JEE Main 2022 · 30 Jun · Shift 1 · Physics Q32

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PhysicsSingle correct+4 / -1Question 32 of 90

If n main scale divisions coincide with $(n+1)$ vernier scale divisions. The least count of vernier callipers, when each centimetre on the main scale is divided into five equal parts, will be :

\dfrac2{n+1}

\dfrac5{n+1}

\dfrac1{2n}

\dfrac1{5n}

Answer and worked solution

Official answer: A

Divisions. 1 MSD $=2$ mm and $n$ MSD $=(n+1)$ VSD, so 1 VSD $=\frac{2n}{n+1}$ mm.

Least count $2-\frac{2n}{n+1}=\frac2{n+1}$ mm.

(A)

Topics:Vernier callipers and screw gauge

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JEE Main 2022 · 30 Jun · Shift 1, Physics question 32