Question

The length of cylinder is measured with a metre rod having least count $0.1\text{cm}$. Its diameter is measured with vernier caliper having least count $0.01\text{cm}$. Given that length is $5.0\text{cm}$ and diameter is $4.0\text{cm}$. The percentage error in the calculated value of the volume will be

• A. $1.5\%$
• B. $2.5\%$
• C. $3.5\%$
• D. $4\%$

Answer 1

The correct option is B $2.5\%$

Volume of cylinder,
$V=\pi r^{2}l=\pi \frac{d^2}{4}l.........\left(1\right)$
From the data given in the question,
Least count of metre rod $\Delta l=0.1\text{cm}$
Fractional Error in measuring length, $\frac{\Delta l}{l}=\frac{0.1}{5}$
$\Rightarrow \%$ error in measuring length $=\frac{\Delta l}{l}\times 100=\frac{0.1}{5}\times 100=2.0\%$
From the data given in the question,
Least count of vernier calipers $\Delta d=0.01\text{cm}$
Fractional Error in measuring diameter by vernier calipers, $\frac{\Delta d}{d}=\frac{0.01}{4}$
$\Rightarrow \%$ error in measuring diameter $=\frac{\Delta d}{d}\times 100=\frac{0.01}{4}\times 100=0.25\%$
Now, from $\left(1\right)$ we can write that,
$\frac{\Delta V}{V}=\frac{2\Delta d}{d}+\frac{\Delta l}{l}$
$\Rightarrow \frac{\Delta V}{V}\times 100=\frac{2\Delta d}{d}\times 100+\frac{\Delta l}{l}\times 100$
$\Rightarrow \%$ error in volume $=(2\times 0.25)+\left(2.0\right)=2.5\%$
Thus, option (b) is the correct answer.