Physics · Work, Power & Energy

JEE Main 2025 — 23 January, Morning Shift — Question 66

A force $f=x^{2} y \hat{i}+y^{2} \hat{j}$ acts on a particle in a plane $x+y=10$ . The work done by this force during a displacement from $(0,0)$ to $(4 \mathrm{~m}, 2 \mathrm{~m})$ is $\qquad$ Joule (round off to the nearest integer)

Answer: 152

Numerical answer — enter this value.

Step-by-step solution

$\int_{0}^{4} x^{2}(10-x) d x+\int_{0}^{2} y^{2} d y$

$=\left[\frac{10 \mathrm{x}^{3}}{3}-\frac{\mathrm{x}^{4}}{4}\right]_{0}^{4}+\left[\frac{\mathrm{y}^{3}}{3}\right]_{0}^{2}=\frac{640}{3}-64+\frac{8}{3}=152$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 23 January, Morning Shift
Subject: Physics
Chapter: Work, Power & Energy
Topic: Work Done by a Constant and Variable Force