Experimental Physics

JEE Main level, single correct, medium

13 of 70

Question 13 of 70

Least count of a vernier caliper is 120 N cm\frac{1}{20 \mathrm{~N}} \mathrm{~cm}. The value of one division on the main scale is 1 mm1 \mathrm{~mm}. Then the number of divisions of main scale that coincide with N\mathrm{N} divisions of vernier scale is :

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Answer: (C) (2 N−12)\left(\frac{2 \mathrm{~N}-1}{2}\right)

Least count of vernier calipers =120 N cm=\frac{1}{20 \mathrm{~N}} \mathrm{~cm}
∵\because Least count =1MSD−1VSD=1 \mathrm{MSD}-1 \mathrm{VSD}
let x\mathrm{x} no. of divisions of main scale coincides with N\mathrm{N} division of vernier scale, then
1VSD=x×1 mm N∴120 N cm=1 mm−x×1 mm N12 N mm=1 mm−xNmmx=(1−12 N)Nx=2 N−12\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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