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 diameterI2 = M.I. of solid cylinder about its axisI3 = M.I. of solid circular disc about its diameterI4 = M.I. of thin circular ring about its diameterIf 2(I2 + I3) + I4 = x. I1 then the value of x will be ________

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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 _________ .

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answer is 5.

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Detailed Solution

$\begin{aligned} I_{1} = \frac{2}{5} M (2 R)^{2} = \frac{8}{5} {MR}^{2} \\ I_{2} = \frac{1}{2} M (2 R)^{2} = 2 {MR}^{2} \\ I_{3} = \frac{M (2 R)^{2}}{4} = {MR}^{2} \\ I_{4} = \frac{M (2 R)^{2}}{2} = 2 {MR}^{2} \\ 2 (I_{2} + I_{3}) + I_{4} = {xI}_{1} \\ 8 {MR}^{2} = x \frac{8}{5} {MR}^{2} \\ x = 5 \end{aligned}$