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(D) The kinetic energy $E$ of a body is given by $E = \frac{1}{2}mv^2$.Since $m$ is constant,the relative change in kinetic energy is related to the v

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$125$

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(D) The kinetic energy $E$ of a body is given by $E = \frac{1}{2}mv^2$.Since $m$ is constant,the relative change in kinetic energy is related to the velocity $v$ by the relation $\frac{\Delta E}{E} \approx 2 \frac{\Delta v}{v}$.Given the percentage error in velocity is $\frac{\Delta v}{v} 	imes 100 = 50\%$,which means the new velocity $v' = v + 0.5v = 1.5v$.The new kinetic energy $E'$ is $E' = \frac{1}{2}m(v')^2 = \frac{1}{2}m(1.5v)^2 = 2.25 	imes (\frac{1}{2}mv^2) = 2.25E$.The percentage error in kinetic energy is $\frac{E' - E}{E} 	imes 100 = \frac{2.25E - E}{E} 	imes 100 = 1.25 	imes 100 = 125\%$.

If there is a positive error of $50\%$ in the measurement of velocity of a body,then the error in the measurement of kinetic energy is .............. $\%$

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