Q. If there is a positive error of 50% in the measurement of speed of a body, then the error in the measurement of kinetic energy is
A) 25%
B) 50%
C) 100%
D) 125%
Explanation
Kinetic energy, $E = \frac{1}{2}mv^{2}$ $\therefore\ \ \ \frac{\mathrm{\Delta}E}{E} \times 100 = \frac{v^{'2} - v^{2}}{v^{2}} \times 100$ $= \lbrack(1.5)^{2} - 1\rbrack \times 100$ $= 125\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.