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Physics-

General

Easy

Question

Four balls each of radius 10 cm and mass 1 kg, 2kg, 3 kg and 4 kg are attached to the periphery of massless plate of radius 1 m. What is moment of inertia of the system about the centre of plate?

12.04 kg- $m^{2}$
10.04 kg- $m^{2}$
11.50 kg- $m^{2}$
5.04 kg- $m^{2}$

The correct answer is: 10.04 kg- $m to the power of 2 end exponent$

Moment of inertia of the system about the centre of plane is given by
$I equals open square brackets fraction numerator 2 over denominator 5 end fraction cross times 1 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 1 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets$
$plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 2 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 2 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets$
$plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 3 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 3 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets$
$plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 4 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 4 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets$
$equals 1.004 plus 2.008 plus 3.012 plus 4.016$
$equals 10.04 blank k g minus m to the power of 2 end exponent$

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