Question

The length of a cylinder is measured with a meter rod having least count $0.1cm$ . Its diameter is measured with Vernier calipers having least count $0.01cm$ . Given that length is $5.0$ $cm$ and radius is $2.0cm$ . What will be the percentage error in the calculated value of the volume ?

• A. $1\%$
• B. $2\%$
• C. $3\%$
• D. $4\%$

Answer 1

Answer: (C)

Assuming $r$ as the radius of cylinder , $d$ as the diameter and $l$ as its length
since $d=2r\Rightarrow \Delta r=\Delta d$
since least count is also the maximum permissible error
$\Rightarrow \frac{\Delta r}{r}=\frac{0.01}{2.0};\frac{\Delta l}{l}=\frac{0.1}{5.0}$
Volume of cylinder $V=\pi r^{2}l$
Percentage error in volume
$\Rightarrow \frac{\Delta V}{V}\times 100=\frac{2\Delta r}{r}\times 100+\frac{\Delta l}{l}\times 100$
$=\left(2\times \frac{0.01}{2.0}\times 100+\frac{0.1}{5.0}\times 100\right)=\left(1+2\right)\%=3\%$