If there is a positive error of 50% in the me | vedclass

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Question

(d) Kinetic energy $E = \frac{1}{2}m{v^2}$

$	herefore \,\,\,\frac{{\Delta E}}{E} 	imes 100 = \frac{{{{v'}^2} - {v^2}}}{{{v^2}}} 	imes 100$

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(d) Kinetic energy $E = \frac{1}{2}m{v^2}$

$	herefore \,\,\,\frac{{\Delta E}}{E} 	imes 100 = \frac{{{{v'}^2} - {v^2}}}{{{v^2}}} 	imes 100$

$ = [{(1.5)^2} - 1] 	imes 100$

$	herefore \,\,\frac{{\Delta E}}{E} 	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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