Moment of Inertia (M.I.) of four bodies having same mass 'M' and… | JEE Main 2022 (25 Jun, Shift 2)

Q116JEE Main 2022 · 25 JUN · SHIFT 2Medium

Moment of Inertia & Radius of Gyration

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 __________.

Numerical answer type. Work it out, then open the answer.

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Answer: 5

Radius $2R$ , so $r^2 = 4R^2$ .
$I_1 = \frac{2}{5}M(4R^2) = \frac{8}{5}MR^2$ , $I_2 = \frac{1}{2}M(4R^2) = 2MR^2$
$I_3 = \frac{1}{4}M(4R^2) = MR^2$ , $I_4 = \frac{1}{2}M(4R^2) = 2MR^2$
$2(I_2 + I_3) + I_4 = 6MR^2 + 2MR^2 = 8MR^2 = x \cdot \frac{8}{5}MR^2$
So $x = 5$ .

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