Physics · 2021
JEE Main · 25 July 2021, Shift 2 · Q21
A force of F =(5 y+20) j N acts on a particle. The workdone by this force when the particle is moved from y=0 m to y=10 m is \ \ \ \ J.
A force of $\displaystyle \mathrm{F}=(5 y+20) \hat{\mathrm{j}} \mathrm{N}$ acts on a particle. The workdone by this force when the particle is moved from $\displaystyle y=0 \mathrm{~m}$ to $\displaystyle y=10 \mathrm{~m}$ is \_\_\_\_J.
Official answer
450
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JEE Main 2021 Physics question, with the answer from NTA’s final answer key. Where our answers come from.