Q. The length of a cylinder is measured with a meter rod having least count 0.1 $cm.$ Its diameter is measured with vernier callipers having least count $0.01$ $cm$. Given that length is $5.0\ cm$. and radius is $2.0$ $cm$. The percentage error in the calculated value of the volume will be
A) 1%
B) 2%
C) 3%
D) 4%
Explanation
Volume of cylinder $V = \pi r^{2}l$ Percentage error in volume $\frac{\mathrm{\Delta}V}{V} \times 100 = \frac{2\mathrm{\Delta}r}{r} \times 100 + \frac{\mathrm{\Delta}l}{l} \times 100$ $= \left( 2 \times \begin{array}{r} \begin{array}{r} \frac{0.01}{2.0} \\ \ \end{array} \times 100 + \frac{0.1}{5.0} \\ \ \end{array} \times 100 \right) = (1 + 2)\% = 3\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.