JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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 \(............\).
- A \(1\)
- B \(3\)
- C \(4\)
- D \(2\)
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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