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(C) The force exerted on the wall is given by the rate of change of momentum.Initial momentum of $N$ balls per second = $Nmu \hat{i}$.Since the collis

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$2mNu$

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(C) The force exerted on the wall is given by the rate of change of momentum.Initial momentum of $N$ balls per second = $Nmu \hat{i}$.Since the collision is elastic,the balls return with the same speed in the opposite direction.Final momentum of $N$ balls per second = $-Nmu \hat{i}$.Change in momentum per second = $	ext{Final momentum} - 	ext{Initial momentum} = -Nmu \hat{i} - (Nmu \hat{i}) = -2Nmu \hat{i}$.The magnitude of the change in momentum per second is $2Nmu$.Since force is the rate of change of momentum,$F = \frac{\Delta p}{\Delta t} = \frac{2Nmu}{1} = 2Nmu \ N$.Therefore,the correct option is $C$.

$N$ number of balls,each of mass $m \ kg$,moving along the positive direction of the $x$-axis,strike a wall per second and return elastically. The velocity of each ball is $u \ m/s$. The force exerted on the wall by the balls in newtons is:

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