From a disc of radius R a smaller disc of radius R/4 is removed as shown in figure. Distance of centre of mass of the remaining part from centre of the bigger disc is:-

Question

From a disc of radius R a smaller disc of radius R/4 is removed as shown in figure. Distance of centre of mass of the remaining part from centre of the bigger disc is:- <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-l1zvsx6bawknjlxc.jpg" /> (A) (R/10) (B) (R/20) (C) (R/16) (D) (R/5)

Tardigrade Admin · Accepted Answer

m1=m,m2=(m/π R2)× π ((R/4))2=(M/16) xcm=(m1 x1 - m2 x2/ m1 - m2)=(m (0) - (m/16) ((3 R/4))/m - (m)16)

Q. From a disc of radius $R$ a smaller disc of radius $R /4$ is removed as shown in figure. Distance of centre of mass of the remaining part from centre of the bigger disc is:-

$\frac{R}{10}$

$\frac{R}{20}$

$\frac{R}{16}$

$\frac{R}{5}$

$m_{1} = m, m_{2} = \frac{m}{π R ^{2}} \times π (\frac{R}{4})^{2} = \frac{M}{16}$
$x_{c m} = \frac{m _{1} x _{1} - m _{2} x _{2}}{m _{1} - m _{2}} = \frac{m ( 0 ) - \frac{m}{16} ( \frac{3 R}{4} )}{m - \frac{m}{16}}$

Q. From a disc of radius R a smaller disc of radius R/4 is removed as shown in figure. Distance of centre of mass of the remaining part from centre of the bigger disc is:-

10R​

20R​

16R​

5R​