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(C) The kinetic energy $(K.E.)$ of a body of mass $m$ moving with velocity $v$ is given by the formula: $K.E. = \frac{1}{2}mv^2$.From this relation,it

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$4$ times

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(C) The kinetic energy $(K.E.)$ of a body of mass $m$ moving with velocity $v$ is given by the formula: $K.E. = \frac{1}{2}mv^2$.From this relation,it is clear that $K.E. \propto v^2$.If the new velocity $v' = 2v$,then the new kinetic energy $K.E.'$ will be:$K.E.' = \frac{1}{2}m(v')^2 = \frac{1}{2}m(2v)^2 = \frac{1}{2}m(4v^2) = 4 	imes (\frac{1}{2}mv^2)$.Therefore,$K.E.' = 4 	imes K.E.$Thus,the kinetic energy becomes $4$ times the previous value.

If the velocity of a body is twice its previous velocity,then its kinetic energy will become:

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