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

# 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

1. A

2. B

3. C

4. D

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### Solution:

Here, Area of a full lamina, ${\mathrm{A}}_{1}=\mathrm{ab}$

Area of the lamina's darkened portion = $\frac{\mathrm{a}}{2}×\frac{\mathrm{b}}{2}=\frac{\mathrm{ab}}{4}$

Using centre of mass,

${\mathrm{x}}_{\mathrm{cm}}=\frac{{\mathrm{A}}_{1}{\mathrm{x}}_{1}-{\mathrm{A}}_{2}{\mathrm{x}}_{2}}{{\mathrm{A}}_{1}-{\mathrm{A}}_{2}}$

${\mathrm{x}}_{\mathrm{cm}}=\frac{\mathrm{ab}\left(\frac{\mathrm{a}}{2}\right)-\frac{\mathrm{ab}}{4}\left(\frac{3\mathrm{a}}{4}\right)}{\mathrm{ab}-\frac{\mathrm{ab}}{4}}$

${\mathrm{x}}_{\mathrm{cm}}=\frac{\left(\frac{8{\mathrm{a}}^{2}\mathrm{b}-3{\mathrm{a}}^{2}\mathrm{b}}{16}\right)}{\frac{3\mathrm{ab}}{4}}=\frac{5\mathrm{a}}{12}$

${\mathrm{y}}_{\mathrm{cm}}=\frac{{\mathrm{A}}_{1}{\mathrm{y}}_{1}-{\mathrm{A}}_{2}{\mathrm{y}}_{2}}{{\mathrm{A}}_{1}-{\mathrm{A}}_{2}}$

${\mathrm{y}}_{\mathrm{cm}}=\frac{\mathrm{ab}\left(\frac{\mathrm{b}}{2}\right)-\frac{\mathrm{ab}}{4}\left(\frac{3\mathrm{b}}{4}\right)}{\mathrm{ab}-\frac{\mathrm{ab}}{4}}=\frac{5\mathrm{b}}{12}$

Hence the correct answer is $\left(\frac{5a}{12},\frac{5b}{12}\right).$

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