A particle moves in a straight line with retardation proportional to square of its displacement. Its loss of kinetic energy for any displacement ‘x’ is proportional to

x^{2}

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x^{3}

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l o g_{e} x

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e^{x}

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Solution

The correct option is B $x^{3}$
$\frac{d v}{d t} = V \frac{d v}{d x} = - K x^{2}$
$\Rightarrow V d v = - K x^{2} d x$
Let $V_{1}$ and $V_{2}$ be the velocities of the particle at locations $x_{1}$ and $x_{2}$ respectively. Then
$\frac{V_{2}^{2} - V_{1}^{2}}{2} = - K [\frac{x_{2}^{3} - x_{1}^{3}}{3}]$
So, loss in K.E $α x^{3}$

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A particle moves in a straight line with retardation proportional to square of its displacement. Its loss of kinetic energy for any displacement ‘x’ is proportional to

The correct option is B x3dvdt=Vdvdx=−Kx2 ⇒Vdv=−Kx2dx Let V1 and V2 be the velocities of the particle at locations x1 and x2 respectively. Then V22−V212=−K[x32−x313] So, loss in K.E α x3