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Standard XII

Physics

Variation of G Due to Height and Depth

The pulley ar...

Question

The pulley arrangements shown in figure are identical, the mass of the rope being negligible. In case I, the mass m is lifted by attaching a mass 2m to the other end of the rope. In case II, the mass m is lifted by pulling the other end of the rope with a constant downward force F = 2mg, where g is acceleration due to gravity. The acceleration of mass in case I is:

Zero

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More then that in case II

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Less than that in case II

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Equal to that in case II

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Solution

The correct option is B Less than that in case II
Case-1 :
Acceleration of mass $m$ is, $a_{1} = \frac{2 m g - m g}{m + 2 m} = \frac{g}{3}$

Case-2:
Acceleration of mass $m$ is, $a_{2} = \frac{F}{m} = \frac{2 m g}{m} = 2 g$

So, $a_{1} < a_{2}$
Option C is correct.

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The correct option is B Less than that in case IICase-1 :Acceleration of mass m is, a1=2mg−mgm+2m=g3Case-2:Acceleration of mass m is, a2=Fm=2mgm=2gSo, a1<a2Option C is correct.

The correct option is B Less than that in case II
Case-1 :
Acceleration of mass $m$ is, $a_{1} = \frac{2 m g - m g}{m + 2 m} = \frac{g}{3}$

Case-2:
Acceleration of mass $m$ is, $a_{2} = \frac{F}{m} = \frac{2 m g}{m} = 2 g$

So, $a_{1} < a_{2}$
Option C is correct.