KVPY SX SB Exam 2016 Question Paper

Question : 100 of 120

Marks: +1, -0

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.

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

Validate

Solution:

Centre of mass of remaining cube x coordinate = b

X_\CM={ \ρa^3×a/2-\ρb^3×b/2}/{\ρa^3-\ρb^3}

We will consider removed mass as a negative mass

b={{\ρa^4 }/2-{\ρb^4}/2}/{\ρa^3-\ρb^3}

a^3b-b^4=a^4/2-b^4/2

2a^3b - 2b^4 = a^4- b^4

put

a = bx

⇒ 2b^4x^3b - 2b^4 = b^4x^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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