One main scale division of a vernier caliper is equal to m units. If n^ th division of main scale coincides with (n+1)^ th division of vernier scale, the least count of the vernier caliper is :

Correct answer: C. (m)/((n+1)) aligned & n M S D=(n+1) VSD & 1 VSD= n n+1 MSD & L · C=1 MSD-1 VSD & L · C=m-m( n n+1 ) & L · C=m( n+1-n n+1 ) & L · C=( m n+1 ) aligned

JEE Main 20249 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 9 April
Shift: Morning Shift
Subject: Physics
Chapter: Experimental Physics

JEE Main 2024 · 9 April · Morning Shift · Physics · Experimental Physics

One main scale division of a vernier caliper is equal to m units....

One main scale division of a vernier caliper is equal to

m

units. If

\mathrm{n}^{\text{th }}

division of main scale coincides with

(n+1)^{\text{th }}

division of vernier scale, the least count of the vernier caliper is :

Options

A $\frac{n}{(n+1)}$
B $\frac{1}{(n+1)}$
C $\frac{m}{(n+1)}$
D $\frac{m}{n(n+1)}$

Correct answer

C. $\frac{m}{(n+1)}$

Step-by-step solution

\begin{aligned} & \mathrm{n} M S D=(\mathrm{n}+1) \mathrm{VSD} \\ & \Rightarrow 1 \mathrm{VSD}=\frac{\mathrm{n}}{\mathrm{n}+1} \mathrm{MSD} \\ & \mathrm{L} \cdot \mathrm{C}=1 \mathrm{MSD}-1 \mathrm{VSD} \\ & \mathrm{L} \cdot \mathrm{C}=\mathrm{m}-\mathrm{m}\left(\frac{\mathrm{n}}{\mathrm{n}+1}\right) \\ & \mathrm{L} \cdot \mathrm{C}=\mathrm{m}\left(\frac{\mathrm{n}+1-\mathrm{n}}{\mathrm{n}+1}\right) \\ & \Rightarrow \mathrm{L} \cdot \mathrm{C}=\left(\frac{\mathrm{m}}{\mathrm{n}+1}\right)\end{aligned}

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