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PPhhyyssiiccss Time: 2 hours Note: Question number 1 to 8 carries 2 marks each, 9 to 16 carries 4 marks each and 17 to 18 carries

6 marks each. Q1. A whistling train approaches a junction. An observer standing at junction observes the frequency to be

2.2 KHz and 1.8 KHz of the approaching and the receding train. Find the speed of the train (speed of sound = 300 m/s)

Sol. While approaching

0s

vf ' fv v

= −

2200 = f0 s

300300 v

−

While receding

0s

vf '' fv v

= +

0s

3001800 f300 v

= +

On solving velocity of source (train) vs = 30 m/s Q2. A conducting liquid bubble of radius a and thickness t (t <<a) is charged to potential V. If the bubble

collapses to a droplet, find the potential on the droplet.

Sol. Potential of the bubble 0

1 q(V)4 a

=πε

by conservation of volume

2 344 a t R3

π = π

( )1 32R 3a t=

Hence, potential on the droplet

0

1 qV '4 R

=πε

(as charge is conserved)

1 3aV ' .V

3t ⇒ =

a

t

Q3. The potential energy of a particle of mass m is given by

( ) 0E 0 x 1V x

0 x 1≤ ≤

= >

λ1 and λ2 are the de−Broglie wavelengths of the particle, when 0 ≤ x ≤ 1 and x > 1 respectively. If the total energy of particle is 2E0, find λ1/λ2.

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Sol. K.E. = 2E0 − E0 = E0 (for 0 ≤ x ≤ 1)

10

h2mE

λ =

KE = 2E0 (for x > 1)

20

h4mE

λ =

1

2

λλ

= 2

Q4. A U tube is rotated about one of it’s limbs

with an angular velocity ω. Find the difference in height H of the liquid (density ρ) level, where diameter of the tube d << L.

H

ω

L

Sol. L

2

0

PA dmx∆ = ω∫

2 2L AgHA

2ω ρ

ρ =

2 2LH2gω

=

H

ω

x dx

Q5. A wooden log of mass M and length L is hinged by a

frictionless nail at O. A bullet of mass m strikes with velocity v and sticks to it. Find angular velocity of the system immediately after the collision about O.

LM

O

m

v Sol. Apply conservation of angular momentum about O

(mv)L = (mL2 + 2ML

3)ω

ω = 3mv(3m M)L+

Q6. What will be the minimum angle of incidence such that the

total internal reflection occurs on both the surfaces?

µ3 = √3

µ1 = √2

µ2 = 2

Sol. For first surface 2 sinc1 = 2 sin90° ⇒ c1 = 450 For second surface 2 sinc2= 3 sin90° ⇒ c2 = 60°

θ

µ1=√2

µ2=2

µ3= √3

θ

θ θ

∴ Minimum angle of incidence = Max {c1, c2} = 60°

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Q7. The side of a cube is measured by vernier callipers (10 divisions of a vernier scale coincide with 9 divisions of main scale, where 1 division of main scale is 1 mm). The main scale reads 10 mm and first division of vernier scale coincides with the main scale. Mass of the cube is 2.736 g. Find the density of the cube in appropriate significant figures.

Sol. Least count of vernier callipers = (1 – 910

)mm = 0.1mm

Side of the cube = 10 mm + 1 × 0.1 mm = 10.1 mm = 1.01 cm

Density = 33

2.736 2.66g/ cm(1.01)

=

Q8. An unknown resistance X is to be determined using

resistances R1, R2 or R3. Their corresponding null points are A, B and C. Find which of the above will give the most accurate reading and why?

R X

A B C

G

R = R1 or R2 or R3

Sol. 1

2

r RX

r=

1 2

1 2

r rXX r r

δ δδ= +

1 2r rδ = δ = ∆

1 2

1 2

r rXX r r

+δ= ∆

R X

A B C

Gr1 r2

For XX

δ to be minimum, 1 2r r should be maximum and as 1 2r r+ is constant.

This is true for 1 2r r= . So R2 gives most accurate value. Q9. A transverse harmonic disturbance is produced in a string. The maximum transverse velocity is 3 m/s

and maximum transverse acceleration is 90 m/s2. If the wave velocity is 20 m/s then find the waveform.

Sol. If amplitude of wave is A and angular frequency is ω,

2A 3

90Aω

=ω

⇒ ω = 30 rad/s

v = kω ⇒ k = 13 m

2−

A = 10 cm Considering sinusoidal harmonic function

∴ y = (10 cm) sin(30t ± 3 x )2

+ φ

Q10. A cylinder of mass m and radius R rolls down an inclined plane of inclination θ. Calculate the linear

acceleration of the axis of cylinder. Sol. axismgsin f maθ− = (1) fR = Iaxis α (2) aaxis = Rα (3)

axis2gsina

3θ

= θ

f

N mg

aaxis

α

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Q11. A long solenoid of radius a and number of turns per unit length n is enclosed by cylindrical shell of radius R, thickness d (d<<R) and length L. A variable current i = i0sin ωt flows through the coil. If the resistivity of the material of cylindrical shell is ρ, find the induced current in the shell.

R a

L

d

Sol. 2

0 0( ni sin t) aφ = µ ω π

20 0

d ( ni cos t) adtφ

ε = = µ ω ω π

Resistance = 2 RLd

ρ π

2

0 0( ni cos t) a (Ld)I

2 Rµ ω ω π

=ρ π

`

× × ×

× × × × × × × × a × × ×

R

d

I

Induced current

Q12. Two identical ladders, each of mass M and

length L are resting on the rough horizontal surface as shown in the figure. A block of mass m hangs from P. If the system is in equilibrium, find the magnitude and the direction of frictional force at A and B.

θ θ

mL

P

A B Sol. For equilibrium of whole system, ΣFy = 0

⇒ N 2M m g2+ =

For rotational equilibrium of either ladder Calculating torque about P

NL cos θ − Mg L2

cosθ − fL sinθ = 0

⇒ f = (M+m)g cot2

θ

f f

N N mg

Mg Mg

P

B A

y

x

Q13. Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. The ratio

of radii of nucleus to that of Helium nucleus is (14)1/3. Find (a) atomic number of the nucleus. (b) the frequency of Kα line of the X−ray produced. (R = 1.1 × 107 m−1 and c = 3 ×108 m/s) Sol. r = r0 A1/3

1 3

1 3

He

r A 14r 4

= =

⇒ A = 56 and Z = (56 − 30) = 26 for Kα−line,

( )3Rcv Z 14

= −

⇒ ν =1.546 × 1018 Hz

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Q14. A small body attached to one end of a vertically hanging spring is performing SHM about it’s mean position with angular frequency ω and amplitude a. If at a height y* from the mean position the body gets detached from the spring, calculate the value of y* so that the height H attained by the mass is maximum. The body does not interact with the spring during it’s subsequent motion after detachment. (aω2 > g)

y0

m

Sol. At position B as the potential energy of the spring will be zero, the total energy (Gravitational potential energy + Kinetic energy) of the block at this point will be maximum and therefore if the block gets detached at this point, it will rise to maximum height,

∴ y* = 2mg g ak

= <ω

m

Position B

Position A

Mean Position

Natural length

Q15. In the given circuit, the switch S is closed at time t = 0. The charge Q on the capacitor at any instant t is given by Q(t) = Q0 (1−e−αt). Find the value of Q0 and α in terms of given parameters shown in the circuit.

+ − V

C R2

R1

S

Sol. Applying KVL in loop 1 and 2,

V − i1R1 − qC

= 0 . . . (1)

qC

− i2R2 = 0 . . . (2)

i1 − i2 = dqdt

. . . (3)

+ − V

C R2

R1 i1

i2 i1 − i2

−q

+q 2 1

On solving we get,

q = 1 2

1 2

t(R R )CR R2

1 2

CVR1 e

R R

+−

− +

⇒ 20

1 2

CVRQ

R R=

+ and 1 2

1 2

R RCR R

+α =

Q16. Two identical prisms of refractive index 3 are kept as shown in the figure. A light ray strikes the first prism at face AB. Find,

(a) the angle of incidence, so that the emergent ray from the first prism has minimum deviation.

(b) through what angle the prism DCE should be rotated about C so that the final emergent ray also has minimum deviation.

600

600

600

600

B

A C E

D

Sol. (a) For minimum deviation

r1 = r2 = B2

∠

0sini 3

sin30=

⇒ i = 600

r1 i r2

A

B

C

(b) Prism DCE should be rotated about C in anticlockwise direction through 60° so that the final

emergent ray is parallel to the incident ray and angle of deviation is zero (minimum)

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Q17. A cylinder of mass 1 kg is given heat of 20000J at atmospheric pressure. If initially temperature of cylinder is 200C, find

(a) final temperature of the cylinder. (b) work done by the cylinder. (c) change in internal energy of the cylinder. (Given that Specific heat of cylinder= 400 J kg–1 0C–1, Coefficient of volume expansion

= 9 × 10–5 °C−1, Atmospheric pressure = 105 N/m2 and Density of cylinder = 9000 kg/m3) Sol. (a) ∆Q = ms∆T

⇒ ∆T = 020000J

1kg (400J/kg C)× = 50°C

Tfinal = 70°C (b) W = Patm ∆V = Patm V0γ∆T

= (105 N/m2) 33

1 m9 10

× (9 × 10−5 /°C) (50°C) = 0.05 J

(c) ∆U = ∆Q – W = 20000 J – 0.05 J = 19999.95 J Q18. In a moving coil galvanometer, torque on the coil can be expressed as τ = ki, where i is current

through the wire and k is constant. The rectangular coil of the galvanometer having numbers of turns N, area A and moment of inertia I is placed in magnetic field B. Find

(a) k in terms of given parameters N, I, A and B. (b) the torsional constant of the spring, if a current i0 produces a deflection of π/2 in the coil. (c) the maximum angle through which coil is deflected, if charge Q is passed through the coil almost

instantaneously. (Ignore the damping in mechanical oscillations) Sol. (a) τ = iNAB sinα For a moving coil galvanometer α = 90° ki = iNAB ⇒ k = NAB (b) τ = Cθ

i0NAB = Cπ/2 ⇒ C = 02i NABπ

(c) Angular impulse = ∫τ. dt = ∫ NABidt = NABQ ⇒ NABQ = Iω0

⇒ ω0 = NABQI

Using energy of conservation

2 20 max

1 1I C2 2

ω = θ

⇒ θmax = ω00

I NABQC 2Ii

π=

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IIT-JEE2005-M-1

FFIIIITTJJEEEE SSoolluuttiioonnss ttoo IIIITTJJEEEE––22000055 MMaaiinnss PPaappeerr

MMaatthheemmaattiiccss Time: 2 hours Note: Question number 1 to 8 carries 2 marks each, 9 to 16 carries 4 marks each and 17 to 18 carries

6 marks each.

Q1. A person goes to office either by car, scooter, bus or train probability of which being 1 3 2, ,7 7 7

and 17

respectively. Probability that he reaches office late, if he takes car, scooter, bus or train is 2 1 4, ,9 9 9

and 19

respectively. Given that he reached office in time, then what is the probability that he travelled

by a car. Sol. Let C, S, B, T be the events of the person going by car, scooter, bus or train respectively.

Given that P(C) = 17

, P(S) = 37

, P(B) = 27

, P(T) = 17

Let L be the event of the person reaching the office in time.

⇒ L 7PC 9

=

, L 8P

S 9

=

, L 5PB 9

=

, L 8PT 9

=

⇒ ( )

( )

L 1 7P .P CCC 17 9P

1 7 3 8 2 5 8 1 7L P L7 9 7 9 7 9 9 7

×

= = = × + × + × + ×

.

Q2. Find the range of values of t for which 2 sin t = 2

21 2x 5x3x 2x 1− +

− −, t ∈ ,

2 2π π −

.

Sol. Let y = 2 sin t

so, y = 2

21 2x 5x3x 2x 1− +

− −

⇒ (3y − 5)x2 − 2x(y − 1) − (y + 1) = 0

since x ∈ R − 11,3

−

, so D ≥ 0

⇒ y2 − y − 1 ≥ 0

or y ≥ 1 52+ and y ≤ 1 5

2−

or sin t ≥ 1 54+ and sin t ≤ 1 5

4−

Hence range of t is 3, ,2 10 10 2π π π π − − ∪

.

Q3. Circles with radii 3, 4 and 5 touch each other externally if P is the point of intersection of tangents to

these circles at their points of contact. Find the distance of P from the points of contact.

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Sol. Let A, B, C be the centre of the three circles. Clearly the point P is the in–centre of the ∆ABC, and

hence

r = ( ) ( ) ( )s s a s b s c (s a)(s b)(s c)

s s s− − − − − −∆

= =

Now 2s = 7 + 8 + 9 = 24 ⇒ s = 12.

Hence r = 5.4.3 512

= .

A

C

B

3

3

4 4

5

5 •P

Q4. Find the equation of the plane containing the line 2x – y + z – 3 = 0, 3x + y + z = 5 and at a distance of

16

from the point (2, 1, – 1).

Sol. Let the equation of plane be (3λ + 2)x + (λ − 1)y + (λ + 1)z − 5λ – 3 = 0

⇒ 2 2 2

6 4 1 1 5 3 16(3 2) ( 1) ( 1)

λ + + λ − − λ − − λ −=

λ + + λ − + λ +

⇒ 6(λ – 1)2 = 11λ2 + 12λ + 6 ⇒ λ = 0, − 245

.

⇒ The planes are 2x – y + z – 3 = 0 and 62x + 29y + 19z − 105 = 0. Q5. If |f(x1) – f(x2)| < (x1 – x2)2, for all x1, x2 ∈R. Find the equation of tangent to the curve y = f(x) at the

point (1, 2). Sol. |f (x1) – f (x2)| < (x1 – x2)2

⇒ 1 2 1 2

1 21 2x x x x1 2

f(x ) f(x )lim lim | x x |

x x→ →

−< −

− ⇒ |f′ (x)| < δ ⇒ f′ (x) = 0.

Hence f (x) is a constant function and P (1, 2) lies on the curve. ⇒ f (x) = 2 is the curve. Hence the equation of tangent is y – 2 = 0.

Q6. If total number of runs scored in n matches is n 14+

(2n+1 – n – 2) where n > 1, and the runs scored

in the kth match are given by k. 2n+1–k, where 1 ≤ k ≤ n. Find n.

Sol. Let Sn = n

n 1 k

k 1

k.2 + −

=∑ =

nn 1 k

k 1

2 k.2+ −

=∑ = n 1

n n 11 n2 .2 1

2 2+

+ − −

(sum of the A.G.P.)

= 2[2n+1 – 2 – n]

⇒ n 1 24+

= ⇒ n = 7.

Q7. The area of the triangle formed by the intersection of a line parallel to x-axis and passing through

P (h, k) with the lines y = x and x + y = 2 is 4h2. Find the locus of the point P.

Sol. Area of triangle = 12

. AB. AC = 4h2

and AB = 2 |k – 1| = AC

⇒ 4h2 = 12

. 2. (k – 1)2

⇒ k – 1 = ± 2h. ⇒ locus is y = 2x + 1, y = – 2x + 1.

A(1,1)

P(h,k)B(k,k) C(2−k,k) y =k

x+y=2

y = x

O X

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Q8. Evaluate |cos x|

0

1 1e 2 sin cos x 3 cos cos x2 2

π +

∫ sin x dx.

Sol. I = |cos x|

0

1 1e 2 sin cos x 3 cos cos x2 2

π +

∫ sin x dx

= 6/ 2

cos x

0

1e sin x cos cos x dx2

π ∫ ( ) ( )

2aa

00

0, if f(2a x) f(x)

f x dx2 f x dx, if f(2a x) f(x)

− = − = − =

∫ ∫∵

Let cos x = t

I = 61

t

0

te cos dt2

∫

= 24 1 e 1ecos sin 15 2 2 2 + −

.

Q9. Incident ray is along the unit vector v and the reflected ray is

along the unit vector w . The normal is along unit vector a outwards. Express w in terms of a and v .

v

a w

Sol. v is unit vector along the incident ray and w

is the unit vector along the reflected ray. Hence a is a unit vector along the external bisector of v and w . Hence

ˆˆ ˆw v a− = λ ⇒ 1 + 1 – 2ˆ ˆw v⋅ = λ or 2 – 2 cos 2θ = λ2 or λ = 2 sin θ where 2θ is the angle between v and w .

a

v

w(90-θ)

2θmirror

Hence 0ˆ ˆ ˆ ˆˆ ˆ ˆw v 2sin a 2cos(90 )a (2a v)a− = θ = − θ = − ⋅ ⇒ ˆ ˆˆ ˆ ˆw v 2(a v)a= − ⋅ .

Q10. Tangents are drawn from any point on the hyperbola 22 yx 1

9 4− = to the circle x2 + y2 = 9. Find the

locus of mid–point of the chord of contact.

Sol. Any point on the hyperbola 22 yx 1

9 4− = is (3 secθ, 2 tanθ).

Chord of contact of the circle x2 + y2 = 9 with respect to the point (3 sec θ, 2tan θ) is 3 secθ.x + 2 tanθ.y = 9 ….(1) Let (x1, y1) be the mid–point of the chord of contact. ⇒ equation of chord in mid−point form is xx1 + yy1 = x1

2 + y12 ….(2)

Since (1) and (2) represent the same line,

2 2

1 1 1 1

3 sec 2 tan 9x y x y

θ θ= =

+

⇒ secθ = ( )

12 2

1 1

9x

3 x y+, tanθ =

( )1

2 21 1

9y

2 x y+

Hence ( ) ( )

2 21 1

2 22 2 2 21 1 1 1

81x 81y1

9 x y 4 x y− =

+ +

⇒ the required locus is 22 2 22 y x yx

9 4 9 +

− =

.

IIT JEE QUESTION PAPERS PAPER I2005-2015

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