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A particle of unit mass undergoes one-di...

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v (x) = `beta x^(-2n)` , where `beta` and n are constants and x is the position of the particle. The acceleration of the particle as a function of x is given by

A

`-2nbeta^(2)x^(-2n-1)`

B

`-2nbeta^(2)x^(-4n-1)`

C

`-2nbeta^(2)x^(-2n+1)`

D

`-2nbeta^(2)e^(-4n+1)`

Text Solution

Verified by Experts

The correct Answer is:
B
Acceleration ,
`a(dv)/(dt)=(dv)/(dx).(dx)/(dt)=(dv)/(dx)v=v(dv)/(dx)`
`= betax^(-2n){(-2n)betax^(-2n-1)}=-2nbeta^(2)x^(-4n-1)`
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