A smaller cube with side b (depicted by dashe | vedclass

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Question

(a)

We choose origin at point $P$.

Now, given centre of mass of remaining solid is at point $O$. Coordinates of $O$ as per axes choosen are $O \

Vedclass · Accepted Answer

$X^3-X^2-X-1=0$

Vedclass · Answer

(a)

We choose origin at point $P$.

Now, given centre of mass of remaining solid is at point $O$. Coordinates of $O$ as per axes choosen are $O \equiv(b, b, b)$

i.e. $x=b, y=b$ and $z=b$

Centre of mass of complete large cube lies at its centre.

$	herefore$ Coordinates of centre of mass of large cube are $x_1=\frac{a}{2}, y_1=\frac{a}{2}, z_1=\frac{ a }{2}$

And centre of mass of removed cube of side $b$ is $x_2=\frac{b}{2}, y_2=\frac{b}{2}, z_2=\frac{b}{2}$.

Treating removed mass as negative mass, we have

$x_{ CM }\left(ight.$ of remaining part) $=\frac{m_1 x_1-m_2 x_2}{m_1-m_2}$

$\Rightarrow \quad b=\frac{ho a^3 	imes \frac{a}{2}-ho b^3 	imes \frac{b}{2}}{ho a^3-ho b^3}$

where, $ho=$ mass density.

$\Rightarrow \quad a^3 b-b^4=\frac{a^4}{2}-\frac{b^4}{2}$

$\Rightarrow \quad \frac{a^3}{b^3}-1=\frac{a^4}{2 b^4}-\frac{1}{2}$

As $X=\frac{ a }{b}$, we have

$X^{-3}-1=\frac{X^4}{2}-\frac{1}{2}$

$\Rightarrow \quad 2 X^3-1=X^4$

$\Rightarrow \quad 2\left(X^3-1ight)=X^4-1$

$\Rightarrow 2(X-1)\left(X^2+1+Xight)$

$=(X-1)(X+1)\left(X^2+1ight)$

$\Rightarrow \quad X^3-X^2-X-1=0$

A smaller cube with side $b$ (depicted by dashed lines) is excised from a bigger uniform cube with side $\alpha$ as shown below, such that both cubes have a common vertex $P$. Let $X=a / b$. If the centre of mass of the remaining solid is at the vertex $O$ of smaller cube, then $X$ satisfies

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