Physics 2006

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MOMENTUM, 1525, Wright Town, Jabalpur (M.P.), Ph.(0761) 4005358, 4035241

IIT-JEE 2006

IIT-JEE 2006 Solutions by MOMENTUM(questions based on memory of students)

PHYSICS

INSTRUCTIONS:

(i) Question no. 1 to 12 has only one correct option. You will be awarded 3 marks for rightanswer and −1 mark for wrong answer.

(ii) Question no. 13 to 20 has one or more than one correct option(s). You will be awarded 5

marks if you answer all correct options and only correct option(s) and −1 mark will beawarded for wrong answer.

(iii) Question no. 21 to 32 are based on small write up first go through it then answer these

questions. You will be awarded 5 marks for right answer and −2 marks for wrong answer.(iv) Question no. 33 to 36 are subjective problems. Circle their correct answers. There is no

negative marking for it. Each question carries 6 marks.(v) Question no. 37 to 40 carry 6 marks each. These may have more than one correct options.

There is no negative marking for these.

1. 2

2

T L4g π=

A student measures g N number of times. The error is ΔL in L and ΔT in T. Which will give bestresult?(a) ΔL = 0.5, ΔT = 0.1, N = 20 (b) ΔL = 0.5, ΔT = 0.1, N = 50(c) ΔL = 0.5, ΔT = 0.01, N = 20 (d) ΔL = 0.1, ΔT = 0.05, N = 50

Sol.TT

2LL

gg Δ

+Δ

=Δ

In option (d) error in Δg is minimum and number of repetition of measurement are maximum. Inthis case the error in g is minimum.∴ (d)

2. Parallel rays of light from Sun falls on a biconvex lens of focal length f and the circular image of radius r is formed on the focal plane of the lens. Then which of the following statement is correct?(a) Area of image is πr2 and area is directly proportional to f (b) Area of image is πr2 and area is directly proportional to f 2

(c) Intensity of image increases if f is increased.(d) If lower half of the lens is covered with black paper area will become half.

Sol.(b)

3. The string between mass m and 2m isinextensible. If the string is cut find themagnitudes of accelerations of mass 2m and

m(a) g, g (b) g, g/2(c) g/2, g (d) g/2, g/2 2m

m

Sol. Initially the spring force Kx = 3 mg (before the string is cut)Just after the string is cut, for block 2m

Kx - 2mg = 2ma1 ⇒ 3mg - 2mg = 2ma1 ⇒ a1 = g/2For block of mass m,

mg = ma2 or a2 = g ∴ (c)

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4. If current I is flowing through ABC. Then the correct relationship is

A BC

I

2r r

l /2l /2

(a) VAB = 2VBC

(b) Power across BC is 4 times the power across AB.(c) Current densities in AB and BC are equal.(d) Electric field due to current inside AB and BC are equal.

Sol.2

r

l

A

l

πρ=ρ=R

1:41

R:R BCAB = = 1 : 4

4:1IR:IRV:V BCABBCAB ==

4:1RI:RIP:P BC2

AB2

BCAB ==

4:1A

I:A

I j: jBCAB

BCAB ==

4:1V:VE:E BCABBCAB ==

∴ (b) 5. Focal length of the plano-convex lens is 15

cm. The object is at A as shown in thefigure. The plane side is silvered. Theimage is(a) 60 cm to the left of lens(b) 12 cm to the left of lens

(c) 60 cm to the right of lens(d) 30 cm to the left of lens

A

20 cm

Sol. Focal length F of equivalent mirror is given by

∞+=+=

1152

f 1

f 2

F1

me

or2

15F = cm

The equivalent mirror will concave mirror of focal length2

15cm.

Now for a spherical mirror,

F

1

u

1

v

1=+

⇒ 15

2

20

1

v

1 −=

−+ ⇒

121

605

6038

201

152

v1

−=−=+−

=+−

=

v = -12 cm∴ (b)

6. Double star system consists of two stars A and B which have time period TA and TB, Radius RB A and RBB and mass MA and MB. Choose the correct option.B

(a) If TA > TB then RB A > RBB (b) If TA > TB then MB A > MBB

(c)

3

B

A2

B

A

R

R

T

T⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =⎟⎟

⎠

⎞⎜⎜⎝

⎛ (d) TA = TB B

Sol. For binary star the time period of two stars are equal.∴ (d)


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7. In the arrangement shown, calculate from where the mass shouldbe hanged so that wire (2) vibrates in second harmonic and wire(1) in first harmonic?

(a)5L

(b)4L

(c)5L4

(d)4L3

l

L

A

B D

C

M

Sol. f v λ=

Lf Lf 2

T

T

2

1

=

μ

μ ⇒ 4

T

T

2

1 =

T1 + T2 = Mg

T2 = Mg/5ΣτB = 0 ⇒ Mg × l = (Mg/5) × LB

l = L/5∴ (a)

l

Mg

T 1T 2

8. The graph between object distance (u) and imagedistance (v) is as shown in the figure. What is thefocal length of the convex lens?(a) 5.00 cm, ±0.05 cm (b) 5.00 cm, ±0.10 cm(c) 0.5 cm, ± 0.05 cm (d) 0.5 cm, ±0.10 cm

(v)

10

11

12

13

(-9, 9)(-u) -10-11-12-13

Sol.v2

u1

v1

f 1

=+= ⇒ 2v

f =

Δf = 05.021.0

2v

==Δ

cm

∴ (a) 9. The circular scale of a screw gauge has 50 divisions and pitch of 0.5 mm. Find the diameter of

sphere. Main scale reading is 2.

50


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252

(a) 1.2 (b) 1.25 (c) 2.20 (d) 2.25Ans. (a) 10. Find the time constant for the given RC circuits in correct order.

C1

V

C2

C1

C2

V R1

R2 R1

R2

C1

C2

R1

R2

V

R1 = 1 Ω, R2 = 2 Ω, C1 = 4 μF, C2 = 2 μF(a) 18, 4, 8/9 (b) 18, 8/9, 4 (c) 4, 18, 8/9 (d) 4, 8/9, 18

Sol. τeq = Req.Ceq τ1 = (2 + 1)(2 + 4) = 18 μs

τ2 =98

4242

.1212

=+×

+×

μs

τ3 = ( ) 424.1212

=++×

μs

∴ (b) 11. undergoes radioactive decay with half life 4 days. What is the probability that a nucleus

undergoes decay in two half lives

Ra22187

(a) 1 (b) 1/2 (c) 3/4 (d) 1/4

Sol. Required probability = ( )( )

( )

43

41

1e1e1e1e1 2ln2T2T

2lnT2t =−=−=−=−=− −−λ−λ−

∴ (c) 12. A solid sphere of mass M and radius R having moment of inertia I about its diameter is recast into

a solid disc of radius r and thickness t. The moment of inertia of the disc about an axis about theedge and perpendicular to the plane is I. The relation between radius R and r

(a) R152

r = (b) R15

2r = (c) R15

2r = (d) R15

2r =

Sol. Moment of inertia of sphere I = 2MR52

Moment of inertia of disc about an axis passing through its edge and perpendicular to its plane

I = 22

2cm Mr

2Mr

MhI +=+ = 2Mr23

⇒ 22 Mr23

MR52

= or R15

2r =

∴ (b)


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13. Current I1 is flowing out from the plane of paper. Asteady state current I2 is flowing in the loop ABCD.(a) The net force is zero(b) The net torque is zero(c) As seen from O, the loop will rotate in clockwisealong OO′ axis(d) As seen from O, the loop will rotate inanticlockwise direction along OO′ axis.

OO ′

A

B

C

D

I 1

I 2

Sol. The magnetic force on AB and CD part of the loop will be zero because on this part ld and B areparallel and anti-parallel. The magnetic force on BC part will act upward and on DA part the forceis of equal magnitude but in downward direction. These two forces will produce a torque inclockwise sense when viewed from O.∴ (a) & (c)

14. A solid cylinder is rolling down the inclined plane without slipping.Which of the following is/are correct?

(a) The friction force is dissipative(b) The friction force is necessarily changing(c) The friction force will aid rotation but hinder translation(d) The friction force is reduced if θ is reduced

θ

Sol. Acceleration of center of mass of cylinder

3sing2

MR

2 / MR1

sing

MR

I1

singa

2

2

2cm

cmθ

=

+

θ=

+

θ=

Since3sing2

MMaf sinMg cmθ

==−θ

θ

f mg sin θ

⇒ 3sinMgf θ=

So, if θ is reduced friction f will also decrease.The work done by friction will be zero since point of application of friction force isinstantaneously at rest. Hence frictional force is not dissipative. Since frictional force is static, itsvalue can lie from 0 to μmg cos θ but not necessarily equal to μmg cos θ.The frictional force will provide torque for rotation but it will oppose translation.∴ (c), (d)

15. The following field line do not represent(a) induced electric field(b) magnetostatic field(c) Gravitational field of a mass at rest(d) electrostatic field

Sol. (c), (d)


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16. The graph between the stopping potential(Vo) and wave number (1/ λ) is as shownin the figure. φ is the work function, then(a) φ1 : φ2 : φ3 = 1 : 2 : 4(b) φ1 : φ2 : φ3 = 4 : 2 : 1(c) tan θ ∝ hc/e where θ is the slope

(d) ultraviolet light can be used to lightphotoelectrons from metal 2 and metal 3only

λ1

Vo

0.001 0.002 0.004

θ θ θ

Metal 1 Metal 2 Metal 3

φ1 φ2 φ3

Sol. φ1 : φ2 : φ3 = 4:2:11

:1

:1hc

:hc

:hc

321321 oooooo

=λλλ

=λλλ

From Einstein's photoelectric equation

seVhc

+φ=λ

⇒ ee

hcVs

φ−

λ=

Therefore slope tan θ = ehc

001.01

1o

=λ

nm-1 ⇒ 1oλ = 1000 nm = 10000 Å

002.01

2o

=λ

nm-1 ⇒ 2oλ = 500 nm = 5000 Å

004.01

3o

=λ

nm-1 ⇒ 3oλ = 250 nm = 2500 Å

Hence UV light can be used to eject photoelectrons from all the metals 1, 2 and 3.∴ (a), (c)

17. The ball rolls down without slipping (which is at restat a) along ab having friction. It rolls to a maximumheight hc where bc has no friction. Ka, Kb and Kc arekinetic energies at a, b and c.(a) Ka = Kc, ha = hc (b) Kb > Kc, ha = hc (c) Kb > Kc, ha < hc (d) Kb > Kc, ha > hc

ac

hc ha b

Sol. Since part bc is frictionless, torque on the ball on this part is zero and hence its angular velocity will beconstant on this part. Further from conservation of mechanical energyhc < ha.∴ (d)

18. A black body of temperature T is inside chamber of To

temperature initially. Now the closed chamber is slightly openedto sun such that temperature of black body (T) and chamber (To)remains constant.(a) Black body will absorb more radiation.(b) Black body will absorb less radiation(c) Black body emit more energy(d) black body emit energy equal to energy absorbed by it

ToT

Sol. Since the temperature of the black body is constant. So heat absorbed = heat radiated.∴ (d)

19. y = A sin2ωt + B cos2ωt + C sin ωt cos ωt.For what value of A, B and C it will represent SHM

(a) for all value of A, B and C (C ≠ 0) (b) A = B, C = 2B(c) A = -B, C = 2B (d) A = B, C = 0


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Sol. y = A sin2ωt + B cos2ωt + C sin ωt cos ωt.

y = ( ) ( ) t2sin2C

t2cos12B

t2cos12A

ω+ω++ω−

t2sin2C

t2cos2

AB2B

2A

y ω+ω⎟ ⎠

⎞⎜⎝

⎛ −++=

By the above equation, we can find that option (a), (b) and (c) are correct for represent SHM.∴ (a), (b) & (c)

20. Spherical symmetric charge system centered at origin.

Electric potential φ =ooR4

Qπε

r ≤ Ro

φ =r4

Q

oπεr > Ro

(a) A spherical symmetry at r = 2Ro encloses a net charge Q(b) Electric field is discontinued at r = Ro (c) Change is only present at r = Ro (d) Electrostatic energy is zero for r < Ro

Ro r

φ

Sol. For r > Ro, E =drdφ

− =2

or4

Q

πε

Charge enclosed by the concentric spherical surface of r = 2Ro

= Qr4r4

Qr4E 2

2o

o2

oCo =ππε

ε=×π×ε=φε

For r < Ro, E = 0drd

=φ

− (Θ c = constant)

For r > Ro, E =2or4

Q

dt

d

πε

=φ

−

Since for r < Ro, E = 0, hence charge will be only on the spherical surface of r = Ro.

For r < Ro, E = 0 ⇒ energy density 0E21 2

o =ε

∴ (a), (b), (c) and (d)

∴Passage I

y1 = A cos (0.5πx - 100πt)y2 = A cos (0.46πx - 92πt)

21. How many time is a second does a stationary observer hears loud sound (maximum intensity)(a) 4 (b) 8 (c) 10 (d) 12

Sol. ω1 = 100 π

502

f 1 =πω

= Hz , 462

92f 2 =

ππ

= Hz

Beat frequency = f 1 - f 2 = 4∴ (a)

22. What is velocity of sound wave?(a) 200 m/s (b) 180 m/s (c) 100 m/s (d) 194 m/s

Sol. v = ω /K = 100π /(0.5 π) = 200∴ (a)

23. At x = 0, how many times does the net amplitude between 0 in 1 s(a) 46 (b) 42 (c) 50 (d) 100


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Sol. At x = 0,y = y1 + y2 = 2A cos 96 πt cos 4πt

For y = 0, cos 96 πt = 0 or cos 4πt = 0⇒ 96 πt = (2n + 1) (π /2) and 4πt = (2m + 1) π /2For 0 < t < 1

-

2

1< n < 95.5 and -

2

1< m < 3.5

Here n and m are integers, therefore net amplitude becomes zero 100 times.∴ (d)

Passage II

The density of block is ρ /3 and is used to block a hole H.24. If liquid of density ρ is filled in the apparatus find out h1 (h1 is

the height of upper surface of water level from the top of theblock) so that cylinder just starts rising.

(a) h32

h1 = (b) h35

h1 =

(c) h25h1 = (d) h

52h1 =

2r

4r

Po

h1

h2 ρ/3 h

Po

2r 2r

Sol. The net upward force on block due to pressureP2 × 12 πr2 - P1 × 16πr2

= ρg(h1 + h)12 πr2 - ρgh1 × 16πr2 = ρgh × 12πρ2 −ρgh1 × 4πr2

The block will rise if ρg × 4πr2 (3h - h1) > 16πr2h(ρ /3)g

(3h - h1) > h34

1h3h5

>

h1 =3h5

∴ (b)

h1 P1 × 16πr2

16πr2h(ρ /3)g

P2 × 12πr2

25. If the water level is lowered such that it comes below the top of the block keeping the blockconstant at which height h2 (h2 is depth of the upper surface of water level from the top of theblock) will the block be at a stable condition?

(a)9h4

h 2 = (b)3h

h 2 = (c)9h2

h 2 = (d)3h2

h2 =

Sol. The upward force on block due to pressureP3 × 12 πr2 + (2h2) = ρgh2 × 12πr2 ρgh2 × 12πr2 = 16πr2 h(ρ /3)g

9h4

312h16

h 2 =×

=

∴ (a) h2

16πr2h(ρ /3)g

P3 × 12πr2

26. If water level is lowered below h2, then(a) the block will not rise (b) the block will start rising again at h2 = h/3(c) the block will start rising again at h2 = h/2 (d) the block will start rising again at h2 = h/5


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Sol. If water level is further decrease below h2 the upward force on block due to pressure willbecome less that the weight of block, therefore, the block will not rise.∴ ∴ (a)

Passage III

27. In the given circuit, initially S1 is closed and S2 is open thenif maximum charge is Qo

(a) Charge Q = 2Qo in time t = τ

(b) Charge Q = Qo(1 - e-2) in time t = 2τ

(c) work done by battery =21

energy dissipated in resistor.

(d) Charge Q = Qo(1 - e-1) in time t = 2τ

R S1

S2

V

C

L

Sol. In one time constant the charge on capacitor is 63% of maximum charge.At t = 2τ, Q = Qo (1 - e-2τ / τ) = Qo(1 - e-2)∴ (b)

28. In the above L-C circuit the charge Q on capacitor at any time can hold a value

(a) ⎟ ⎠

⎞⎜⎝

⎛ ω−π

= t2

cosQQ o (b) ⎟ ⎠

⎞⎜⎝

⎛ ω+π

= t2

cosQQ o

(c)2

2

dt

QdLCQ −= (d)

2

2

dt

Qd

LC

1Q =

Sol. Q = Qo cos ωt

andLCQ

dt

Qd2

2

−= (from KVL)

∴(c)

29. After S2 is closed, then which one is correct?(a) The current always flows in one direction(b) The energy is exchanged between the capacitor and the inductor(c) The energy initially is magnetic

(d) The value of a maximum instantaneous current I =CL

V

Sol. From energy conservation,

22 CV21

LI21

= ⇒ VLC

I =

∴ (b)

Passage IV Passage was very lengthy and could not be retrieved.

33. If 0.05 kg steam at 373 K is mixed with 0.45 kg ice at -20oC then find the resultant temperature.Sol. Heat required to melt the entire ice

= msΔT × mL = 0.45 × 0.5 × 20 + 0.45 × 80 = 40.5 k cal.Maximum heat which can be supplied by steam

= mL + msΔT = 0.05 × 540 + 0.05 × 1 × 100 = 27 + 5 = 32 kcalTherefore the entire ice will not melt hence the resultant temperature of mixture will be 0oC.

34. Determine the value of n for which the de-Broglie wavelength corresponding to nth orbit is equalto wavelength of nth line of Lymann series. Given Z = 11.


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Sol. de Broglie wavelength of electron in the nth orbit is given by2πr = nλ1

⇒ λ1 = 102

10Z

n53.0

n2

nr2 −××

π=

πm

Wavelength of nth line of Lymann series is given by

( ) =⎟⎟ ⎠

⎞⎜⎜⎝

⎛

+−=λ 222 1n 11RZ1 1.097 × 107 × 112 × ( )( )2

2

1n n2n ++

Since λ1 = λ2

⇒ n

2π × 0.53 ×

Zn 2

× 10-10 =1 / { 1.097 × 107 × 112 ( ) ⎟

⎟ ⎠

⎞⎜⎜⎝

⎛

+

+2

2

1n

n2n}

Solving this we get n = 25

35. A disc is kept on a smooth horizontal plane with its planeparallel to horizontal plane. A groove is made in the disc asshown in the figure. The coefficient of friction between a

mass m inside the groove and the surface of the groove is 2/5and sin θ = 3/5. Find the acceleration of mass with respect tothe frame of reference of the disc. θ

ao = 25 m/s2

Sol. ( )'NNf +μ=

( )gsinam52

f +θ=

andmf

cosa'a −θ=

= 25 × ⎟ ⎠

⎞⎜⎝

⎛ +×− 105

3

255

2

5

4

= 20 - 6 - 4 = 10 m/s2

θ

ao = 25 m/s2 mao

N

μ N+N ′

36. If n number of balls each of mass m collide elastically withthe plate of mass M per second with a velocity v elasticallyand reflect back. Find the velocity of balls such that plate isat equilibrium.Given: n = 100, m = 0.01 kg, M = 3 kg

b

a

v

Sol. Force applied by striking balls = rate of change of momentum = n × 2mv

For equilibrium net torque should be zero.⇒ Mg ×

2b

= n × 2mv × 4b3

⇒ v = 1001.01003

103nm3

Mg=

×××

= m/s

Match the following

37. Column 1 Column 2

(a) Angular magnification (P) Dispersion of lenses

(b) Sharpness of the image (Q) f o and f e(c) Light gathering power (R) Aperture of lenses


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