The dimensions of a cone are measured using a scale with a least count of $2 \mathrm{~mm}$. The diameter of…
Solution
$\mathrm{V}=\frac{1}{3} \pi\left(\frac{\mathrm{D}}{2}\right)^2 \mathrm{H}$
$\therefore \%$ Error in $\mathrm{V}=2$ (% error in $\mathrm{D})+\%$ error in $\mathrm{H}$.
$\because$ Least count is $2 \mathrm{~mm}$.
$\therefore \%$ error in D $=\frac{2 \mathrm{~mm}}{20 \mathrm{~cm}} \times 100 \%=1 \%$
& % error in $\mathrm{H}=\frac{2 \mathrm{~mm}}{20 \mathrm{~cm}} \times 100 \%=1 \%$
So $\%$ error in $\mathrm{V}=2 \times 1 \%+1 \%=3 \%$Asked in: JEE Advanced 2024 (Paper 2)