JEE Main 2025 — Experimental Physics Question with Solution

Since least count of the instrument can be calculated as $\begin{aligned} & \text { Least count }=\frac{\text { pitch length }}{\text { No. of division on circular scale }} \\ & =\frac{0.75}{15}=0.05 \mathrm{~mm} . \end{aligned}$ 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 $\mathrm{dL}=\mathrm{dW}=0.05\mathrm{mm}$ $\begin{aligned} & \frac{\mathrm{dA}}{\mathrm{~A}}=\frac{0.05}{5}+\frac{0.05}{2.5} \\ & \Rightarrow \frac{\mathrm{dA}}{\mathrm{~A}}=\frac{1}{100}+\frac{2}{100}=\frac{3}{100} \end{aligned}$ So, $x=3$