Answer:
Suppose
mass per unit area of disc = m \[\therefore \]Mass
of disc. \[M=\pi {{\left( d/2 \right)}^{2}}\] \[m=\frac{\pi m{{d}^{2}}}{4}\]
Mass of
portion removed \[M'=\pi {{\left( d/12 \right)}^{2}}m=\frac{\pi
{{d}^{2}}m}{144}\]
Let
centre O of disc be taken as the origin. Suppose masses M and M' are
concentrated at O and O' respectively. Fig. 5(HT).9. As the portion is removed,
taking M' as negative, c.m of remaining portion is at P, at a distance x from
O, where
\[x=\frac{M{{x}_{1}}-M'{{x}_{2}}}{M-M'}\]
\[\frac{(\pi
m{{d}^{2}}/4)\times 0-\left( \pi m{{d}^{2}}/144 \right)\left( d/4 \right)}{\pi
m{{d}^{2}}/4-\pi m{{d}^{2}}/144}\]
\[=-\frac{\pi
m{{d}^{2}}}{144}\times \frac{d}{4}\times \frac{144}{35\pi m{{d}^{2}}}=-d/140\]
Negative
sign indicates c.m is to left of origin O.
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