Question

Error in measurement of volume of a sphere is $9 \%$, then

• A. the error in measurement of radius is $2 \%$
• B. the error in measurement of radius is $3 \%$
• C. the error in measurement surface area is $4 \%$
• D. the error in measurement surface area is $6 \%$

Answer 1

Answer: (B), (D)

$\because V=\frac{4}{3} \pi R^{3}$
$\therefore \frac{\Delta V}{V}=3 \frac{\Delta R}{R}$
or $\frac{\Delta V}{V} \times 100=\frac{\Delta R}{R} \times 100$
or $9 \%=3 \frac{\Delta R}{R} \times 100$
$\therefore \frac{\Delta R}{R} \times 100=3 \%$
(4) $\because A=4 \pi R ^{3}$
$\therefore \frac{\Delta A}{A}=\frac{2 \Delta R}{R}$
$\therefore \frac{\Delta A}{A} \times 100=2 \frac{\Delta R}{R} \times 100$
$=2 \times 3=6 \%$