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JEE Physics Solved Question Paper 2009

Multiple Choice Questions

The number of significant figures in the numbers 4.8000 x 10⁴ and 48000.50 are respectively

5 and 6
5 and 7
2 and 7
2 and 6

A body of mass m moving along a straight line covers half the distance with a speed of 2ms^-1. The remaining half of the distance is covered in two equal time intervals with a speed of 3 ms^-1 and 5 ms^-1 respectively. The average speed of the particle for the entire journey is

$\frac{3}{8} {ms}^{- 1}$
$\frac{8}{3} {ms}^{- 1}$
$\frac{4}{3} {ms}^{- 1}$
$\frac{16}{3} {ms}^{- 1}$

The moment of inertia ofa circular ring of radius r and mass M about diameteris

$\frac{2}{5} {Mr}^{2}$
$\frac{{Mr}^{2}}{4}$
$\frac{{Mr}^{2}}{2}$
$\frac{{Mr}^{2}}{12}$

A body of mass 0.05 kg is observed to fall with an acceleration of 9.5 ms^-2. The opposing force of air on the body is (g = 9.8 ms^-2)

0.015 N
0.15 N
0.030 N
zero

Three concurrent co-planar forces 1 N, 2 N and 3 N acting along different directions on a body

can keep the body in equilibrium if 2 N and 3 N act at right angle
can keep the body in equilibrium if 1 N and 2 N act at right angle
cannot keep the body in equilibrium
can keep the body in equilibrium in 1 N and 3 N act at an acute angle

6.
Two rectangular blocks A and B of masses 2 kg and 3 kg respectively are connected by a spring of spring constant 10.8 Nm^-1 and are placed on a frictionless horizontal surface. The block A was given an initial velocity of 0.15 ms^-1. in the direction shown in the figure. The maximum compression of the spring during the motion is

0.01 m

0.02 m

0.05 m

0.03 m

0.05 m

As the block A moves with velocity 0.15 ms^-1, it compresses the spring which pushes B towards right. A goes on compressing the spring till the velocity acquired by B becomes equal to the velocity of A, ie, 0.15 ms^-1. Let this velocity be v. Now, spring is in a state of maximum compression. Let x be the maximum compression at this stage.

According to the law of conservation of linear momentum, we get

$m_{A} u = (m_{A} + m_{B}) v or v = \frac{m_{A} u}{m_{A} + m_{B}} = \frac{2 \times 0.15}{2 + 3} = 0.06 {ms}^{- 1}$

According to the law of conservation of energy.

$\frac{1}{2} m_{A} u^{2} = \frac{1}{2} (m_{A} + m_{B}) v^{2} + \frac{1}{2} {kx}^{2} \frac{1}{2} m_{A} u^{2} - \frac{1}{2} (m_{A} + m_{B}) v^{2} = \frac{1}{2} {kx}^{2} \frac{1}{2} \times 2 \times {(0.15)}^{2} - \frac{1}{2} (2 + 3) {(0.06)}^{2} = \frac{1}{2} {kx}^{2} 0.0225 - 0.009 = \frac{1}{2} {kx}^{2} or 0.0135 = \frac{1}{2} {kx}^{2} or x = \sqrt{\frac{0.027}{k}} = \sqrt{\frac{0.027}{10.8}} = 0.05 m$

The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of 20 m is

225 J
200 J
400 J
175 J

A door 1.6 m wide requires a force of 1 N to be applied at the free end to open or close it. The force that is required at a point 0.4 m distance from the hinges for opening or closing the door is