• # question_answer If the percentage error in the measurement of radius of a sphere is 2%, then the maximum percentage errors in the measurements of its diameter, surface area and volume are respectively A)  2%, 4% and 4 %B)  6%, 4% and 2%C)  6%, 6% and 6%D)  2%, 4% and 6%

$\because$  $D=2R$$\therefore$ $\frac{\Delta D}{D}\times 100=\frac{\Delta R}{R}\times 100=2%$ $\because$$A=4\pi {{R}^{2}}$$\therefore$$\frac{\Delta A}{A}\times 100=2\frac{\Delta R}{R}\times 100=4%$ $\because$$V=\frac{4}{3}\pi {{R}^{3}}$$\therefore$$\frac{\Delta V}{V}\times 100=3\frac{\Delta R}{R}\times 100=6%$