Exercise - 1
1
🧠 Concept
Friction
Which one of the following statements is incorrect?
2018
A
Rolling friction is smaller than sliding friction.
B
Limiting value of static friction is directly proportional to normal reaction.
C
Frictional force opposes the relative motion.
D
Coefficient of sliding friction has dimensions of length.
✅ Show Answer
✔ Correct Answer:
D
(Coefficient of sliding friction has dimensions of length.)
💡 Explanation
The coefficient of sliding friction is the ratio of frictional force to normal reaction. Since both are forces, their ratio is dimensionless. Therefore, it does not have the dimensions of length.
🎯 Conclusion
Statement D is incorrect because the coefficient of sliding friction is dimensionless.
2
🧠 Concept
Apparent Weight
If rope of lift breaks suddenly, the tension exerted by the surface of lift (a = acceleration of lift):
2002
A
mg
B
m(g + a)
C
m(g - a)
D
0
✅ Show Answer
✔ Correct Answer:
D
(0)
💡 Explanation
When the rope of the lift breaks, the lift and the person inside it fall freely under gravity with acceleration g. Therefore, the normal reaction exerted by the surface of the lift becomes zero.
🎯 Conclusion
The tension/normal reaction exerted by the surface of the lift is zero.
3
🧠 Concept
Newton’s Third Law
Rocket engines lift a rocket from the earth surface because hot gas with high velocity:
1998
A
Push against the earth
B
Push against the air
C
React against the rocket and push it up
D
Heat up the air which lifts the rocket
✅ Show Answer
✔ Correct Answer:
C
(React against the rocket and push it up)
💡 Explanation
Rocket engines eject hot gases downward at high velocity. According to Newton’s third law, the gases exert an equal and opposite force on the rocket, producing an upward thrust.
🎯 Conclusion
Therefore, rocket engines lift a rocket because the hot gases react against the rocket and push it upward.
4
🧠 Concept
Newton’s Third Law
A bird weighs 2 kg and is inside a closed cage of 1 kg. If it starts flying, then what is the weight of the bird and cage assembly:
1997
A
1.5 kg
B
2.5 kg
C
3 kg
D
4 kg
✅ Show Answer
✔ Correct Answer:
C
(3 kg)
💡 Explanation
When the bird starts flying inside the closed cage, it pushes air downward. The air exerts an equal and opposite force on the bird, and this force is ultimately transmitted to the cage. Therefore, the total weight of the bird and cage assembly remains unchanged.
🎯 Conclusion
The weight of the bird and cage assembly remains 3 kg.
5
🧠 Concept
Inertia
Newton's first law of motion describes the following:
1996
A
Energy
B
Work
C
Inertia
D
Moment of inertia
✅ Show Answer
✔ Correct Answer:
C
(Inertia)
💡 Explanation
Newton's first law states that a body remains at rest or continues to move with uniform velocity in a straight line unless acted upon by an external unbalanced force. The tendency of a body to resist any change in its state of rest or motion is called inertia.
🎯 Conclusion
Therefore, Newton's first law of motion describes the concept of inertia.
6
🧠 Concept
Apparent Weight
A man is standing at a spring platform. Reading of spring balance is 60 kgwt. If man jumps outside platform, then reading of spring balance:
1996
A
First increases then decreases to zero
B
Decreases
C
Increases
D
Remains same
✅ Show Answer
✔ Correct Answer:
A
(First increases then decreases to zero)
💡 Explanation
Initially, the spring balance reads the man's weight. When the man jumps, he pushes downward on the platform, producing an additional downward force and hence the reading of the spring balance first increases. As the man leaves the platform, he no longer exerts any force on it, so the reading decreases to zero.
🎯 Conclusion
Therefore, the spring balance reading first increases and then decreases to zero.
7
🧠 Concept
Newton’s Second Law
A 10 N force is applied on a body to produce in it an acceleration of 1 m/s². The mass of the body is:
1996
A
15 kg
B
20 kg
C
10 kg
D
5 kg
✅ Show Answer
✔ Correct Answer:
C
(10 kg)
💡 Explanation
According to Newton’s second law, F = ma. Therefore, m = F/a = 10/1 = 10 kg.
🎯 Conclusion
Therefore, the mass of the body is 10 kg.
8
🧠 Concept
Rocket Propulsion
In a rocket, fuel burns at the rate of 1 kg/s. This fuel is ejected from the rocket with a velocity of 60 km/s. This exerts a force on the rocket equal to:
1994
A
6,000 N
B
60,000 N
C
60 N
D
600 N
✅ Show Answer
✔ Correct Answer:
B
(60,000 N)
💡 Explanation
The thrust force is given by F = ṁv. Here, ṁ = 1 kg/s and v = 60 km/s = 60,000 m/s. Therefore, F = 1 × 60,000 = 60,000 N.
🎯 Conclusion
Therefore, the force exerted on the rocket is 60,000 N.
9
🧠 Concept
Rocket Propulsion
A 600 kg rocket is set for a vertical firing. If the exhaust speed is 1,000 m s⁻¹, the mass of the gas ejected per second to supply the thrust needed to overcome the weight of rocket is:
1990
A
117.6 kg s⁻¹
B
58.6 kg s⁻¹
C
6 kg s⁻¹
D
76.4 kg s⁻¹
✅ Show Answer
✔ Correct Answer:
C
(6 kg s⁻¹)
💡 Explanation
The thrust required to just overcome the weight of the rocket is mg. Therefore, ṁv = mg, giving ṁ = mg/v = (600 × 9.8)/1000 = 5.88 kg s⁻¹, approximately 6 kg s⁻¹.
🎯 Conclusion
Therefore, the mass of gas ejected per second is approximately 6 kg s⁻¹.
10
🧠 Concept
Impulse and Momentum
A ball of mass 0.5 kg moving with a velocity of 2 m/sec strikes a wall normally and bounces back with the same speed. If the time of contact between the ball and the wall is one millisecond, the average force exerted by the wall on the ball is:
1990
A
2,000 N
B
1,000 N
C
5,000 N
D
125 N
✅ Show Answer
✔ Correct Answer:
A
(2,000 N)
💡 Explanation
The change in momentum is Δp = m(v - u) = 0.5(-2 - 2) = -2 kg m s⁻¹. The magnitude of average force is F = Δp/Δt = 2/0.001 = 2000 N.
🎯 Conclusion
Therefore, the average force exerted by the wall on the ball is 2,000 N.