Moment of Inertia (M.I.) of four bodies having same mass 'M' and radius '$2 \mathrm{R}$' are as follows: $\mathrm{I}_{1}=$ M.I. of solid sphere about its diameter $\mathrm{I}_{2}=$ M.I. of solid cylinder about its axis $\mathrm{I}_{3}=$ M.I. of solid circular disc about its diameter $\mathrm{I}_{4}=$ M.I. of thin circular ring about its diameter If $2\left(\mathrm{I}_{2}+\mathrm{I}_{3}\right)+\mathrm{I}_{4}=\mathrm{x}$. $\mathrm{I}_{1}$ then the value of x will be

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Rotational Mechanics

Moment of Inertia (M.I.) of four bodies having same mass 'M' and radius '

2 R

' are as follows:

I_{1} =

M.I. of solid sphere about its diameter

I_{2} =

M.I. of solid cylinder about its axis

I_{3} =

M.I. of solid circular disc about its diameter

I_{4} =

M.I. of thin circular ring about its diameter
If

2 (I_{2} + I_{3}) + I_{4} = x

I_{1}

then the value of x will be

Answer: 5

Using standard formulas with radius

2 R

I_{1} = \frac{2}{5} M (2 R)^{2} = \frac{8}{5} M R^{2}

(solid sphere)

I_{2} = \frac{1}{2} M (2 R)^{2} = 2 M R^{2}

(solid cylinder)

I_{3} = \frac{M ( 2 R ) ^{2}}{4} = M R^{2}

(disc about diameter)

I_{4} = \frac{M ( 2 R ) ^{2}}{2} = 2 M R^{2}

(ring about diameter)

Substituting into

2 (I_{2} + I_{3}) + I_{4} = x I_{1}

2 (2 M R^{2} + M R^{2}) + 2 M R^{2} = x \cdot \frac{8}{5} M R^{2}

8 M R^{2} = \frac{8 x}{5} M R^{2}

Therefore

x = 5

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Integer answer