Physics · Units, Dimensions & Error Analysis

JEE Main 2024 — 4 April, Shift 1 — Question 34

In an experiment to measure focal length (f) of convex lens, the least counts of the measuring scales for the position of object ( $u$ ) and for the position of image ( $v$ ) are $\Delta u$ and $\Delta v$ , respectively. The error in the measurement of the focal length of the convex lens will be :

Option A:
$\frac{\Delta u}{u}+\frac{\Delta v}{v}$
Option B:
$\mathrm{f}^{2}\left[\frac{\Delta \mathrm{u}}{\mathrm{u}^{2}}+\frac{\Delta \mathrm{v}}{\mathrm{v}^{2}}\right]$
Correct
Option C:
$2 f\left[\frac{\Delta u}{u}+\frac{\Delta v}{v}\right]$
Option D:
$f\left[\frac{\Delta \mathrm{u}}{\mathrm{u}}+\frac{\Delta \mathrm{v}}{\mathrm{v}}\right]$

Answer: B

Step-by-step solution

$\mathrm{f}^{-1}=\mathrm{v}^{-1}-\mathrm{u}^{-1}$

$-f^{-2} d f=-v^{-2} d v-u^{-2} d u$

$\frac{\mathrm{df}}{\mathrm{f}^{2}}=\frac{\mathrm{dv}}{\mathrm{v}^{2}}+\frac{\mathrm{du}}{\mathrm{u}^{2}}$

$\mathrm{df}=\mathrm{f}^{2}\left[\frac{\mathrm{dv}}{\mathrm{v}^{2}}+\frac{\mathrm{du}}{\mathrm{u}^{2}}\right]$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 1
Subject: Physics
Chapter: Units, Dimensions & Error Analysis
Topic: Significant Figures and Error Analysis

The equation of stationary wave is : y=2 a sin (2 pi n t/lambda ) cos (2 pi x/lambda ) . Which of the following is NOT correct?

Answer: B

Step-by-step solution

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