Question

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. 25%
• B. 50%
• C. 100%
• D. 125%

Answer 1

Answer: (C)

100%

Answer 2

Kinetic energy $E = \frac{1}{2}mv^{2}$

$\therefore\frac{\Delta E}{E} \times 100 = \left( \frac{\Delta m}{m} + \frac{2\Delta v}{v} \right) \times 100$

Here$\Delta m = 0$ and $\frac{\Delta v}{v} \times 100 = 50\%$

$\therefore\frac{\Delta E}{E} \times 100 = 2 \times 50 = 100\%$