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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^(-2 n)`
where `beta` and `n` are constant and `x` is the position of the particle. The acceleration of the particle as a function of `x` is given by.

A

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

B

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

C

3) `-2nbeta^(2)x^(-2n+1)` 4) `-2nbeta^(2)e^(-4n+1)`

D

3) `-2nbeta^(2)x^(-2n+1)` 4) `-2nbeta^(2)e^(-4n+1)`

Text Solution

Verified by Experts

The correct Answer is:
B
We are given velocity of the particle
`v(oversetrightarrowx)=betax^(-2n)`
We know acceleration `a=v(dv)/(dx)`
`a=betax^(-2n)(d)/(dx)(betax^(-2)n)`
`beta^(2)x^(-2n) (-2n)x^(-2n-1)=2nbeta^(2)x^(-2n-1-2n)`
`a= 2nbeta^(2x^-4n-1)`
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