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 _____ .

Question

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 _____ .

Aicharya · Accepted Answer

&#8757; &#160; 2 I 2 + I 3 + I 4 = x I 1 Now For solid sphere about its diameter I 1 = 2 5 M R 2 , For solid cylinder about its axis I 2 = 1 2 M R 2 , For circular disc about its diameter I 3 = 1 4 M R 2 And for circular ring about its diameter I 4 = 1 2 M R 2 &#8658; 2 M R 2 2 + M R 2 4 + M R 2 2 = x &#215; 2 5 M R 2 &#8658; 2 3 4 + 1 2 = x &#215; 2 5 &#8658; 2 = x &#215; 2 5 &#8658; x = 5

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…

Solution