1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A solid sphere of mass M and radius R is divided into two unequal parts. The first part has a mass of 7M8 and is converted into a uniform disc of radius 2R. The second part is converted into a uniform solid sphere. Let I1 be the moment of inertia of the disc about its axis and I2 be the moment of inertia of the new sphere about its axis. The ratio I1I2 is given by

Open in App
Solution

## The given situation is shown in the figure given below Density of given sphere of radius R is ρ=MV=M43πR3 Let radius of sphere formed from second part is r, then mass of second part = V×ρ 18M=43πr3×M43πR3 ∴r3=R38⇒r=R2 Now, I1= moment of inertia of disc (radius is 2R) =Mass×R22=78M×(2R)22=74MR2 and I2 = moment of inertia of sphere (radiusR2and mass18M)about its axis=25×Mass×(Radius)2 =25×18M×(R2)2=MR280 ∴I1I2=74MR2180MR2=140

Suggest Corrections
30
Join BYJU'S Learning Program
Related Videos
Moment of Inertia of Solid Bodies
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program