01-iit Screening-2006 (physics)

IIT-JEE - 2006 MEMORY BASED QUESTIONS

PHYSICS SECTION-A

Single choice: questions 1 to 12, 3 marks for each correct answer and –1 mark for incorrect answer. 1.

Two blocks are attached as shown in the figure. When the string connecting the two blocks is being cut, the acceleration both the blocks are (a) g, g

(b)

g 2

(d)

(c) g,

g ,g 2 g g , 2 2

2m m

Sol: Ans [a] When string is cut

kx 0 = 3 m g

a1

a2 2m g

mg a1 = g, a2 = g 2.

A mass less rod is hanged at two ends by identical strings A and B as shown in the figure. At what distance from wire A, a mass is hanged on the rod, so that wire A vibrates in first harmonic while wire B in second harmonic

A

B L

(a)

L 4

(b)

L 5

(c)

3L 4

(d)

4L 5

Sol: Ans [b] TA + TB = mg

...(i)

mgx = TBL

...(ii)

TA / µ 2 L′

=

Solving, x =

2 TB / µ

TA

TB L x mg

2L′ L 5

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3.

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Figure shows graph between image distance v and object distance u. What is focal length of lens

v (c m ) 31

(a) (0.5 ± 0.05) cm

30

(b) (5 ± 0.05) cm 20

(c) (0.5 ± 0.10) cm (d) (5 ± 0.10) cm

10

u (c m ) – 3 1 – 3 0

Sol: Ans [b] 1 1 1 − = − 10 10 f ⇒

–20

– 11 –10

f = 5 cm

dv df du  = ± 2 + 2  2 f u   v  0.1 df 0.1 = ± 2 + 2 2 (5) 10  10

⇒

  

⇒ df = ±0.05cm

4.

A convex lens of focal length is used to focus the image of Sun. Area of image formed by lens is directly proportional to (a) f

2

(b) f

(c) f

3

(d) 1/f

Sol: Ans [a] d′ d = f D d′ =

f

α

d ·f D

⇒ Area ∝ f

5.

d

D

has half life of 4 days. Find the probability that a nuclei of exactly (i.e., 8 days) 1 4

d′

2

222 82Ra

(a)

F

α

(b)

2 3

(c)

3 4

222 82Ra

decays in two half lives

(d)

1 3

Sol: Ans [c] Probability of not-decayiong in one half life = 1/2 2

1 3 ⇒ Total probability of decaying in two half lives = 1 −   = 4 2

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6.

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Two wires PQ and QR of radius 2r and r respectively having equal length l/2 and equal resistivity ρ are joined as shown. Current I is flowing through the wires. (a) Power loss in PQ is four times power loss in QR

P

l/2

2r

Q

l/2

(b) Current density will be equal in both wire (c) P.D. across wire PQ is four time compare to wire QR (d) Electricfield is equal in both wires Sol: Ans [c] 4RPQ = PQR ⇒ PQR = 4PPQ VPQ R PQ 1 = = VQR R QR 4

⇒ VPQ = 4VPQ

2 µF 1Ω 7.

2Ω

(a)

4 µF

1Ω 2Ω

(b)

2 µF

4 µF

2 µF 4 µF

1Ω 2Ω

(c)

Time constant (in micro seconds) of circuits in order are (a) 18, 4, 8/9

(b) 18, 8/9, 4

(c) 4, 8, 8/9

(d) 4, 8/9. 18

Sol: Ans [b] τ a = 18 × 10–6 sec, τb =

8 × 10–6 sec, τ c = 4 × 10–6 sec 9

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R r

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8.

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A solid sphere of radius R, and mass m has moment of Inertia I about its diameter. It is melted and recausted into a disc with thickness t and radius r, whose moment of inertia perpendicular to its plane and passing through one of its edge is also I, the value of radius r is 2

(a)

(c)

15

R

(b)

2 R 3

(d)

2 R 15 3 R 4

Sol: Ans [a] 2 3 m R 2 = mr 2 5 2

⇒ r=

9.

2 15

R

A plano convex lens of focal length 15 cm is polished at plane surface. A object is placed at distance 25 cm from the lens as shown in the figure. Distance of image from lens is (a) 12 cm to the left

(b) 60 cm to the left

(c) 30 to the right

(d) 20 to the right

20 cm

Sol: Ans [a] Effective focal length of system is 7.5 cm and nature will be converging 1 1 2 − =− V 20 15

⇒ V = –12 cm

10. A system of binary stars of masses mA and mB are moving in circular orbits of radii rA and rB respectively. If TA and TB are the time periods of masses mA and mB respectively, then (a)

r  TA =  A  TB  rB 

3/ 2

(b) TA > TB (if rA > rB) (c) TA = TB (d) TA > TB (if mA > mB) Sol: Ans [c] Since both the stars are rotating about their common c.m. which remain stationary. Their angular velocity is same therefore period of rotation of both the stars is same.

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11.

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  

  , l ≈ 1m, and he commits an   error of ∆l . For T he takes the time of n oscillations with the stop watch of least count ∆ T and he

A student performs an experiment for determination of g  =

4π 2 l T2

commits a human error of 0.1 sec. For which of the following data, the measurement of g will be most accurate? ∆l

∆T

n

Amplitude of oscillation

(a)

5 mm

0.2 sec

10

5 mm

(b)

5 mm

0.2 sec

20

5 mm

(c)

5 mm

0.1 sec

20

1 mm

(d)

1 mm

0.1 sec

50

1 mm

Sol: Ans [d] Smaller the amplitude closer the motion of pendulum to be S.H.M. and larger number of oscillation will reduce human error in measuring time period, hence (d) is most accurate. 12. The number of circular divisions on the shown screw gauge is 50. It moves 0.5 mm on main scale for one complete rotation. The diameter of the ball is :

0

5

(a) 2.25 mm

2

(b) 2.20 mm

(c) 1.20 mm

25

(d) 1.25 mm

Sol: Ans [c] According to figure there is zero error which is equal to 5 × least count =5×

0.5 = 0.05 mm 50

Using second figure, diameter of ball  0.5   mm – 0.05 mm = 1.20 mm = (2 × 0.5) mm +  25 × 50  

SECTION-B Multiple choice: questions 13 to 20, 5 marks for each corect answer and –1 mark for incorrect answer. 13. Figure shows line of the field which of the following field(s) is(are) not shown by the figure (a) electric field

(b) magnetic field

(c) induced electric field

(d) gravitational field

Sol: Ans [a,d] Electric and gravitational field lines dot make closed loop. Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42 -5-

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14. Part AB of fixed curved path is rough while part BC is smooth. A ball rolls down without slipping on part AB (K A and K C are kinetic energies at A and C respectively) (a) hA > hC, KB < KC

(b) hA > hC, KC > KA

(c) hA = hC, KB = KC

(d) hA < hC, KB > KC

C

A hA

hC B

Sol: Ans [a,b] Since part BC is smooth, it will provide no torque to effect rotational kinetic energy ⇒ At C ball will have pure rotation motion ⇒ hA > hC and KC > KA

15. Light is falls on three mettalic surface M1, M2 and M3 and graph is drawn between their stopping potential and

1 (where λ is wavelength of light) if φ1 , φ 2 and φ3 are their work functions λ

respectively (a)

φ 3 : φ 2 : φ1 = 4 : 2 : 1

(b)

φ 3 : φ 2 : φ1 = 1 : 2 : 4

(c)

tan θ ∝

V (Volt)

M1

hc e

θ

⇒

VS =

hc 1 · −φ e λ

φ1 =

e × 0.001 e × 0.002 e × 0.004 , φ2 = , φ3 = hc hc hc

M3

θ

θ

0 .0 0 1 0 .0 0 2

0 .0 0 4

(d) voilet light will cause photo emision in M2 and M3 Sol: Ans [a,c]

M2

φ 3 : φ 2 : φ1 = 4 : 2 : 1

Slope =

hc (i.e., independent of material) e

16. For spherical symmetrical charge distribution, variation of electric potential with distance from centre is given in diagram. Given that V V=

q for r ≤ r0 and 4πε0 r0

V=

q for r ≥ r0 4πε0 r

Then which option(s) are correct (a) Total charge within 2r0 is q (b) Total electrostatic energy for r ≤ r0 is zero (c) At r = r0 electric field is discontinuous (d) There will be no charge anywhere except at r = r0 Sol: Ans [a,b,d]

r0

The graphs shown is for uniformly charged hollow spherical shell.

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r

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17. A solid cylinder of mass m and radius r is rolling on a rough inclined plane of inclination θ . The coefficient of friction between the cylinder and incline is µ. Then (a) frictional force is always µmg cos θ (b) friction is a dissipative force (c) by decreasing θ , frictional force decreases (d) friction opposes translation & supports rotation Sol: Ans [c,d]

f

In case of rolling frictional force is static which is not dissipative force. fs =

mg sin θ 2   1 + mr   I c  

VC ≤ µmg cos θ

θ

18. Function x = A sin2 ωt + B cos2 ωt + C sin ωt cos ωt represent SHM (a) for any value of A, B and C (except C = 0)

(b) if A = –B; C = 2B, amplitude = | B 2 |

(c) if A = B; C = 0

(d) if A = B; C = 2B, amplitude = | B |

Sol: Ans [a,b,d] x = A sin2 ωt + B cos2 ωt + C sin ωt cos ωt Simplifying x–

B A A C = sin 2ωt +  2 − 2  cos 2ωt 2 2   ≡ A1sin 2ωt + A2cos 2ωt

if

C ≠ 0

⇒

A1 ≠ 0

for all values of A2, motion will be SHM. Hence (a) is correct. If A = –B, C = 2B, x–

A  π = B sin 2ωt + B cos 2ωt = B 2 sin  2ωt +  2 4 

Hence (b) is correct If

A = B, C = 0,

If

A = B, C = 2B x–

⇒

x=

A , Hence (c) is incorrect. 2

A = B sin 2ωt 2

Hence (d) is correct.

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19. In a dark room with ambient temperature T0, a black body is kept at a temperature T. Keeping the temperature of the black body constant (at T), Sun rays are allowed to fall on the black body through a hole in the roof of the dark room. Assuming that there is no change in the ambient temperature of the room, which of the following statement(s) is/are correct? (a) The quantity of radiation absorbed by the black body in unit time will increase (b) Since emissivity = absorptivity, hence the quantity of radiation emitted by black body in unit time will increase (c) Black body radiates more energy in unit time in the visible spectrum (d) The reflected energy in unit time by the black body remains same Sol: Ans [a,d] Rate of radiation depends upon temperature of body and enclosure. There is no reflection of energy by black body which are constant respectively. 20. An infinite current carrying wire passes through point O and is perpendicular to the plane containing a current carrying loop ABCD as shown in the figure. Choose the correct option (s).

C B

(a) Net force on the loop is zero

O′

O

(b) Net torque on the loop is zero (c) As seen from O, the loop rotates clockwise

A D

(d) As seen from O, the loop rotates anticlockwise Sol: Ans [a,c]

Since magnetic field is along the current in AB and CD part so force on AB and CD will be zero. Magnetic force on wire BC would be perpendicular to the plane of the loop along the outward direction and on wire DA the magnetic force would be along the inward normal, so net force on the wire loop is zero and torque on the loop would be along the clockwise sense as seen from O.

SECTION-C Compreshension: (questions 21 to 32) 4 Comprehension, with 3 question in each comprehension each question has 5 marks for correct anwers and –2 marks for incorerct Comprehension-I A cylindrical block of density ρ /3, radius 2r and height h is placed on a hole of radius r made in cylindrical container as shown in the figure. Liquid of density ρ is filled in the container 21. Height of liquid in container is reduced till the block is just about to lift. What is h′ ? (a)

5h 3

(b)

h 3

h′

4r h

(c) h

(d)

4h 3

2r

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Sol: Ans [a] ρ gh + ρgh′ × g × 4πr 2 = ρg (h′ + h)·ρ·3πr 2 3 5h h′ = 3 4πr 2

⇒

22. Maintaing the block with external force the height of liquid is further reduced, what is new height of liquid from the base of the block ( h′′ ) when block is just about to lift (a)

4h 9

(b)

5h 9

(c)

h 3

(d)

4h 3

Sol: Ans [a] 4πr 2

⇒

ρ gh = ρgh × 3πr 2 3

4h h′′ = 9

h ′′ 2r

23. If height of liquid ( h′′ ) is further reduced then (a) cylinder will not move up and remains at its original position (b) for h′′ = h/3, cylinder again starts moving up (c) for h′′ = h/4, cylinder again starts moving up (d) for h′′ = h/5, cylinder again starts moving up Sol: Ans [a] As we get unique value of h′′ in above question, therefore answer is (a).

V

Comprehension-II The capacitor of capacitance C can be charged (with the help of a resistance R) by a voltage source V, by closing switch S1 while keeping switch S2 open. The capacitor can be connected in series with an inductor L by closing switch S2 and opening S1.

S1 R

C S2 L

24. Initially, the capacitor was uncharged. Now, switch S1 is closed and S2 is kept open. If time constant of this circuit is τ , then (a) after time interval τ , charge on the capacitor is CV/2 (b) after time interval 2 τ , charge on the capacitor is CV(1 – e–2) (c) the work done by the voltage source will be half of the heat dissipated when the capacitor is fully charged (d) after time interval 2 τ , charge on the capacitor is CV(1 – e–1) Sol: Ans [b] Q = Q0(1 – et/RC), τ = RC at

t = 2 τ , Q = Q0(1 – e–2)

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25. After the capacitor gets fully charged, S1 is opened and S2 is closed so that the inductor is connected in series with the capacitor. Then, (a) at t = 0, energy stored in the circuit is purely in the form of magnetic energy (b) at any time t > 0, current in the circuit is in the same direction (c) at t > 0, there is no exchange of energy between the inductor and capacitor (d) at any time t > 0, instantaneous current in the circuit may V

C L

Sol: Ans [d] When S1 is open and S2 is closed circuit will act as L-C oscillator, in which energy keeps changing 1

between electric and magnetic form with angular frequency

LC

.

q = q0cos( ωt ) while

i=

dq = q0ω sin(ωt ) dt

imax = CV×

1 LC

=V

C L

26. If S1 is opened and S2 is closed at t = 0 and total charge stored in the LC circuit is Q0, then for t ≥ 0 π

  LC 

π

  LC 

 (a) the charge on the capacitor is Q = Q0 cos  2 +   (b) the charge on the capacitor is Q = Q0 cos  2 − 

(C) the charge on the capacitor is Q = − LC (D) the charge on the capacitor is Q = −

t

t

d2 Q dt 2

d2 Q 2 LC dt 1

Sol: Ans [c] d2 Q 1 + Q=0 When S2 is closed and S1 is open circuit will be LC oscillator and LC dt 2

Comprehension-III Two waves y1 = A cos(0.5 πx – 100 πt ) and y2 = A cos(0.46 πx – 92 πt ) are travelling in a pipe placed along x-axis. 27. Find the number of times intensity is maximum in time interval of 1 sec. (a) 4

(b) 6

(c) 8

(d) 10

Sol: Ans [a] It is beat frequency f b = f 1 ~ f 2 = 50 – 46 = 4 Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42 - 10 -

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28. Find wave velocity of louder sound (a) 96 m/s

(b) 100 m/s

(c) 192 m/s

(d) 200 m/s

Sol: Ans [d] Wave velocity =

ω 100π = = 200 m/s 0.5π k

29. Find the number of times y1 + y2 = 0 at x = 0 in 1 sec (a) 100

(b) 46

(c) 192

Sol: Ans [a] At x = 0

y1 + y2

(d) 96

co s 4 πt

y1 + y2 = A cos 100 πt + A cos 92 πt = 0 ⇒ 2A cos (96 πt ) cos (4 πt ) = 0

According to graph shown y1 + y2 will be zero times in one second

0 .5

co s 96 πt

96 + 4 = 100, because cos 96 πt and cos 4 πt can not be zero simultaneously. Comprehension-IV Modern trains are based on Maglev technology in which trains are magnetically levitated, which runs its EDS Maglev system. There are coils on both sides of wheels. Due to motion of train current induces in the coil of track which levitate it. This is in accordance with lenz.s law. If trains lower down then due to lenz.s law a repulsive force increases due to which train gets uplifted and if it goes much high then there is a net downward force due to gravity. The advantage of maglev train is that there is no friction between the train and the track, thereby reducing power consumption and enabling the train to attain very high speeds. Disadvantage of maglev train is that as it slows down the electromagnetic forces decreases and it becomes difficult to keep it levitated and as it moves forward according to Lenz law there is an electromagnetic drag force. 30. What is the advantage of this system? (a) No friction hence no power consumption (b) No electric power is used (c) Gravitation force is zero (d) Electrostatic force draws the train Sol: Ans [a] 31. Which force causes the train to elevate up? (a) Electrostatic force (b) Time varying electric field (c) Magnetic force (d) Induced electric field Sol: Ans [c] Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42 - 11 -

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32. What is the disadvantage of this system? (a) Train experiences upward force according to Lenz’s law (b) Friction force create a drag on the train (c) Retardation (d) By Lenz’s law train experience a drag Sol: Ans [d]

SECTION-D Question 33 to 36, no negative marking, each question of 6 marks. 33. A rectangle plate (a × b) of mass 3 kg and length a = 2 m is hinged at one end so that it can rotate freely about horizontal axis AB. 100 balls of mass 0.01 kg each hit the plate at shaded portion per second from below. (Assuming collision is elastic and ball hit the shaded portion uniformly). What is velocity of balls so that plate is in equilibrium.

A a b a /2

B

Sol: No of ball striking at strip of width dx and parallel to axis  n  dn =  a / 2 · dx  Assuming balls are hitting the shaded portion uniformly.   torque due to impulsive forces a

τ=

3   n  2m0 v · x · m vna · dx  = 2 0 a/2  a/2

∫

a /2 dx

a 3 To be in equilibrium mg = m vna 2 2 0

⇒ v=

mg = 10 m/sec. 3m0 n

(Putting m = 3kg, m0 = 0.01 kg, n = 100)

34. A groove is cut in disc placed on horizontal plane and a mass of 1 kg is put in the groove as shown in the figure. What is the accleration of block w.r.t. disc if disc is provided acceleration of 25 m/sec2 (co-efficient of friction between block and disc is 2/5) 2 a = 2 5 m /sec sin θ =

3 4 , cos θ = 5 5

θ

F r ic

fmax = µ N + µ mg = 10 N 3

ma cos θ = 20 N > fmax ⇒

20 − 10 = 10 m/sec2 and a′ = 1

tio n

block will slide

N1

N ×

ma s

N = ma sin θ = 15 N

in θ

Sol: Using F.B.D. of block w.r.t. disc

ma ma

mg

cos

θ

a′

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35. 0.45 kg Ice at –20°C is mixed with steam of 100°C (mass = 0.05 kg). What will be final temperature of mixture, given that Latent heat of vaporization = 540 cal/gm = 2268 J/g Latent heat of fusion = 80 cal/gm = 336 J/g Specific heat of ice = 0.5 cal/g-K = 2100 J/kg Specific heat of water = 1 cal/g-K = 4200 J/kg Sol: Heat required by ice ot become water at 0°C Hr = 0.45 × 0.5[0 – (–20)] + 0.45 × 80 = 4.5 + 4.5 × 8 = 40.5 Kcal Heat that steam can provide, when become water at 0°C Hs = 0.05 × 540 + 0.05 × 1 × (100 – 0) = 27.0 + 5 = 32 Kcal Hs < Hr, hence whole of the ice will not be melted and final temperature will remain 0°C. 36. In hydrogen-like atom (z = 11), nth line of Lyman series has wavelength λ equal to the de-Broglie’s wavelength of electron in the level from which it is originated. What is the value of n? Sol: For Lyman series  1 1 1 = R(11) 2  2 −  λ (n + 1) 2   1

De Broglie wavelength of electron λd =

= 3

h 2πr = mv (n + 1)

2π(0.529 × 10 −10 )(n + 1) 2 11(n + 1)

λ = λd

⇒

n = 24

SECTION-E Match the followings: question 37 to 40, no negative marking, each question of 6 marks 37. Match the following An ideal gas taken in cyclic process as shown in the figure. ÄW is positive if work is done by the system and ÄQ is positive if heat is absorbed in the process. Now match the following

J

(a) JK

(i)

ÄW > 0

(b) KL

(ii)

ÄQ < 0

(c) LM

(iii) ÄW < 0

(d) MJ

(iv) ÄQ > 0

M

P K

L V

Sol: (a)–(ii); (b)–(i), (iv); (c)–(iv); (d)–(ii), (iii)

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38. Match the following (a) Nuclear fusion

(i)

Mass is converted in to energy

(b) Nuclear fission

(ii) Generally light nuclei take part

(c) b emission

(iii) Generally heavy nuclei undergo the process

(d) Exothermic nuclear reaction

(iv) The process takes place through weak nuclear forces

Sol: (a)–(i), (iii); (b)–(i), (iii); (c)–(i), (iv); (d)–(i), (ii), (iii) 39. A simple telescope used to distant objects has eyepiece and objective lens of focal length fe, and f0 respectively. Then Column-I

Column-II

(a) Intensity of light received by lens

(i)

radius of aperture (R)

(b) Angular magnification

(ii) dispersion of lens

(c) Length of telescope

(iii) focal length f0, fe

(d) Sharpness of image

(iv) spherical aberration

Sol: (a)–(i); (b)–(iii); (c)–(iii); (d)–(i), (ii), (iv) 40. Match the following Column-I

Column-II

(a) Dielectric ring uniformly charged

(i)

Time independent electrostatic field out of system

(b) Dielectric ring uniformly charged rotating with angular velocity ω

(ii) Magnetic field

(c) Constant current in ring i0

(iii) Induced electric field

(d) i = i0 cos ωt

(iv) Magnetic moment

Sol: (a)–(i); (b)–(ii), (iv); (c)–(ii), (iv); (d)–(ii), (iii), (iv)

>=?@

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