OnlineCBT
Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
हिन्दीA circular disc of radius \( R \) has a uniform mass distribution. A circular hole of radius \( R/3 \) is cut out such that the rim of the hole touches the outer edge of the disc. The distance of the centre of mass of the remaining portion from the centre of the original disc is:
Correct answer · Option 4
\( \frac{R}{12} \)
Explanation
Let centre of disc be at origin. Hole centre is at \( x_1 = R - R/3 = 2R/3 \). Mass of original disc \( M_0 = \pi R^2 \sigma \), mass of hole \( m = \pi (R/3)^2 \sigma = M_0/9 \). For remaining disc: \( X_{\text{cm}} = \frac{M_0(0) - (M_0/9)(2R/3)}{M_0 - M_0/9} = \frac{-2R/27}{8/9} = -\frac{R}{12} \). Distance is \( R/12 \).
