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  4. 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

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Q. 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 $2\left( I _{2}+ I _{3}\right)+ I _{4}= x . I _{1}$ then the value of $x$ will be ___

JEE MainJEE Main 2022System of Particles and Rotational Motion

Solution:

$I _{1}=\frac{2}{5} M (2 R )^{2}=\frac{8}{5} MR ^{2}$
$I _{1}=\frac{1}{2} M (2 R )^{2}=2 MR ^{2}$
$I _{3}=\frac{ M (2 R )^{2}}{4}= MR ^{2}$
$I _{4}=\frac{ M (2 R )^{2}}{2}=2 MR ^{2}$
$2\left( I _{2}+ I _{3}\right)+ I _{4}=x I _{1}$
$8 MR ^{2}= x \frac{8}{5} MR ^{2}$
$x =5$