JEE Main 2024 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 9 April 2024, Shift 1, Physics Q35

Question 35 of 90 in this shift, Physics question 5 of 30, Section A.

A particle of mass m moves on a straight line with its velocity increasing with distance according to the equation

v=\alpha\sqrt{x}

, where

\alpha

is a constant. The total work done by all the forces applied on the particle during its displacement from

x=0

x=d

, will be :

(1) $\frac{md}{2\alpha^2}$
(2) $\frac{m\alpha^2d}{2}$ Official answer
(3) $\frac{m}{2\alpha^2d}$
(4) $2m\alpha^2d$

Official answer

Option 2

NTA final key.

Chapter: Work, Energy and Power
Topic: Work and Kinetic Energy
Idea tested: Work-Energy Theorem

Same idea in other shifts

Asked 7× in all

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.