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Two block of masses m(1) and m(2) are co...

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))`

Text Solution

Verified by Experts

The correct Answer is:
A

Let `a=` acceleration of `m_(1)`
then acceleration 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_(1)a`
`:. A=(m_(2)g)/(m_(1)+m_(2))`
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