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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Motion in One Dimension

NTA Abhyas 2020 · Physics · Motion in One Dimension

A particle undergoes a one-dimensional motion such that its...

A particle undergoes a one-dimensional motion such that its velocity varies according to v(x)=\betax−2nv\left(x\right) = \betax^{- 2n}, where β\beta and n\text{n} are constants and x\text{x} is the position of the particle. The acceleration of the particle as a function of x\text{x}& is given by

Options

  1. A−2nβ2x−2n−1- 2n{β}^{2}x^{- 2n - 1}
  2. B−2nβ2x−4n−1- 2n{β}^{2}x^{- 4n - 1}
  3. C−2nβ2x−2n+1- 2n{β}^{2}x^{- 2n + 1}
  4. D−2nβ2x−4n+1- 2n{β}^{2}x^{- 4n + 1}

Correct answer

B. −2nβ2x−4n−1- 2n{β}^{2}x^{- 4n - 1}

Step-by-step solution

According to the question

a=vdvdx=v(ddx\betax−2n)<br/><br/>a = v\dfrac{dv}{dx} = v \left(\dfrac{d}{dx}\betax^{-2n}\right)<br /><br />= \left(-2n\betax^{-2n - 1}\right)\betax^{- 2n}<br/><br/><br /><br />= - 2n{\beta}^{2}x^{- 4n - 1}
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