Question

A circular hole of radius $1cm$ is cut off from a disc of radius $6cm$. The centre of hole is $3m$ from the centre of the disc. The position of centre of mass of the remaining disc from the centre of disc is:

• A. $-\frac{3}{35}cm$
• B. $\frac{1}{35}cm$
• C. $\frac{3}{10}cm$
• D. $Noneofthese$

Answer 1

The correct option is A $-\frac{3}{35}cm$

Given,

radius of big disc is $6cm$

radius of small disc is $1cm$

Distance from the Centre of disc is $3cm$

Let mass of big disc is $M$

Let mass of small disc is $m$

Density of disc is

$d=\frac{M}{area}=\frac{M}{\pi \times 6\times 6}=\frac{M}{36\pi }$

Mass of small disc

$M_1=\pi 1^2\times \frac{M}{36\pi }=\frac{M}{36}$

After cutting of small disc from bigger disc mass is

$M_2=M-\frac{M}{36}=\frac{35M}{36}$

Now the disc is symmetry about X-axis.

$\overline{X}=\frac{M_1{X}_1+M_2{X}_2}{M_1+M_2}$

Here,

$X_1=3cm$

Substitute all value in above equation

$\overline{X}=\frac{M_1{X}_1+M_2{X}_2}{M_1+M_2}$

$0=\frac{\frac{M}{36}\times 3+\frac{35M}{36}{X}_2}{\frac{M}{36}+\frac{35M}{36}}3+35X_2=0X_2=-\frac{3}{35}cm$

correct option is A.