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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 in fig. The string and the pulley are light and the surface of the table is smooth. The acceleration of `m_(1)` is `:`

A

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

B

`( m_(2)g)/( 2(m_(1)-m_(2)))` towards left

C

`( m_(2)g)/( 2(m_(2)-m_(1)))` towards right

D

`( m_(2)g)/( 2(m_(2)-m_(1)))` towards left

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