Physics · Units, Dimensions & Error Analysis

JEE Main 2025 — 24 January, Evening Shift — Question 61

The energy E and momentum p of a moving body of mass $m$ are related by some equation. Given that c represents the speed of light, identify the correct equation.

Option A:
$\mathrm{E}^{2}=\mathrm{pc}^{2}+\mathrm{m}^{2} \mathrm{c}^{4}$
Option B:
$\mathrm{E}^{2}=\mathrm{pc}^{2}+\mathrm{m}^{2} \mathrm{c}^{2}$
Option C:
$E^{2}=p^{2} c^{2}+m^{2} c^{2}$
Option D:
$E^{2}=p^{2} c^{2}+m^{2} c^{4}$
Correct

Answer: D

Step-by-step solution

$[E]=M^{1} L^{2} T^{-2}$

$[\mathrm{Pc}]=\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-1} \cdot \mathrm{~L}^{1} \mathrm{~T}^{-1}=\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}$

$\left[\mathrm{mc}^{2}\right]=\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}$

$\mathrm{E}^{2}=\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}$

$\mathrm{E}^{2}=\mathrm{P}^{2} \mathrm{c}^{2}+\mathrm{m}^{2} \mathrm{c}^{4}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 24 January, Evening Shift
Subject: Physics
Chapter: Units, Dimensions & Error Analysis
Topic: Units and Dimensions Analysis