General Physics Ques 26
- The diameter of a cylinder is measured using a vernier calipers with no zero error. It is found that the zero of the vernier scale lies between $5.10 \mathrm{~cm}$ and $5.15 \mathrm{~cm}$ of the main scale. The vernier scale has 50 division equivalent to $2.45 \mathrm{~cm}$. The 24 th division of the vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is
(2013 Adv.)
(a) $5.112 \mathrm{~cm}$
(b) $5.124 \mathrm{~cm}$
(c) $5.136 \mathrm{~cm}$
(d) $5.148 \mathrm{~cm}$
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Answer:
Correct Answer: 26.( b )
Solution:
$\begin{aligned} & \text { 6. 1 MSD }=5.15 \mathrm{~cm}-5.10 \mathrm{~cm}=0.05 \mathrm{~cm} \\ & 1 \mathrm{VSD}=\frac{2.45 \mathrm{~cm}}{50}=0.049 \mathrm{~cm} \\ & \therefore \quad L C=1 \mathrm{MSD}-1 \mathrm{VSD}=0.01 \mathrm{~cm} \\ & \text { Hence, diameter of cyclinder }=(\text { Main scale reading }) \\ & +(\text { Vernier scale reading })(\mathrm{LC}) \\ & =5.10+(24)(0.001)=5.124 \mathrm{~cm}\end{aligned}$
BETA
