363164
A lift moving upwards with a velocity 10 \(m\)/\(s\) is stopped by uniform retardation in 5 \(s\). The apparent weight of a man of mass 60 \(kg\) before stopping will be (take \(g = 10\,m/{s^2}\))
1 72 \(kgwt\)
2 60 \(kgwt\)
3 48 \(kgwt\)
4 50 \(kgwt\)
Explanation:
The deceleration \(a = \frac{{10}}{5} = 2m/{s^2}\) The normal force or apparent weight is \(N = m\left( {g - a} \right)\) \(N = 60 \times 8\) \(R = \frac{N}{{10}} = 48kg\;wt\)
PHXI05:LAWS OF MOTION
363165
A person is standing in an elevator. In which situation, he finds himself weightless?
1 When the elevator moves upward with constant acceleration
2 When the elevator moves downward with constant acceleration
3 When the elevator moves upward with uniform velocity
4 When the elevator moves downward with uniform velocity.
Explanation:
When elevator is moving downward with constant acceleration a, then net weight of the person is given by \(w^{\prime}=m(g-a)=0 \quad(\text { at } a=g)\) i.e. the person finds him weightless
AIIMS - 2005
PHXI05:LAWS OF MOTION
363166
Assertion : The apparent weight of a body in an elevator moving with some downward acceleration is less than the actual weight of a body. Reason : Pseudo force acts on the body in upward direction which reduces its apparent weight.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The Apparent weight of a body in an elevator moving with downward acceleration ' \(a\) ' is given by \(W^{\prime}=m g-m a\). Pseudo force is responsible for the lesser apparent weight. So correct option is (1).
PHXI05:LAWS OF MOTION
363167
A \(60\;kg\) man stands on a spring scale in the lift. At some instant he finds, scale reading has changed from \(60\;kg\) to \(50\;kg\) for a while and then comes back to the original make. What should we conclude?
1 The lift was in constant motion upwards.
2 The lift was in constant motion downwards.
3 The lift while in constant motion upwards, is suddenly stopped.
4 The lift while in constant motion downwards is suddenly stopped.
Explanation:
For upward accelerations, apparent weight \( = m\,(g + a)\) If lift suddenly stops during upward motion, then apparent weight \( = m\,(g - a)\) because instead of acceleration, we will consider retardation. In the problem, it is given that scale reading initially was \(60\;kg\) and due to sudden jerk reading decreases and finally comes back to the original mark, \(i.\,e.\) \(60\;kg.\) So, we can conclude that lift was moving upwards with constant speed and suddenly stops.
363164
A lift moving upwards with a velocity 10 \(m\)/\(s\) is stopped by uniform retardation in 5 \(s\). The apparent weight of a man of mass 60 \(kg\) before stopping will be (take \(g = 10\,m/{s^2}\))
1 72 \(kgwt\)
2 60 \(kgwt\)
3 48 \(kgwt\)
4 50 \(kgwt\)
Explanation:
The deceleration \(a = \frac{{10}}{5} = 2m/{s^2}\) The normal force or apparent weight is \(N = m\left( {g - a} \right)\) \(N = 60 \times 8\) \(R = \frac{N}{{10}} = 48kg\;wt\)
PHXI05:LAWS OF MOTION
363165
A person is standing in an elevator. In which situation, he finds himself weightless?
1 When the elevator moves upward with constant acceleration
2 When the elevator moves downward with constant acceleration
3 When the elevator moves upward with uniform velocity
4 When the elevator moves downward with uniform velocity.
Explanation:
When elevator is moving downward with constant acceleration a, then net weight of the person is given by \(w^{\prime}=m(g-a)=0 \quad(\text { at } a=g)\) i.e. the person finds him weightless
AIIMS - 2005
PHXI05:LAWS OF MOTION
363166
Assertion : The apparent weight of a body in an elevator moving with some downward acceleration is less than the actual weight of a body. Reason : Pseudo force acts on the body in upward direction which reduces its apparent weight.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The Apparent weight of a body in an elevator moving with downward acceleration ' \(a\) ' is given by \(W^{\prime}=m g-m a\). Pseudo force is responsible for the lesser apparent weight. So correct option is (1).
PHXI05:LAWS OF MOTION
363167
A \(60\;kg\) man stands on a spring scale in the lift. At some instant he finds, scale reading has changed from \(60\;kg\) to \(50\;kg\) for a while and then comes back to the original make. What should we conclude?
1 The lift was in constant motion upwards.
2 The lift was in constant motion downwards.
3 The lift while in constant motion upwards, is suddenly stopped.
4 The lift while in constant motion downwards is suddenly stopped.
Explanation:
For upward accelerations, apparent weight \( = m\,(g + a)\) If lift suddenly stops during upward motion, then apparent weight \( = m\,(g - a)\) because instead of acceleration, we will consider retardation. In the problem, it is given that scale reading initially was \(60\;kg\) and due to sudden jerk reading decreases and finally comes back to the original mark, \(i.\,e.\) \(60\;kg.\) So, we can conclude that lift was moving upwards with constant speed and suddenly stops.
363164
A lift moving upwards with a velocity 10 \(m\)/\(s\) is stopped by uniform retardation in 5 \(s\). The apparent weight of a man of mass 60 \(kg\) before stopping will be (take \(g = 10\,m/{s^2}\))
1 72 \(kgwt\)
2 60 \(kgwt\)
3 48 \(kgwt\)
4 50 \(kgwt\)
Explanation:
The deceleration \(a = \frac{{10}}{5} = 2m/{s^2}\) The normal force or apparent weight is \(N = m\left( {g - a} \right)\) \(N = 60 \times 8\) \(R = \frac{N}{{10}} = 48kg\;wt\)
PHXI05:LAWS OF MOTION
363165
A person is standing in an elevator. In which situation, he finds himself weightless?
1 When the elevator moves upward with constant acceleration
2 When the elevator moves downward with constant acceleration
3 When the elevator moves upward with uniform velocity
4 When the elevator moves downward with uniform velocity.
Explanation:
When elevator is moving downward with constant acceleration a, then net weight of the person is given by \(w^{\prime}=m(g-a)=0 \quad(\text { at } a=g)\) i.e. the person finds him weightless
AIIMS - 2005
PHXI05:LAWS OF MOTION
363166
Assertion : The apparent weight of a body in an elevator moving with some downward acceleration is less than the actual weight of a body. Reason : Pseudo force acts on the body in upward direction which reduces its apparent weight.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The Apparent weight of a body in an elevator moving with downward acceleration ' \(a\) ' is given by \(W^{\prime}=m g-m a\). Pseudo force is responsible for the lesser apparent weight. So correct option is (1).
PHXI05:LAWS OF MOTION
363167
A \(60\;kg\) man stands on a spring scale in the lift. At some instant he finds, scale reading has changed from \(60\;kg\) to \(50\;kg\) for a while and then comes back to the original make. What should we conclude?
1 The lift was in constant motion upwards.
2 The lift was in constant motion downwards.
3 The lift while in constant motion upwards, is suddenly stopped.
4 The lift while in constant motion downwards is suddenly stopped.
Explanation:
For upward accelerations, apparent weight \( = m\,(g + a)\) If lift suddenly stops during upward motion, then apparent weight \( = m\,(g - a)\) because instead of acceleration, we will consider retardation. In the problem, it is given that scale reading initially was \(60\;kg\) and due to sudden jerk reading decreases and finally comes back to the original mark, \(i.\,e.\) \(60\;kg.\) So, we can conclude that lift was moving upwards with constant speed and suddenly stops.
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PHXI05:LAWS OF MOTION
363164
A lift moving upwards with a velocity 10 \(m\)/\(s\) is stopped by uniform retardation in 5 \(s\). The apparent weight of a man of mass 60 \(kg\) before stopping will be (take \(g = 10\,m/{s^2}\))
1 72 \(kgwt\)
2 60 \(kgwt\)
3 48 \(kgwt\)
4 50 \(kgwt\)
Explanation:
The deceleration \(a = \frac{{10}}{5} = 2m/{s^2}\) The normal force or apparent weight is \(N = m\left( {g - a} \right)\) \(N = 60 \times 8\) \(R = \frac{N}{{10}} = 48kg\;wt\)
PHXI05:LAWS OF MOTION
363165
A person is standing in an elevator. In which situation, he finds himself weightless?
1 When the elevator moves upward with constant acceleration
2 When the elevator moves downward with constant acceleration
3 When the elevator moves upward with uniform velocity
4 When the elevator moves downward with uniform velocity.
Explanation:
When elevator is moving downward with constant acceleration a, then net weight of the person is given by \(w^{\prime}=m(g-a)=0 \quad(\text { at } a=g)\) i.e. the person finds him weightless
AIIMS - 2005
PHXI05:LAWS OF MOTION
363166
Assertion : The apparent weight of a body in an elevator moving with some downward acceleration is less than the actual weight of a body. Reason : Pseudo force acts on the body in upward direction which reduces its apparent weight.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The Apparent weight of a body in an elevator moving with downward acceleration ' \(a\) ' is given by \(W^{\prime}=m g-m a\). Pseudo force is responsible for the lesser apparent weight. So correct option is (1).
PHXI05:LAWS OF MOTION
363167
A \(60\;kg\) man stands on a spring scale in the lift. At some instant he finds, scale reading has changed from \(60\;kg\) to \(50\;kg\) for a while and then comes back to the original make. What should we conclude?
1 The lift was in constant motion upwards.
2 The lift was in constant motion downwards.
3 The lift while in constant motion upwards, is suddenly stopped.
4 The lift while in constant motion downwards is suddenly stopped.
Explanation:
For upward accelerations, apparent weight \( = m\,(g + a)\) If lift suddenly stops during upward motion, then apparent weight \( = m\,(g - a)\) because instead of acceleration, we will consider retardation. In the problem, it is given that scale reading initially was \(60\;kg\) and due to sudden jerk reading decreases and finally comes back to the original mark, \(i.\,e.\) \(60\;kg.\) So, we can conclude that lift was moving upwards with constant speed and suddenly stops.
