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Physics · 2025

JEE Main · 23 January 2025, Shift 1 · Q26

The position of a particle moving on x -axis is given by x(t)=A sin t+B cos^2 t+C t^2+D, where t is time. The dimension of (A B C)/D is

The position of a particle moving on $\displaystyle x$-axis is given by $\displaystyle x(t)=A \sin t+B \cos ^2 t+C t^2+D$, where $\displaystyle t$ is time. The dimension of $\displaystyle \frac{A B C}{D}$ is
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JEE Main 2025 Physics question, with the answer from NTA’s final answer key. Where our answers come from.