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JEE-MAIN 2021
17-03-2021 S2
Question
The disc of mass $M$ with uniform surface mass density $s$ is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position $\frac{\mathrm{x}}{3} \frac{\mathrm{a}}{\pi}, \frac{\mathrm{x}}{3} \frac{\mathrm{a}}{\pi}$ where x is $\_\_\_\_$ .
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Question Tags
JEE Main
Physics
Easy
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