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

Correct answer: C. ( 2 ~N-1 2 ) Least count of vernier calipers = 1 20 ~N ~cm Least count =1 MSD-1 VSD let x no. of divisions of main scale coincides with N division of vernier scale, then aligned & 1 VSD= x × 1 ~mm ~N & 1 20 ~N ~cm=1 ~mm- x × 1 ~mm ~N & 1 2 ~N ~mm=1 ~mm- x N mm & x=(1- 1 2 ~N ) N & x= 2 ~N-1 2 aligned

JEE Main 20248 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 8 April
Shift: Evening Shift
Subject: Physics
Chapter: Experimental Physics

JEE Main 2024 · 8 April · Evening Shift · Physics · Experimental Physics

Least count of a vernier caliper is 1 20 ~N ~cm . The value of one...

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 :

Options

A $(2 \mathrm{~N}-1)$
B $\left(\frac{2 \mathrm{~N}-1}{2 \mathrm{~N}}\right)$
C $\left(\frac{2 \mathrm{~N}-1}{2}\right)$
D $\left(\frac{2 \mathrm{~N}-1}{20 \mathrm{~N}}\right)$

Correct answer

C. $\left(\frac{2 \mathrm{~N}-1}{2}\right)$

Step-by-step solution

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}

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