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JEE MAIN 2025
23-01-2025 SHIFT-1
Question
A force $\text{f}={{\text{x}}^{2}}\text{y}\widehat{\text{i}}+{{\text{y}}^{2}}\widehat{\text{j}}$ acts on a particle in a plane $\text{x}+\text{y}=10$ . The work done by this force during a displacement from (0,0) to (4m, 2m) is _______ Joule (round off to the nearest integer)
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Solution
ANSWER 152
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Question Tags
JEE Main
Physics
Medium
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