QJEEM22Phaa04: JEE Main 2022 Physics - Class 11 JEEM22Phaa04 Solution: Moment of Inertia (M.I.) of four bodies having same mass 'M'…

Question

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Moment of Inertia (M.I.) of four bodies having same mass 'M' and radius '2R' are as follows:

I₁ = M.I. of solid sphere about its diameter

I₂ = M.I. of solid cylinder about its axis

I₃ = M.I. of solid circular disc about its diameter

I₄ = M.I. of thin circular ring about its diameter

If 2(I₂ + I₃) + I₄ = x . I₁, then the value of x will be __________.

Accepted Answer

$2\left( {{1 \over 2} + {1 \over 4}} \right) \times M{(2R)^2} + {1 \over 2}M{(2R)^2} = x{2 \over 5}M{(2R)^2}$

$\Rightarrow 1 + {1 \over 2} + {1 \over 2} = x \times {2 \over 5}$

$\Rightarrow x = 5$

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