﻿ JEE Physics Solved Question Paper 2014 | Previous Year Papers | Zigya

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# JEE Physics Solved Question Paper 2014

#### Multiple Choice Questions

1.

A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?

• A meter scale.

• A vernier calliper where the 10 divisions in vernier scale match with 9 division in the main scale and main scale have 10 divisions in 1 cm.

• A screw gauge having 100 divisions in the circular scale and pitch as 1 mm.

• A screw gauge having 100 divisions in the circular scale and pitch as 1 mm.

B.

A vernier calliper where the 10 divisions in vernier scale match with 9 division in the main scale and main scale have 10 divisions in 1 cm.

If student measures 3.50 cm, it means that there is an uncertainty of order 0.01 cm
For vernier scale with 1 MSD = 1mm
and 9 MSD = 10 VSD
LC of vernier scale with 1 MSD - 1 VSD

450 Views

2.

From the tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle to hit the ground is n times that taken by it to reach the highest point of its path. The relation between H,u and n is

• 2gH = n2u2

• gH = (n-2)2u2

• 2gH = nu2(n-2)2u2

• gH = (n-2)2u2

C.

2gH = nu2(n-2)2u2

Time is taken to reach the maximum height t1 = u/g

If t2 is the time taken to hit the ground,
i.e,

740 Views

3.

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d1 and d2 are filled in the tube. Each liquid subtends 90o angle at centre. Radius joining their interface makes an angle α with vertical. ratio d1/d2 is

C.

1237 Views

4.

A block of mass m is placed on a surface with a vertical cross-section given by y = x3/6. If the coefficient of friction is 0.5, the maximum height above ground at which the block can be placed without slipping is

A.

A block of mass m is placed on a surface with vertical cross section, then

At limiting equilibrium, we get
μ = tan θ
0.5 = x2/2
⇒ x2 =1
⇒ x = ±1
Now, putting the value of x in y = x3/6, we get
When x =1

When x =-1

So, the maximum height above the ground at which the block can be placed without slipping is 1/6m.

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5.

A bob of mass m attached to an inextensible string of length l is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed ω rad/s about the vertical. About the point of suspension

• angular momentum is conserved

• angular momentum changes in magnitude but not in the direction

• angular momentum changes in direction but not in magnitude

• angular momentum changes both in direction and magnitude

C.

angular momentum changes in direction but not in magnitude

542 Views

6.

The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100°C is (For steel Young’s modulus is 2×1011 N m–2 and coefficient of thermal expansion is 1.1×10–5 K–1 )

• 2.2 x 108 Pa

• 2.2 x 109 Pa

• 2.2 x 107 Pa

• 2.2 x 106 Pa

A.

2.2 x 108 Pa

According to Hooke's law, i.e,

If the rod is compressed, then compressive stress and strain appear. Their ratio Y is same as that for the tensile case.
Give, length of a steel wire (L) = 10 cm,
Temperature (θ) = 100oC
As length is constant,
∴
Now, pressure = stress =y x strain
[Given, Y = 2 x 1011N/M2 and α = 1.1 x 10-5 K-1]
= 2 x 1011 x1.1 x 10-5 x 100
=2.2 x 108

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7.

Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

D.

Net force acting on any one particle M,
=

This force will equal to centripetal force
So,

658 Views

8.

On the heating water, bubbles being formed at the bottom of the vessel detach and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the vessel. If r < < R and the surface tension of water is T, value of r just before bubbles detach is (density of water is ρw)

A.

The bubble will detach if,
Buoyant force ≥ Surface tension force

526 Views

9.

A mass ‘m’ is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?

• 2g/3

• g/2

• 5g/6

• g

B.

g/2

For the mass m,
mg-T = ma

As we know, a = Rα    ... (i)
So, mg-T = mRα
T x R = mR2α ...... (ii)
From Eqs. (i) and (ii), we get
a = g/2
Hence, the acceleration of the mass of a body fall is g/2.

455 Views

10.

When a rubber band is strecthed by a distance x, it exerts a restoring force of magnitude F = ax +bx2, where a and b are constants. The work done in stretching are unstretched rubber-band by L is

• aL2 +bL2

C.

Given, F = ax +bx2
We know that work done in stretching the rubber band by L is
|dW|= |Fdx|
368 Views