Question 1

A light string passing over a smooth light pulley connects two blocks of masses m _1 and m _2 (where m_2 m_1). If the acceleration of the system is g 2 , then the ratio of the masses m_1 m_2 is:

Accepted Answer

Acceleration is given as: a= ( m_2-m_1 m_1+m_2 ) g g 2 = ( m _2- m _1 ~m _1+ m _2 ) g 2 (m_2-m_1 )=m_1+m_2 m _1 ~m _2 = ( 2 -1 2 +1 )

Question 2

A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 60^ by a force of 10 N parallel to the inclined surface as shown in figure. When the block is pushed up by 10 m along

Accepted Answer

Mass of block, m=1 ~kg Force of parallel inclined surface, F =10 ~N Work done against frictional force = _2 ~N 10= Mg 60 10=0.1 5 10=5 ~J

Question 3

A person of mass 60 kg is inside a lift of mass 940 kg and presses the button on control panel. The lift starts moving upwards with an acceleration 1.0 ~m / s ^2. If g =10 ~ms ^ -2 , the tension in th

Accepted Answer

Total mass =(60+940) kg =1000 ~kg Let T be the tension in the supporting cable, then T-1000 g=1000 1 T =1000 11=11000 ~N

Question 4

At any instant the velocity of a particle of mass 500 ~g is (2 t i +3 t^2 j ) ms ^ -1 . If the force acting on the particle at t=1 s is ( i +x j ) N. Then the value of x will be:

Accepted Answer

Mass of particle, m =500 ~g =0.5 ~kg velocity of a particle, v =2 t i +3 t^2 j a = d v dt =2 i +6 t j at t =1, a =2 i +6 j Force acting on the particle, F = ma =0.5(2 i +6 j )= i +3 j F = i +x j Hence x=3

Question 5

In the diagram shown below, both the strings A B and C D are made of same material and have same cross-section. The pulleys are light and fictionless. If the speed of wave in string A B is v_1 and in

Accepted Answer

Let the speed of wave in string be v and tension in string be T. gathered v T \ v_ A B v_ C D = T 2 T = 1 2 gathered

Question 6

What will be the acceleration due to gravity at a depth d, where g is acceleration due to gravity on the surface of earth?

Accepted Answer

Acceleration due to gravity at the surface of earth, g= G M R^2 = 4 3 G R [ M= 4 3 R^3 ] Where, = density of earth. Acceleration due to gravity at depth d from the surface of earth, g^ = 4 3 G(R-d) From Eqs. (i) and (ii), we get g^ =g [1- d R ]

Question 7

A particle of mass m is projected with velocity v at an angle with the horizontal. At its highest point, it explodes into two pieces of equal mass, one of the piece continue to move on the original tr

Accepted Answer

According to the law of conservation of linear momentum at the highest point, m v = m 2 (v )+ m 2 v^ m v - m 2 v = m 2 v^ v^ =2 (v - v 2 )=v

Question 8

A body of weight 2 ~kg is suspended as shown in the figure. The tension T _ 1 in the horizontal string (in kg wt ) is

Accepted Answer

The tension in the above string can be divided into two components where the component opposite to weight will balance the weight while the component of this tension opposite to $ T _ 1 $ will be equal to $ T _ 1 $. $ T =2 ~kg wt $ $ T = T _ 1 $ The first equation gives $ T =4 ~kg wt $ So $ T _ 1 $ will be, $ T _ 1 =2 3 ~kg wt $

Question 9

A ball of mass 0.2 ~kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 ~m while applying the force and the ball goes upto 2 ~m height further, find the magnitude of the

Accepted Answer

Let the velocity of the ball just when it leaves the hand is u then applying, array l v^ 2 -u^ 2 =2 a for upward journey \ -u^ 2 =2(-10) 2 \ u^ 2 =40 array Again applying v^ 2 -u^ 2 =2 a s for the upward journey of the ball, when the ball is in the hands of the thrower, array l v^ 2 -u^ 2 =2 a s \ 40-0=2(a) 0.2 \ a=100 ~m / s ^ 2 \ F =m a=0.2 100=20 ~N \ N -m g=20 \ N =20+2=22 ~N array

Laws of Motion — Previous Year Questions & Solutions