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Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
हिन्दीFrom a uniform circular disc of radius \( R \), a circular hole of radius \( R/2 \) is cut out such that the rim of the hole touches the centre of the original disc and passes through its outer edge. The distance of the centre of gravity of the remaining flat body from the centre of the original disc is:
Correct answer · Option 2
\( R/6 \)
Explanation
Mass of original disc \( M_0 = \sigma \pi R^2 = M \). Mass of hole \( m = \sigma \pi (R/2)^2 = M/4 \). Centre of hole is at \( x = R/2 \). For the remaining portion, \( X_{\text{cm}} = \frac{M(0) - (M/4)(R/2)}{M - M/4} = \frac{-MR/8}{3M/4} = -\frac{R}{6} \). Magnitude is \( R/6 \).
