Interpret the figure

For a rectangular lamina of length $a$ and breadth $b$, let the origin $O$ be at one corner.

The shaded portion $HBGO$ is the small rectangle cut off from the corner at $O$, formed by joining the midpoints along the sides. Hence its dimensions are: $\frac{a}{2} \times \frac{b}{2}$

Masses of the full sheet and removed sheet

Since the sheet is uniform, mass is proportional to area.

• Full rectangle area: $A = ab$
• Removed rectangle area: $A_r = \frac{a}{2}\cdot \frac{b}{2} = \frac{ab}{4}$

Therefore removed mass is $m = \frac{M}{4}$

Remaining mass: $M' = M - \frac{M}{4} = \frac{3M}{4}$

Coordinates of centres of mass

• Centre of mass of the full rectangle: $\left(\frac{a}{2},\frac{b}{2}\right)$

• Centre of mass of the removed corner rectangle $HBGO$: $\left(\frac{a}{4},\frac{b}{4}\right)$

Use subtraction of masses

Let the centre of mass of the remaining portion be $(x,y)$.

Using the formula for composite bodies: $x = \frac{M\left(\frac{a}{2}\right)-\left(\frac{M}{4}\right)\left(\frac{a}{4}\right)}{\frac{3M}{4}}$

Simplify: $x = \frac{\frac{Ma}{2}-\frac{Ma}{16}}{\frac{3M}{4}}$ $x = \frac{\frac{7Ma}{16}}{\frac{3M}{4}}$ $x = \frac{7a}{16}\cdot \frac{4}{3} = \frac{7a}{12}$

Similarly, $y = \frac{M\left(\frac{b}{2}\right)-\left(\frac{M}{4}\right)\left(\frac{b}{4}\right)}{\frac{3M}{4}}$ $y = \frac{\frac{Mb}{2}-\frac{Mb}{16}}{\frac{3M}{4}}$ $y = \frac{\frac{7Mb}{16}}{\frac{3M}{4}} = \frac{7b}{12}$

Derived answer

Thus the centre of mass of the remaining portion is $\boxed{\left(\frac{7a}{12},\frac{7b}{12}\right)}$

Compare with given options

The options listed are:

• A: $\left(\frac{5a}{3},\frac{5b}{3}\right)$
• B: $\left(\frac{2a}{3},\frac{2b}{3}\right)$
• C: $\left(\frac{5a}{12},\frac{5b}{12}\right)$
• D: $\left(\frac{3a}{4},\frac{3b}{4}\right)$

Our result $\left(\frac{7a}{12},\frac{7b}{12}\right)$ is not among the options.

Also, physically this makes sense: removing mass near the origin shifts the centre of mass away from the origin, so coordinates should be greater than $\left(\frac{a}{2},\frac{b}{2}\right)$, not smaller. Therefore option C, $\left(\frac{5a}{12},\frac{5b}{12}\right)$, cannot be correct.

Conclusion

The stored answer appears inconsistent with the standard centre-of-mass calculation for a corner rectangle removed from a uniform rectangular sheet.