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.

Options

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

A

Explanation

`X_(CM)=(rho(a^(3))((a)/(2))-rho(b^(3))((b)/(2)))/(rho(a^(3))-rho(b^(3)))`
We will consider removed mass as a negative mass
`b=(rho((a^(4))/(2))-rho((b^(4))/(2)))/(rho(a^(3))-rho(b^(3)))`
`a^(3)b-b^(4)=(a^(4))/(2)-(b^(4))/(2)`
`2a^(3)b-2b^(4)=a^(4)-b^(4)`
Put `a=bx`
`2b^(4)x^(3)-2b^(4)=b^(4)x^(4)-b^(4)`
`2x^(3)-1=x^(4)`
`2x^(3)-2+1=x^(4)`
`2(x^(3)-1)=(x^(2)+1)(x^(2)-1)`
`2(x-1)(x^(2)+1+x)=(x-1)(x+1)(x^(2)+1)`
`2x^(2)+2+2x=x^(3)+x+x^(2)+1`
`x^(3)-x^(2)-x-1=0`