JEE Main 2022 · 25 Jun · Shift 2 · Physics Q54

PhysicsInteger+4 / -1Question 54 of 90

Moment of Inertia (M.I.) of four bodies having same mass ' $M$ ' and radius ' $2R$ ' 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 $2(I_2+I_3)+I_4=x\cdot I_1$ then the value of $x$ will be ______.

Your answer

Answer and worked solution

Official answer: 5

Radius $2R$ . $I_1=\frac25M(2R)^2=\frac85MR^2$ ; $I_2=\frac12M(2R)^2=2MR^2$ ; $I_3=\frac14M(2R)^2=MR^2$ ; $I_4=\frac12M(2R)^2=2MR^2$ .

Equation. $2(2+1)MR^2+2MR^2=8MR^2=x\cdot\frac85MR^2$ , so $x=5$ .

Answer: 5

JEE Main 2022 · 25 Jun · Shift 2, Physics question 54