ h′ = 2π σg g where h is the height of cylinder of density r and s is the density of a liquid in which cylinder is floating, h′ is the height of the cylinder inside the liquid. ‰‰ Time period of LC oscillations of a circuit february ‘13



v = lu

‰‰ Intensity of a wave :

I = 2p2u2A2rv where u is the frequency, A is the amplitude, v is the velocity of the wave, r is the density of the medium. ‰‰ Energy density of wave u = 2p2A2u2r where r is the density of the medium.

‰‰

Wave velocity, v =

‰‰ Particle velocity,

ω k

 ω  dy dy = ωA cos( kx − ωt + φ) = −    k  dx dt 2 d y Particle acceleration, a = = −ω 2 y dt 2 Relationship between phase difference, path difference and time difference 2π Phase difference = × path difference λ 2π Phase difference = × time difference T Speed of a transverse waves on a stretched string T is given by v = µ where T is the tension in the string, m is the mass per unit length of the string called linear density. Speed of a transverse wave in a solid is given by

vparticle = ‰‰ ‰‰

in a liquid is given by T = 2π

90 physics for you |

MV BA2

WAVES

T = 2π

or T = 2 π

mL YA

T = 2π

‰‰

‰‰

η ρ where h is the modulus of rigidity, r is the density of a solid. v=



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‰‰ Speed of a longitudinal wave in a medium is

E given by v = ρ where E is the modulus of elasticity and r is the density of the medium.



4L

€€ For nth mode, λ n = ( 2n − )

‰‰

υ1 =

‰‰ Speed of a longitudinal wave in a metallic bar is

Y ρ where Y is the Young’s modulus and r is the density of material of a metallic bar. ‰‰ Speed of a longitudinal wave in a fluid is given B by v = ρ where B is the bulk modulus and r is density of a fluid. given by v =

‰‰ Newton’s formula : v =

P ρ

‰‰ Speed of sound in a gas, v =

γ . v 3 rms 

frequency of a closed organ pipe is given by v v υC = = 4[L + e ] 4[L + 0.6r ]

‰‰ Due to the end correction, the fundamental

frequency of an open pipe is given by v v υO = = 2[L + 2 e ] 2[L + 1.2r ]

‰‰ Speed of sound in air at room temperature using

resonance tube is given by v = 2u(L2 – L1)

‰‰ Beat frequency = no. of beats/sec = (u1 – u2)



frequency and frequency of a tuning fork of arm length L and thickness d in the direction of vibration is given by

where v0 is the speed of sound in the gas at 0°C.

€€ Effect of pressure : The speed of sound in a gas

d d υ =  v =  L2  L2

is given by

γP ρ ‰‰ Vibrations in a stretched string of length L fixed at both ends. €€ Fundamental frequency v=



υ1 =

v v 1 T = = λ1 2 L 2 L µ

€€ For the nth mode, λn = 2L/n



Frequency of nth mode

 Y since v =  ρ  

Y ρ

‰‰ According to Doppler’s effect the apparent

frequency heard by the observer is given by  v ± vo  υ′ = υ    v  vs 



where vs, vo and v are the speed of source, observer and sound relative to air. The upper sign on vs (or vo) is used when source (observer) moves towards the observer (source) while lower sign is used when it moves away. ‰‰ If the wind blows with speed vw in the direction of sound, v is replaced by v + vw in the above equation. If the wind blows with speed vw in a direction opposite to that of sound, v is replaced by v – vw in the above equation. ‰‰ A practical and small unit of loudness of sound is

v nv n T υn = where n = 1, 2, 3, .... = = nυ1 = λn 2L 2L µ p T, υp = 2L µ where p = number of loops. ‰‰ Vibrations of a closed organ pipe

= difference in frequencies.

‰‰ Tuning fork is a source of sound of single

t 

γP γRT = ρ M Speed of sound in gas is independent of the pressure of the gas, provided temperature remains constant. €€ Effect of humidity : With increase in humidity, density of air decreases

v v = λ1 4 L

v( 2n − 1) v = = ( 2n − 1)v1 λn 4L

‰‰ Due to the end correction the fundamental

€€ Effect of temperature : vt = v0 1 +  546 

v=

Frequency, υn =

decibel (dB). 1 decibel = 1/10 bel.

‰‰ In decibel the loudness of a sound of intensity I is

I  given by L = 10 log10   .  I0  mmm

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physics for you | february ‘13

91

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92 physics for you |

february ‘13

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