1.Units, Dimensions and Measurement
hard

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 :

A

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

B

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

C

$(2 \mathrm{~N}-1)$

D

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

(JEE MAIN-2024)

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

$1 \mathrm{VSD}=\frac{\mathrm{x} \times 1 \mathrm{~mm}}{\mathrm{~N}}$

$\therefore \frac{1}{20 \mathrm{~N}} \mathrm{~cm}=1 \mathrm{~mm}-\frac{\mathrm{x} \times 1 \mathrm{~mm}}{\mathrm{~N}}$

$\frac{1}{2 \mathrm{~N}} \mathrm{~mm}=1 \mathrm{~mm}-\frac{\mathrm{x}}{\mathrm{N}} \mathrm{mm}$

$\mathrm{x}=\left(1-\frac{1}{2 \mathrm{~N}}\right) \mathrm{N}$

$\mathrm{x}=\frac{2 \mathrm{~N}-1}{2}$

Standard 11
Physics

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