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A constant force F = m(2)g//2 is applied...

A constant force `F = m_(2)g//2` is applied on the block of mass `m_(1)` as shown. The string and the pulley are light and the surface of the table is smooth. Find the acceleration of `m_(1)`.

A

`(m_(2)g)/(2m_(1))`

B

`(m_(2)g)/(2(m_(1)+m_(2)))`

C

`(3 m_(2)g)/(2(m_(1)+m_(2)))`

D

`(3m_(2)g)/(2m_(1))`

Text Solution

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The correct Answer is:
B
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AAKASH SERIES-LAWS OF MOTION-EXERCISE - II
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