General Physics Question 19
Question 19
- The pitch of a screw gauge is $1 \mathrm{~mm}$ and there are 100 divisions on the circular scale. While measuring the diameter of a wire, the linear scale reads $1 \mathrm{~mm}$ and 47th division on the circular scale coincides with the reference line. The length of the wire is $5.6 \mathrm{~cm}$. Find the curved surface area $\left(\right.$ in $^{2}$ ) of the wire in appropriate number of significant figures.
(2004, 2M)
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Answer:
Correct Answer: 20. $\left(2.6 \mathrm{~cm}^{2}\right)$
Solution:
- Least count of screw gauge $=\frac{1 \mathrm{~mm}}{100}=0.01 \mathrm{~mm}$
Diameter of wire $=(1+47 \times 0.01) \mathrm{mm}=1.47 \mathrm{~mm}$
Curved surface area $\left(\right.$ in $\left.\mathrm{cm}^{2}\right)=(2 \pi) \frac{d}{2}(L)$
or
$$ \begin{aligned} S & =\pi d L=(\pi)\left(1.47 \times 10^{-1}\right)(5.6) \mathrm{cm}^{2} \ & =2.5848 \mathrm{~cm}^{2} \end{aligned} $$
Rounding off to two significant digits
$$ S=2.6 \mathrm{~cm}^{2} $$
