## A pumping machine pumps a liquid at a rate of 60 cc per minute at a pressure of 1.5 atmosphere

A pumping machine pumps a liquid at a rate of 60 cc per minute at a pressure of 1.5 atmosphere. The power of the machine is
(A) 9 watts
(B) 6 watts
(C ) 9 kW
(D) None of these

Solution : Power = F.v , where F = force imparted by the machine , v = velocity of the liquid
P = p.A.v, Where p = pressure & A = effective area
= p dV/dt
= (1.5×105 N/m2)(60 ×10-6 N/m2)
( 1 atm ≈ 105 N/m2 )
= 9 watts

## When a man walks on a horizontal surface with constant velocity, work done by

When a man walks on a horizontal surface with constant velocity, work done by

(A) Friction is zero

(B) Contact force is zero

(C) Gravity is zero

(D) Man is zero

Solution :  Since mg & N are perpendicular to velocity and , work done by these forces is zero.   Since no relative sliding occurs during walking, static friction comes into play.  Hence the point of application of static frictional force does not move.  Therefore no work is done by frictional force.

The man has to lose his body’s (internal) energy E , hence performs  work because   W = ΔE

## A body has zero velocity and still be accelerating ?

Q: Which statements can be possible cases in one/two dimensional motion:

(A) A body has zero velocity and still be accelerating

(B) The velocity of an object reverses direction when acceleration is constant

(C) An object be increasing in speed as its acceleration decreases

(D) None of these

Solution : (a) When the body is projected vertically up, at the highest point its speed becomes zero whereas it is accelerating downwards with g = 9.8 m/s2.

(b) When a body is projected up, the velocity during ascent reverses its direction during its descent, whereas the acceleration of the body remains constant, that is .

(c) Acceleration means, increasing speed (magnitude of velocity). Therefore, the speed of a particle increases with an acceleration. When the acceleration is decreased the speed of the particle goes on decreasing till the acceleration reduces to zero.

Ans:  (A) and (B)

## A particle is moving along a circular path of radius 5m and with uniform speed 5 m/s. What will be the average acceleration when the particle completes half revolution

A particle is moving along a circular path of radius 5m and with uniform speed 5 m/s.  What will be the average acceleration when the particle completes half revolution?

(A) zero
(B) 10 m/s2
(C) 10 π m/s2
(D) 10/π m/s2

Solution :  The change in velocity when the particle completes half revolution is given by
Dv = 5 m/s – (-5 m/s) = 10 m/s
Time taken to complete half revolution
t = π r/v = π×5/5 = π
Average acceleration  = Δv/Δt =10/π m/s2

## A driver applies the brakes on seeing traffic signal 400 m ahead. At the time of applying the brakes vehicle was moving with 15 ms-1 and retarding with 0.3 ms-2. The distance of vehicle after 1 min from the traffic light

A driver applies the brakes on seeing traffic signal 400 m ahead. At the time of applying the brakes vehicle was moving with 15 ms-1 and retarding with 0.3 ms-2. The distance of vehicle after 1 min from the traffic light:

(A)  25 m
(B)  375 m
(C)  360 m
(D)  40 m

Solution : The maximum distance covered by the vehicle before coming to rest
= v2/2a = (15)2/2×0.3 = 375 m
The corresponding time  = t = v/a = 15/0.3 = 50sec
Therefore after 50 seconds, the distance covered by the vehicle = 375 m from the instant of beginning of braking.
The distance of the vehicle from the traffic after one minute  = (400 – 375) m = 25 m

Ans: (A)