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A square shaped hole of side $l=\frac{a}{2}$ is carved out at a distance $d=\frac{a}{2}$ from the centre 'O' of a uniform circular disk of radius a. If the distance of the centre of mass of the remaining portion from O is $-\frac{a}{X},$ value of X (to the nearest integer)is .......
Option: 1 2
Option: 2 5
Option: 3 23
Option: 4 56

Answers (1)

best_answer

Position of centre of mass when some mass is removed

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

Where $m_1$ is value of whole mass and $\vec{r_{1}}$ is position of centre of mass for whole mass.Similarly $m_2$ & $\vec{r_{2}}$ are values for mass which has been removed.

From question

Let the density is $\rho$

So

$\mathrm{m}_{1}=\pi \mathrm{a}^{2} \sigma$

And

$m_2=\frac{a^{2}}{4} \cdot \sigma$

And $r_1=0 \ \ and \ \ r_2=\frac{a}{2}$

So

$X_{cm}= \frac{(\sigma \pi a^{2} \times 0)- (\sigma \frac{a^{2}}{4} \times \frac{a}{2})}{\sigma \pi a^{2}-\frac{\sigma a^{2}}{4}}$

$X_{cm}=\frac{-a^{3} / 8}{\left(\pi-\frac{1}{4}\right) a^{2}}=\frac{-a}{2(4 \pi-1)}=\frac{-a}{8 \pi-2}=-\frac{a}{23}$

So X=23

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