Rotational Motion

JEE Main level, single correct, medium

75 of 280

Question 75 of 280

A solid sphere of mass MM and radius RR is divided into two unequal parts. The first part has a mass of 7M8\dfrac{7M}{8} and is converted into uniform disc of radius 2R2R . The second part is converted into a uniform solid sphere. Let I1{I}_{1} be the moment of inertia of the disc about its axis and I2{I}_{2} be the moment of inertia of the new sphere about its axis. The ratio I1/I2{I}_{1}/{I}_{2} is given by:

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Answer: (A) 140140

Moment of inertia of disc about the axis passing its centre and perpendicular to its surface, I1=(7M8)(2R)22=7MR24{I}_{1}=\dfrac{\left(\dfrac{7M}{8}\right){\left(2R\right)}^{2}}{2}=\dfrac{7M{R}^{2}}{4}.
Radius of small sphere rr is related as
M43πR3=M843πr3\dfrac{M}{\dfrac{4}{3}\pi {R}^{3}}=\dfrac{\dfrac{M}{8}}{\dfrac{4}{3}\pi {r}^{3}}
r=R2r=\dfrac{R}{2}
Moment of inertia of sphere about its axis passing through its centre∴I2=25(M8)(R2)2=MR280∴{I}_{2}=\dfrac{2}{5}\left(\dfrac{M}{8}\right){\left(\dfrac{R}{2}\right)}^{2}=\dfrac{M{R}^{2}}{80}
I1I2=74×801=140\dfrac{{I}_{1}}{{I}_{2}}=\dfrac{7}{4}\times \dfrac{80}{1}=140

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