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A particle of mass ‘M’ is moved under a constant power P₀. At some instant after the start, its speed is u and at a later instant, the speed is 2v. Neglecting friction, distance travelled by the particle as its speed increases from v to 2v is :

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$\frac{12 {Mv}^{3}}{5 P_{0}}$

$\frac{9 {Mv}^{3}}{5 P_{0}}$

$\frac{8 P_{0} v^{2}}{M}$

$\frac{7 {Mv}^{3}}{3 P_{0}}$

answer is D.

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Detailed Solution

$Power P_{0} = Fv = M (v \frac{dv}{dx}) v \Rightarrow \int_{0}^{x} P_{0} dx = \int_{v}^{2 v} {Mv}^{2} dv$
$\Rightarrow P_{0} x = \frac{M}{3} [(2 v)^{3} - v^{3}] = \frac{7}{3} {Mv}^{3}$

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