h-001/9d7685f37b92.webp" data-orsrc="https://uploads-pracbee-ramanuj.s3.ap-south-1.amazonaws.com/pyq-question-images/batch-001/9d7685f37b92.webp" loading="lazy" style="max-width:100%;height:auto;display:block;margin:0 auto;max-height:40vh;vertical-align:baseline" alt="JEE Main 2025 (Online) 28th January Morning Shift Physics - Units & Measurements Question 43 English Explanation">

Given, L = 5 mm, B = 2.5 mm

We know, least count of a screw gauge,

$$L.C. = {\text{Pitch length} \over \text{No. of division on circular scale}}$$

$$ \Rightarrow L.C. = {{0.75} \over {15}} = 0.05\,mm$$

We know, $$A = LB$$

By taking $ln$ on both sides,

$$ \Rightarrow \ln A = \ln L + \ln B$$

by differentiating both sides,

$$ \Rightarrow {{dA} \over A} = {{dL} \over L} + {{dB} \over B}$$

Here, $$dL = dB = 0.05\,mm$$ (L.C.)

So fractional error,

$${{dA} \over A} = {{0.05} \over 5} + {{0.05} \over {2.5}}$$

$$ = {1 \over {100}} + {2 \over {100}} = {3 \over {100}} = {x \over {100}}$$ (given)

Hence, $$x = 3$$.

19:T5a8,

JEE Main 2025 (Online) 28th January Morning Shift Physics - Units & Measurements Question 43 Hindi Explanation

दिया है, L = 5 mm, B = 2.5 mm

हम जानते हैं, स्क्रू गेज का अल्पतमांक,

$$L.C. = {\text{Pitch length} \over \text{No. of division on circular scale}}$$

$$ \Rightarrow L.C. = {{0.75} \over {15}} = 0.05\,mm$$

हम जानते हैं, $$A = LB$$

दोनों पक्षों पर $ln$ लेने पर,

$$ \Rightarrow \ln A = \ln L

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