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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Physics
Chapter
Rotational Motion

BITSAT 2016 · Physics · Rotational Motion

A circular disc of radius R and thickness (R)/(6) has moment...

A circular disc of radius R\mathrm{R} and thickness R6\frac{R}{6} has moment inertia I about an axis passing through its centre perpendicular to its plane. It is melted and recasted into a solid sphere. The moment of inertia of the sphere about its diameter is

Options

  1. AII
  2. B2I8\frac{2 I}{8}
  3. CI5\frac{I}{5}
  4. DI10\frac{I}{10}

Correct answer

C. I5\frac{I}{5}

Step-by-step solution

According to problem disc is melted and recasted into a solid sphere so their volume  will be same. VDisc =VSphere ⇒\piRDisc 2t=43\piRSphere 3⇒\piRDisc 2(RDisc 6)=43\piRSphere 3[t=RDisc 6, given ]⇒RDisc 3=8RSphere 3⇒RSphere =RDisc 2\begin{array}{c}\text{ will be same. }\\ V_{\text{Disc }}=V_{\text{Sphere }} \Rightarrow \piR_{\text{Disc }}^{2} t=\frac{4}{3} \piR_{\text{Sphere }}^{3} \\ \Rightarrow \piR_{\text{Disc }}^{2}\left(\frac{R_{\text{Disc }}}{6}\right)=\frac{4}{3} \piR_{\text{Sphere }}^{3}\left[t=\frac{R_{\text{Disc }}}{6}, \text{ given }\right]\\ \Rightarrow R_{\text{Disc }}^{3}=8 R_{\text{Sphere }}^{3} \Rightarrow R_{\text{Sphere }}=\frac{R_{\text{Disc }}}{2} \end{array}Moment of inertia of disc IDisc =12MRDisc 2=I (given) ∴ M (RDisc )2=2I\begin{array}{l}I_{\text{Disc }}=\frac{1}{2}M R_{\text{Disc }}^{2}=I \text{ (given) }\\ \therefore \text{ M }\left(R_{\text{Disc }}\right)^{2}=2 I \end{array} Momentofinertiaofsphereof inertia of sphereISphere =12MRSphere 2=25M(RDisc 2)2=M10(RDisc )2=2I10=I5\begin{array}{l}I_{\text{Sphere }}=\frac{1}{2}MR_{\text{Sphere }}^{2} \\ =\frac{2}{5}M\left(\frac{R_{\text{Disc }}}{2}\right)^{2}=\frac{M}{10}\left(R_{\text{Disc }}\right)^{2}=\frac{2 I}{10}=\frac{I}{5} \end{array}$
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