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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 rope with a constant downward force `F =2mg`, where g is acceleration due to gravity The acceleration of mass m in case I is
.

A

zero

B

more than that is case II

C

less than that in case

D

equal to that in case II

Text Solution

Verified by Experts

The correct Answer is:
C
`F =ma, a =g ((m_(1)-m_(2))/(m_(1)+m_(2))),F -T =0`
and `T =2mg` also `T -mg = ma^(1)`Finally `a lt a^(1)` .
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