Question

The length of a cylinder is measured with a metre 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. $1.5\%$
• B. $2.5\%$
• C. $3\%$
• D. $4\%$

Answer 1

Answer: (C)

$3\%$

$\begin{array}{cc}\text{Given-}& \text{length of the cylinder}=5.0\pm 0.1cm\\ & \text{Radius of the cylinder}=2.0\pm 0.01cm\end{array}$

$\text{we know that volume of cylinder}V-$

$V=\pi r^2\cdot l$

$\begin{array}{c}\text{By error analysis -}\\ \text{Taking logarithm both sides in above eq-}\\ \log v=2\log r+\log l+\log \pi \end{array}$

$\begin{array}{c}\text{differntiating both side}\end{array}$

$\frac{dV}{V}=2\cdot \frac{dr}{r}+\frac{dl}{l}$

$\begin{array}{c}\Rightarrow \frac{dV}{V}=2\cdot \frac{\left(0.01\right)}{2}+\frac{0.1}{5}=0.03=3\%\\ \frac{dV}{V}=3\%\end{array}$