JEE Main 2025 — 28 January, Morning Shift — Question 67

Since least count of the instrument can be calculated as

Least count $=\frac{\text { pitch length }}{\text { No. of division on circular scale }}$

$=\frac{0.75}{15}=0.05 \mathrm{~mm}$.

Here we are provided $\mathrm{L}=5 \mathrm{~mm} \& \mathrm{~W}=2.5 \mathrm{~mm}$

$\mathrm{L}=5 \mathrm{~mm} \& \mathrm{~W}=2.5 \mathrm{~mm}$

$\because$ We know that

$\mathrm{A}=\mathrm{L} . \mathrm{W}$

For calculating fractional error, we can write

$\frac{\mathrm{dA}}{\mathrm{A}}=\frac{\mathrm{dL}}{\mathrm{L}}+\frac{\mathrm{dW}}{\mathrm{W}}$

Here $\mathrm{dL}=\mathrm{dW}=0.05 \mathrm{~mm}$

$\frac{\mathrm{dA}}{\mathrm{A}}=\frac{0.05}{5}+\frac{0.05}{2.5}$

$\Rightarrow \frac{\mathrm{dA}}{\mathrm{A}}=\frac{1}{100}+\frac{2}{100}=\frac{3}{100}$,

So, $x=3$

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