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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2023
  • Hard
  • Numerical

A particle of mass \(10\,g\) moves in a straight line with retarcation \(2x\), where \(x\) is the displacement in \(SI\) units. Its loss of kinetic energy for above displacement is \(\left(\frac{10}{x}\right)^{- n }\, J\). The value of \(n\) will be \(............\).

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Answer & Solution

Correct Answer

(D) \(2\)

Step-by-step Solution

Detailed explanation

Loss of K.E = work done against retarding force .

\(=\int \limits _0^{ x } madx =\int \limits_0^{ x } m 2 xdx = mx ^2\)

\(=\left(10^{-2} kg \right) x ^2 J =\left(\frac{10}{ x }\right)^{-2}\,J\)

So \(n=2\)

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