Two' blocks of masses m, and m, are connected to the two ends of a string passing over a pulley as shown in figure. System hans vertically and blocks are held at rest. Assume that `m_(1) gt m_(2)`. Find the speed attained by the blocks when m, descends through a height h. Moment of inertia of the pulley is I and its radius is r. There is enough friction between string and pulley so that there is no slipping between them.

Options

A. `sqrt((2(m_(1)-m_(2))gh)/(m_(1)m_(2)+(I)/(r^(2))))`
B. `sqrt((2(m_(1)+m_(2))gh)/(m_(1)m_(2)+(I)/(r^(2))))`
C. `sqrt((2m_(1)m_(2)mgh)/(m_(1)+m_(2)+(I)/(r^(2))))`
D. `sqrt(((2m_(1)-m_(2))mgh)/(m_(1)+m_(2)+(I)/(r^(2))))`

Explanation

Since there is no slipping between string and the pulley hence there is no energy loss and total mechanical energy of the system remains conserved. Let v is the speed acquired by the blocks then angular speed of pulley must be v/r because there is no slipping between string and the pulley. For conservation of mechanical energy we can write the following:
Loss in gravitational potential energy of `m_(1)` = Gain in gravitational potential energy of `m_(2)`+ Gain in K.E. of the blocks + gain K.E. of pulley.
`impliesm_(1)gh=m_(2)gh+(1)/(2)(m_(1)+m_(2))v^(2)+(1)/(2)I((v)/(r))^(2)`
`impliesv=sqrt((2(m_(1)-m_(2))gh)/(m_(1)+m_(2)+(I)/(r^(2))))`
Hence option (a) is correct.