1.

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


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