Question

A smaller cube with side $b$ (depicted by dashed lines) is excised from a bigger uniform cube with side a 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

• A. $X^3-X^2-X-1=0$
• B. $X^2-X-1=0$
• C. $X^3+X^2-X-1=0$
• D. $X^3-X^2-X+1=0$

Answer 1

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 $O=(b,b,b)$
i.e. $x=b,y=b$ and $z=b$
Centre of mass of complete large cube lies at its centre.
$\therefore$ 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
$x_2=\frac{b}{2},y_2=\frac{b}{2},Z_2=\frac{b}{2}.$
Treating removed mass as negative mass, we have
$x_{cM}$ (of remaining part) $=\frac{m_1{x}_1-m_2{x}_2}{m_1-m_2}$
$\Rightarrow b=\frac{\rho a^3\times \frac{a}{2}-\rho b^3\times \frac{b}{2}}{\rho a^3-\rho b^3}$
where, $\rho$ = mass density.
$\Rightarrow a^{3}b-b^4=\frac{a^4}{2}-\frac{b^4}{2}$
$\Rightarrow \frac{a^3}{b^3}-1=\frac{a^4}{2b^4}-\frac{1}{2}$
As $X=\frac{a}{b}$, we have
$X^3-1=\frac{X^4}{2}-\frac{1}{2}$
$\Rightarrow 2X^3-1=X^4$
$\Rightarrow 2\left(X^3-1\right)=X^4-1$
$\Rightarrow 2\left(X-1\right)\left(X^2+1+X\right)$
$\left(X-1\right)\left(X+1\right)\left(X^2+1\right)$
$\Rightarrow X^3-X^2-X-1=0$