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Q.

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to υ(x) = βx-2nwhere β 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

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a

-2nβ2x-2n-1

b

-2nβ2x-4n-1

c

-2nβ2x-2n+1

d

-2nβ2e-4n+1

answer is B.

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Detailed Solution

We are given velocity of the particleυ(x) = βx-2nWe know acceleration a = υdυdxa = βx-2nddx(βx-2n)= β2x-2n(-2n)x-2n-1 = -2nβ2x-2n-1-2na = -2nβ2x-4n-1

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