The following simple questions may prompt you to pick out the wrong option if you are in a hurry. So, be cautious and don’t overlook basic points. Here are the questions:

**(1)**Two identical frictionless pulleys carry the same mass 2

*m*at the left ends of the light inextensible strings passing over them. The right end of the string carries a mass 3

*m*in the case of arrangement (i) where as a force of 3

*mg*is applied in the case of arrangement (ii) as shown in the adjoining figure. The ratio of the acceleration of mass 2

*m*in case (i) to the acceleration of mass 2

*m*in case (ii) is

(a) 2:3

(b) 1:1

(c) 5:3

(d) 3:5

(e)** **2:5

In both cases the net driving force is 3*mg *– 2*mg* =* mg*. But in case (i) the total mass moved is 5*m* where as in case (ii) the total mass moved is 2*m*. The acceleration in case (i) is *mg*/5*m* = *g*/5 where as the acceleration in case (ii) is *mg*/2*m* = *g*/2.

The ratio of accelerations = (*g*/5)/* *(*g*/2) = 2/5 [Option (e)].

**(2) **An object of mass 4 kg moving along a horizontal surface with an initial velocity of 2 ms^{–1} comes to rest after 4 seconds. If you want to keep it moving with the velocity of 2 ms^{–1}, the force required is

(a) zero

(b) 1 N

(c) 2 N

(d) 4 N

(e)** **8 N ** **

The acceleration ‘a’* *of the body is given by

0 = 2 + *a*×4, on using the equation, v_{t} = v_{0} + at

Therefore, a = – 0.5 ms^{–2}_{}

The body is retarded (as indicated by the negative sign) because of forces opposing the motion. The opposing force has magnitude ma = 4×0.5 = 2 N. Therefore, a force of 2 N has to be applied opposite to the opposing forces to keep the body moving with the velocity of 2 ms^{–1}.

## No comments:

## Post a Comment