Question

If maximum percentage errors in measurement of length, mass and time are 2%,1% respectively then maximum percentage error in kinetic energy will be

• A. 1 %
• B. 4 %
• C. 7 %
• D. 5 %

Answer 1

The correct option is A 1 %

$kE=\frac{1}{2}m{v}^2=\frac{L}{t}$

For small change we have

$=\frac{\Delta m}{m}+2.\frac{\Delta v}{v}$

$=1+2\times 3$

$=7\%$