Question

A solid sphere of mass \(M\) and radius \(R\) is split into two pieces of masses \( \frac{7M{8} \) and \( \frac{M}{8} \). The piece of mass \( \frac{7M}{8} \) is converted into a disc of radius \(2R\) and thickness \(t\) whose moment of inertia is \(I_1\). The other piece is made into a solid sphere with moment of inertia \(I_2\). Find \( \frac{I_1}{I_2} \).}

• A. \(150\)
• B. \(140\)
• C. \(130\)
• D. \(120\)

Answer 1

The Correct Option is B

Concept:
Moment of inertia formulas: \[ I_{\text{disc}} = \frac{1}{2}MR^2 \] \[ I_{\text{sphere}} = \frac{2}{5}MR^2 \] Mass and volume relations are used to determine the radius of the smaller sphere. Step 1: Find density of the original sphere. \[ M = \frac{4}{3}\pi R^3 \rho \] \[ \rho = \frac{3M}{4\pi R^3} \] Step 2: Find radius of smaller sphere of mass \(M/8\). \[ \frac{M}{8} = \frac{4}{3}\pi r^3 \rho \] Substitute \(\rho\): \[ \frac{M}{8} = \frac{4}{3}\pi r^3 \left(\frac{3M}{4\pi R^3}\right) \] \[ \frac{M}{8} = M\frac{r^3}{R^3} \] \[ r^3 = \frac{R^3}{8} \] \[ r = \frac{R}{2} \] Step 3: Moment of inertia of disc formed from mass \(7M/8\). \[ I_1 = \frac{1}{2}\left(\frac{7M}{8}\right)(2R)^2 \] \[ I_1 = \frac{7MR^2}{4} \] Step 4: Moment of inertia of smaller sphere. \[ I_2 = \frac{2}{5}\left(\frac{M}{8}\right)\left(\frac{R}{2}\right)^2 \] \[ I_2 = \frac{MR^2}{80} \] Step 5: Find the ratio. \[ \frac{I_1}{I_2} = \frac{\frac{7MR^2}{4}}{\frac{MR^2}{80}} \] \[ \frac{I_1}{I_2} = 140 \]