Question

The length of a cylinder is measured with a metre rod having least count $0.1cm$. Its diameter is measured with vernier callipers having least count $0.01cm$. Given that length is $5.0cm$ and radius is $2.00cm$. The percentage error in the calculate value of the volume will be

• A. $11\%$
• B. $2.5\%$
• C. $3\%$
• D. $4\%$

Answer 1

The correct option is B $2.5\%$

$\begin{array}{cc}\text{Given}\to & \text{least count of metre rod}=0.1cm\\ & \text{Diameter measured with vernier caliper}\\ & \text{of least count}=0.01cm\\ & \text{Length of}\mathrm{Rod}=5cm\\ & \text{Radius}=2cm\Rightarrow \text{Diameter}=4cm\end{array}$

$\begin{array}{c}\text{we know volume of cylinder}\to \\ v=\pi r^2\cdot l=\frac{\pi d^2\cdot l}{4}-\left(1\right)\end{array}$

$\begin{array}{c}\text{Take logarithm both sides in eq}1\\ \log v=2\cdot \log \left(d\right)+\log \left(l\right)+\log \left(\pi /4\right)\end{array}$

$\begin{array}{c}\text{Differentiating Both Sides}\\ \frac{dv}{v}=\frac{2\cdot d\left(d\right)}{d}+\frac{dl}{l}\end{array}$

$\begin{array}{c}\frac{dv}{v}=\frac{2\cdot (0,01)}{4}+\frac{0.1}{5}=0.025\\ \%\text{error}=2.5\%\end{array}$