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  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

I 1=(2/5) M (2 R )2=(8/5) MR 2 I 1=(1/2) M (2 R )2=2 MR 2 I 3=( M (2 R )2/4)= MR 2 I 4=( M (2 R )2/2)=2 MR 2 2( I 2+ I 3)+ I 4=x I 1 8 MR 2= x (8/5) MR 2 x =5

I1​=52​M(2R)2=58​MR2 I1​=21​M(2R)2=2MR2 I3​=4M(2R)2​=MR2 I4​=2M(2R)2​=2MR2 2(I2​+I3​)+I4​=xI1​ 8MR2=x58​MR2 x=5

$I_{1} = \frac{2}{5} M (2 R)^{2} = \frac{8}{5} M R^{2}$
$I_{1} = \frac{1}{2} M (2 R)^{2} = 2 M R^{2}$
$I_{3} = \frac{M ( 2 R ) ^{2}}{4} = M R^{2}$
$I_{4} = \frac{M ( 2 R ) ^{2}}{2} = 2 M R^{2}$
$2 (I_{2} + I_{3}) + I_{4} = x I_{1}$
$8 M R^{2} = x \frac{8}{5} M R^{2}$
$x = 5$