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ISRO Mechanical Engineering Mechanics Questions, Answers and Explanation

ISRO Mechanical Engineering Mechanics Questions, Answers

 and Explanation

Kinematics of Motion

2006. 37. A person walks up a stalled escalator in 90 seconds. When standing on the same escalator, now moving, he is carried up in 60 seconds. How much time would it take him to walk up the moving escalator.
     a) 30 sec
     b) 36 sec
     c) 45 sec
     d) 54 sec

Answer
b) 36 sec

Explanation
Let the distance the person has to cover be d.
Let the velocity of the person be vm.
Let the velocity of the moving escalator be ve

When the escalator is stalled the person walks with velocity vm.
Use distance = velocity x time
d = vm x 90  ---- 1
When standing on the moving escalator the person moves with velocity ve.
d = ve x 60 ----- 2

Dividing 1 by 2 we get 1 = vm x 90 /(ve x 60)
vm = 2ve/3 ---- 3
Let the time taken for him to walk up the moving escalator be t seconds.
When walking on a moving escalator, the person moves with velocity = vm + ve
d = (vm + ve)xt
d = (2ve/3+ ve)xt, using 3
d = 5ve/3 x t
t = 3d/(5ve), but d/ve = 60 from 2

t = 3*60/5 = 36 seconds.

2014.18. Tension in the cable supporting a lift of weight 'W' and having an acceleration 'a' while going upward is (g is acceleration due to gravity)
     a) W(1+a/g)
     b) W(1-a/g)
     c) W(2+a/g)
     d) W(1+2a/g)

Answer
a) W(1+a/g)

Explanation
Tension increases when the lift accelerate upwards and decreases when the lift accelerate downwards.
Mass of the lift = W/g
Increase in tension due to acceleration = mass of list * acceleration of lift = W/g*a
Total tension = Tension due to gravity + Tension due to acceleration
 = W + W/g*a = W(1+g/a)

2012.8. An elevator weighing 1000 kg attains an upward velocity of 4m/sec in two seconds with uniform acceleration. The tension (in N) in supporting cable will be (g = 9.8 m/sec2)
a) 1204 N
b) 9800 N
c) 2000 N
d) 11800 N

Answer
d) 11800

Explanation
acceleration = change in velocity/time = 4/2 2 m/s2
Tension increases when the lift accelerate upwards and decreases when the lift accelerate downwards.
So tension in supporting cable = m(g+a) = 1000 (9.8+2) = 1000 x 11.8 = 11800 N

2007.1. Only rocket engines can be propelled to space because
a) They can generate very high thrust
b) They have high propulsion efficiency
c) These engines can work on several fuels
d) They are not air-breathing engines

Answer
d) They are not air-breathing engines

Explanation
Even though all the choices are true, for travel in space the engines must use their own oxidizer. So they should not be dependent on air

Linear Momentum


2017.2.52. A jet engine consumes 1kg of fuel for each 40kg of air intake. Fuel consumption is 1kg/sec. When aircraft travels in still air at 200m/sec, velocity of discharge of gases with respect to engine is 700m/sec. The power developed by engine is
a) 7200kW
b) 5600kW
c) 2070kW
d)4140kW


Answer
d) 4140kW

Explanation
Power developed = Thrust* velocity
A jet engine derives its thrust from compressing incoming air and then mix it with fuel to burn it and then exhausting the hot gases at high velocities through a nozzle.
Thrust = Rate of change of momentum of incoming gases and exhaust gases
Inlet momentum = 40*200 kgm/s
Outlet momentum = (40+1)*700 kgm/s
Power developed = (41*700 - 40*200)*200 = 4140kW

Collision

2017.2.3. Which of the following is NOT true?
a) For direct impact of two bodies, coefficient of restitution is the ratio of relative velocity of approach of the two bodies to their relative velocity of separation.
b) Conservation of angular momentum implies that total angular momentum of a system remains constant unless acted on by an external torque
c) Conservation of linear momentum in a given direction implies that the sum of external forces in that direction is zero.
d) The coefficient of friction is independent of area of contact.

Answer
a) For direct impact of two bodies, coefficient of restitution is the ratio of relative velocity of approach of the two bodies to their relative velocity of separation.

Explanation
Actually, for direct impact of two bodies, coefficient of restitution is the ratio of relative velocity of separation to the their relative velocity of approach. Other options are true
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