Two block of masses `m_(1)` and `m_(2)` are connected as shown in the figure. The acceleration of the block `m_(2)` is:

Options

A. `(m_(2)g)/(m_(1)+m_(2))`
B. `(m_(1)g)/(m_(1)+m_(2))`
C. `(4m_(2)g-m_(1)g)/(m_(1)+m_(2))`
D. `(m_(2)g)/(m_(1)+4m_(2))`

Answer

A

Explanation

Let `a=acc^(n)` of `m_(1)` , `acc^(n)` of pulley `=(a+0)/(2)=(a)/(2)`
If acceleration of `m_(2)=b`
Then `0+(b)/(2)=(a)/(2)`
Hence `a=b`
`T=m_(1)` `a` , `m_(2)g-T=m_(2)a`
:. `a(m_(2)g)/(m_(1)+m_(2))` .