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A smaller cube with side b (depicted by ...

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`

Text Solution

Verified by Experts

The correct Answer is:
A

`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`
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