# A circular disc of radius  R is removed from a bigger circular disc of radius  2R  such that the circumferences of the discs coincide The centre of mass of the new disc is $\alpha R$  from the centre of the bigger disc . The value of is Option 1) Option 2) Option 3) Option 4)

As we learnt in

Position of centre of mass when some mass is removed -

${\vec{r}_{CM}}=\frac{m_{1}\vec{r_{1}}-{m_{2}}\vec{r}_{2}}{m_{1}-m_{2}}$

- wherein

m1 is value of whole mass and $\vec{r_{1}}$ is position of centre of mass for whole mass. m$\vec{r_{2}}$ are values for mass which has been removed.

$(M+m)=M=\pi (2R)^{2}.\sigma$

$where\; \sigma =mass\; per\; unit \; area$

$m=\sigma R^{2}.\sigma\; ,M=3\pi R^{2}\sigma$

$\frac{3\pi R^{2}\sigma .x+\pi R^{2}\sigma .R}{M}=0$

Because for the full disc, the centre of mass is at the centre O

$\Rightarrow \; \; \; x=-\frac{R}{3}=\alpha R.\; \; \; \therefore \; \; \; \left | \alpha \right |=\left | \frac{-1}{3} \right |.$

The centre of mass is at R/3 to the left on the diameter of the original disc.

The question should be at a distance $\alpha R$ and not $\alpha/ R$.

Option 1)

Option 2)

Option 3)

Option 4)

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