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Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
हिन्दीA uniform square sheet of side \( 2a \) has a square of side \( a \) removed from one of its corners. Taking the corner of the original square that is diametrically opposite to the removed square as the origin, what is the distance of the centre of mass of the remaining portion from this origin?
Correct answer · Option D
\( \frac{5\sqrt{2}}{6}a \)
Explanation
Original square of area \( 4a^2 \) has CM at \( (a, a) \). Removed square of area \( a^2 \) has CM at \( (1.5a, 1.5a) \). For remaining area \( 3a^2 \): \( X_{\text{cm}} = \frac{4a^2(a) - a^2(1.5a)}{3a^2} = \frac{2.5a}{3} = \frac{5}{6}a \). By symmetry \( Y_{\text{cm}} = \frac{5}{6}a \). Distance \( R = \sqrt{(5/6 a)^2 + (5/6 a)^2} = \frac{5\sqrt{2}}{6}a \).
