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JEE Main 2019 (8 Apr, Shift 2) · Composite & Cut-Out Bodies

Composite & Cut-Out Bodies question from JEE Main 2019 (8 Apr, Shift 2)

Q11JEE Main 2019 · 8 APR · SHIFT 2Medium

Composite & Cut-Out Bodies

A uniform rectangular thin sheet ABCD of mass M has length a and breadth b, as shown in the figure. If the shaded portion HBGO is cut-off, the coordinates of the centre of mass of the remaining portion will be :-

Figure

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Answer: (c)

  1. Full sheet: mass MM at (a2,b2)\left(\frac{a}{2}, \frac{b}{2}\right). Cut part HBGO is a quarter: mass M4\frac{M}{4} at (3a4,3b4)\left(\frac{3a}{4}, \frac{3b}{4}\right).
  2. xcm=Ma2−M4⋅3a43M4=5a1634=5a12x_{cm} = \frac{M\frac{a}{2} - \frac{M}{4}\cdot\frac{3a}{4}}{\frac{3M}{4}} = \frac{\frac{5a}{16}}{\frac{3}{4}} = \frac{5a}{12}
  3. By the same working ycm=5b12y_{cm} = \frac{5b}{12}.
  4. So the answer is (c) (5a12,5b12)\left(\frac{5a}{12}, \frac{5b}{12}\right).
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