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Least count of a vernier caliper is $\frac{1}{20 \mathrm{~N}} \mathrm{~cm}$. The value of one division on the main scale is $1 \mathrm{~mm}$. Then the number of divisions of main scale that coincide with $\mathrm{N}$ divisions of vernier scale is :
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The correct answer is:
$\left(\frac{2 \mathrm{~N}-1}{2}\right)$
Least count of vernier calipers $=\frac{1}{20 \mathrm{~N}} \mathrm{~cm}$
$\because$ Least count $=1 \mathrm{MSD}-1 \mathrm{VSD}$
let $\mathrm{x}$ no. of divisions of main scale coincides with $\mathrm{N}$ division of vernier scale, then
$\begin{aligned} & 1 \mathrm{VSD}=\frac{\mathrm{x} \times 1 \mathrm{~mm}}{\mathrm{~N}} \\ & \therefore \frac{1}{20 \mathrm{~N}} \mathrm{~cm}=1 \mathrm{~mm}-\frac{\mathrm{x} \times 1 \mathrm{~mm}}{\mathrm{~N}} \\ & \frac{1}{2 \mathrm{~N}} \mathrm{~mm}=1 \mathrm{~mm}-\frac{\mathrm{x}}{\mathrm{N}} \mathrm{mm} \\ & \mathrm{x}=\left(1-\frac{1}{2 \mathrm{~N}}\right) \mathrm{N} \\ & \mathrm{x}=\frac{2 \mathrm{~N}-1}{2}\end{aligned}$
$\because$ Least count $=1 \mathrm{MSD}-1 \mathrm{VSD}$
let $\mathrm{x}$ no. of divisions of main scale coincides with $\mathrm{N}$ division of vernier scale, then
$\begin{aligned} & 1 \mathrm{VSD}=\frac{\mathrm{x} \times 1 \mathrm{~mm}}{\mathrm{~N}} \\ & \therefore \frac{1}{20 \mathrm{~N}} \mathrm{~cm}=1 \mathrm{~mm}-\frac{\mathrm{x} \times 1 \mathrm{~mm}}{\mathrm{~N}} \\ & \frac{1}{2 \mathrm{~N}} \mathrm{~mm}=1 \mathrm{~mm}-\frac{\mathrm{x}}{\mathrm{N}} \mathrm{mm} \\ & \mathrm{x}=\left(1-\frac{1}{2 \mathrm{~N}}\right) \mathrm{N} \\ & \mathrm{x}=\frac{2 \mathrm{~N}-1}{2}\end{aligned}$
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