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Q. The length of a cylinder is measured with a metre rod having least count $0.1\, cm$. Its diameter is measured with vernier calipers having least count $0.01 \,cm$. If the length and diameter of the cylinder are $5.0 \,cm$ and $2.00\, cm$, respectively, then the percentage error in the calculated value of volume will be

Physical World, Units and Measurements

Solution:

Volume of cylinder $(V)=\pi r^{2} l$
$\frac{\Delta V}{V} \times 100=\frac{\Delta r}{r} \times 100+\frac{\Delta l}{l} \times 100$
$=2 \times \frac{0.01}{2} \times 100+\frac{0.1}{5} \times 100$
$=1 \%+2 \%=3 \%$