Physics · 2025
JEE Main · 23 January 2025, Shift 1 · Q47
A force f=x^2 y i +y^2 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 m, 2 m ) is ____…
A force $\displaystyle f=x^2 y \hat{i}+y^2 \hat{j}$ acts on a particle in a plane $\displaystyle x+y=10$. The work done by this force during a displacement from $\displaystyle (0,0)$ to $\displaystyle (4 \mathrm{~m}, 2 \mathrm{~m})$ is $\displaystyle \_\_\_\_$ Joule (round off to the nearest integer)
Official answer
From NTA’s final answer key for this paper.
152
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JEE Main 2025 Physics question, with the answer from NTA’s final answer key. Where our answers come from.