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Two bodies of masses m(1) and m(2) (m(2)...

Two bodies of masses `m_(1) and m_(2) (m_(2) gt m_(1))` are connected by a light inextensible string which passes through a smooth fixed pulley. The instantaneous power delivered by an external agent to pull `m_(1)` with constant velocity v is :

A

`(m_(2) - m_(1))g/v`

B

`(m_(2) - m_(1))v/g`

C

`(m_(2) - m_(1))gv`

D

`(m_(1) - m_(2))gv`

Text Solution

Verified by Experts

The correct Answer is:
`(3)`
`F_(ext) + m_(1)g = T & T = m_(2)g`
`Rightarrow F_(ext) = T - m_(1)g`
`Rightarrow F_(ext) = (m_(2) - m_(1))g`
`therefore P = F_(ext)V = (m_(2) - m_(1) ) gV`
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