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The ratio of the weight of a man in a stationary lift and when it is moving downward with uniform acceleration 'a' is 3:2. The value of 'a' is (g-Acceleration due to gravity of the earth)
- a)3/2 g
- b)g/3
- c)2/3 g
- d)g
Correct answer is 'B'. Can you explain this answer?
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Verified Answer
The ratio of the weight of a man in a stationary lift and when it is m...
The weight in stationary lift = mg
The weight in the lift moving downwards with the acceleration 'a' = (mg-ma)
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The ratio of the weight of a man in a stationary lift and when it is m...
Given:
The ratio of the weight of a man in a stationary lift to when it is moving downward with uniform acceleration 'a' is 3:2.
To find:
The value of 'a'.
Explanation:
When the lift is stationary, the weight of the man is equal to his actual weight, which is given by the formula:
Weight = mass × acceleration due to gravity (W = mg)
Where,
W = Weight of the man
m = Mass of the man
g = Acceleration due to gravity
Let's assume the weight of the man when the lift is stationary is 3W.
Therefore, the mass of the man when the lift is stationary is given by:
3W = m × g
m = 3W/g
Now, let's consider the weight of the man when the lift is moving downward with uniform acceleration 'a'. In this case, the apparent weight of the man is given by:
Apparent Weight = Actual Weight - Inertial Force
The inertial force is given by the formula:
Inertial Force = mass × acceleration (F = ma)
The acceleration of the lift is downward, so the inertial force is in the upward direction.
Therefore, the apparent weight of the man is:
Apparent Weight = Weight - Inertial Force
2W = W - ma
Simplifying the equation, we get:
ma = W - 2W
ma = -W
a = -W/m
Substituting the value of m from earlier:
a = -W/(3W/g)
a = -g/3
Since acceleration cannot be negative, the magnitude of acceleration is:
|a| = g/3
Therefore, the value of 'a' is g/3.
The ratio of the weight of a man in a stationary lift to when it is moving downward with uniform acceleration 'a' is 3:2.
To find:
The value of 'a'.
Explanation:
When the lift is stationary, the weight of the man is equal to his actual weight, which is given by the formula:
Weight = mass × acceleration due to gravity (W = mg)
Where,
W = Weight of the man
m = Mass of the man
g = Acceleration due to gravity
Let's assume the weight of the man when the lift is stationary is 3W.
Therefore, the mass of the man when the lift is stationary is given by:
3W = m × g
m = 3W/g
Now, let's consider the weight of the man when the lift is moving downward with uniform acceleration 'a'. In this case, the apparent weight of the man is given by:
Apparent Weight = Actual Weight - Inertial Force
The inertial force is given by the formula:
Inertial Force = mass × acceleration (F = ma)
The acceleration of the lift is downward, so the inertial force is in the upward direction.
Therefore, the apparent weight of the man is:
Apparent Weight = Weight - Inertial Force
2W = W - ma
Simplifying the equation, we get:
ma = W - 2W
ma = -W
a = -W/m
Substituting the value of m from earlier:
a = -W/(3W/g)
a = -g/3
Since acceleration cannot be negative, the magnitude of acceleration is:
|a| = g/3
Therefore, the value of 'a' is g/3.
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The ratio of the weight of a man in a stationary lift and when it is moving downward with uniform acceleration 'a' is 3:2. The value of 'a' is (g-Acceleration due to gravity of the earth)a)3/2 gb)g/3c)2/3 gd)gCorrect answer is 'B'. Can you explain this answer?
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