Q62 / 90 · JEE Main 2024 (08 Apr Shift 2) · Physics

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

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 $\begin{aligned} & 1 \mathrm{VSD}=\frac{\mathrm{x} \times 1 \mathrm{~mm}}{\mathrm{~N}} \\ & \therefore \frac{1}{20 \mathrm{~N}} \mathrm{~cm}=1 \mathrm{~mm}-\frac{\mathrm{x} \times 1 \mathrm{~mm}}{\mathrm{~N}} \\ & \frac{1}{2 \mathrm{~N}} \mathrm{~mm}=1 \mathrm{~mm}-\frac{\mathrm{x}}{\mathrm{N}} \mathrm{mm} \\ & \mathrm{x}=\left(1-\frac{1}{2 \mathrm{~N}}\right) \mathrm{N} \\ & \mathrm{x}=\frac{2 \mathrm{~N}-1}{2}\end{aligned}$

Hence, the correct answer is option: 'C'.

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