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A particle of mass m moves with a variab...

A particle of mass `m` moves with a variable velocity v, which changes with distance covered x along a straight line as `v=ksqrtx`, where k is a positive constant. The work done by all the forces acting on the particle, during the first t seconds is

A

`(mk^(4))/(t^(2))`

B

`(mk^(4) t^(2))/(4)`

C

`(mk^(4) t^(2))/(8)`

D

`(mk^(4) t^(2))/(16)`

Text Solution

Verified by Experts

The correct Answer is:
C
Given `: v= k sqrt(x) ` or `(dx)/( dt) = k sqrtx` or `x^(1/2) dx=k dt`.
Integrating both sides, we get,
`(x^(1/2) )/(1/2) = kt+C`, assuming `x=0` at `t=0`,
Therefore, `C=0`.
So, `2 sqrt(x) = kt rArr x = ( k^(2) t^(2))/( 4) rArr v= (k^(2) t)/( 2)`
Therefore, work done, `Delta W=` Increase in KE
`=(1)/(2) mv^(2) - (1)/(2) m (0)^(2) = (1)/(2) m [(k^(2) t)/(2)]^(2) = (1)/(8) mk^(4) t^(2)`.
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