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A uniform rectangular thin sheet ABCD of mass M has length a and breadth b. If the top-right quarter (HBGO, where O is the centre) is cut off, the coordinates of the centre of mass of the remaining portion (with D at the origin) are

Asked in JEE Main 8th April 2nd Shift 2019 · Centre of mass by integration

Figure: Centre of mass by integration
Answer: (4) ((5a)/(12), (5b)/(12))

Step-by-step solution

Full sheet: mass M, centre (/a2,/b2). Removed quarter (top-right): area (ab)/4, mass /M4, centre ((3a)/4,(3b)/4).

x_cm=(M·/a2-/M4·(3a)/4)/((3M)/4)=(/a2-(3a)/(16))/(3/4)=(5a)/(12).

By symmetry y_cm=(5b)/(12), so the remaining part has its centre of mass at ((5a)/(12),(5b)/(12)).

Why the other options are wrong

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