• # question_answer 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: A)  $15%$                       B)  50%C)  $100%$                    D)  125%

Kinetic energy $\varepsilon =\frac{1}{2}m{{\upsilon }^{2}}$ $\frac{\Delta \varepsilon }{\varepsilon }\times 100=\frac{V{{'}^{2}}-{{V}^{2}}}{{{V}^{2}}}\times 100$ $=[{{(1.5)}^{2}}-1]\times 100$ $\frac{\Delta E}{E}\times 100=125%$