Question

If the error in measuring the radius of the sphere is $ 2\% $ and that in measuring its mass is $ 3\% $ , then the error in measuring the density of material of the sphere is

• A. $ 5\% $
• B. $ 7\% $
• C. $ 9\% $
• D. $ 11\% $

Answer 1

Answer: (C)

As density $=\frac{\text{mass}}{\text{Volume}}$
$\therefore \rho=\frac{M}{\frac{4}{3}\pi R^{3}}=\frac{3}{4}\frac{M}{\pi R^{3}}$
$\therefore $ The percentage error in density is
$\frac{\Delta\rho}{\rho}\times100\%=\left(\frac{\Delta M}{M}+3\frac{\Delta R}{R}\right)\times100\%$
$=3\%+3\left(2\%\right)=3\%+6\%=9\%$