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

The correct option is C $9\%$

As density = $\frac{mass}{volume}$
$\therefore \rho =\frac{M}{\frac{4}{3}\pi R^3}=\frac{3}{4}\frac{M}{4\pi R^3}$
$\therefore$ The percentage error in density is
$\frac{\Delta \rho }{\rho }\times 100$% $=(\frac{\Delta M}{M}+3\frac{\Delta R}{R})\times 100$%
=$3$% $+3(2$%$)=3$% $+6$% $=9$%