Question

Length of a thin cylinder as measured by vernier callipers having least cound $0.01cm$ is $3.25cm$ and its radius of cross-section is measured by a screw gauge having least count $0.01mm$ as $2.75mm$. The percentage error in the measurement of volume of the cylinder will be

• A. $2\%$
• B. $3\%$
• C. $1\%$
• D. $1.5$

Answer 1

The correct option is B $3\%$

$\begin{array}{c}V=\pi r^{2}h\\ V=\pi \frac{d^2}{2^2}h=\frac{\pi d^{2}h}{4}\\ \Rightarrow V=\frac{\pi d^{2}h}{4}\end{array}$

$\begin{array}{c}\text{taking log;}\\ \Rightarrow \log y=\log \frac{\pi }{4}+2\log d+\log h\end{array}$

$\begin{array}{c}\text{differentiating;}\\ \Rightarrow \frac{dv}{V}=\frac{2d\left(d\right)}{d}+\frac{dh}{h}......\left(i\right)\\ \text{According to question,}\\ \begin{array}{cc}h=& 3.25cm\\ dh=0.01cm& \end{array}\\ r=2.75mm\\ dr=0.01mm\end{array}$

$\begin{array}{c}\text{from}\left(i\right)\\ \frac{dy}{V}=\frac{2d\left(2r\right)}{r}+\frac{dh}{h}\\ \frac{dy}{V}=4\frac{dr}{r}+\frac{dh}{h}=4(\frac{0.01}{2.75}+\frac{0.01}{3.25})\\ =0.03\end{array}$

$\begin{array}{cc}\text{\% error in}y& =0.03\times 100\%.\\ & =3\%.\end{array}$