The moment of inertia of hollow sphere (mass M) of inner radius R and outer radius 2R, having material of uniform density, about a diametric axis is

Options

A. `31MR^(2)//70`
B. `43MR^(2)//90`
C. `19MR^(2)//80`
D. None of these

Explanation

To find the moment of inertia of a hollow sphere with mass \( M \), inner radius \( R \), and outer radius \( 2R \) about a diametric axis, we can follow these steps:

### Step 1: Determine the Volume of the Hollow Sphere
The volume of the hollow sphere can be calculated by subtracting the volume of the inner sphere from the volume of the outer sphere.

\[ V = V_{\text{outer}} - V_{\text{inner}} = \frac{4}{3} \pi (2R)^3 - \frac{4}{3} \pi R^3 \]

Calculating the volumes:

\[ V_{\text{outer}} = \frac{4}{3} \pi (8R^3) = \frac{32}{3} \pi R^3 \]
\[ V_{\text{inner}} = \frac{4}{3} \pi R^3 \]

Thus,

\[ V = \frac{32}{3} \pi R^3 - \frac{4}{3} \pi R^3 = \frac{28}{3} \pi R^3 \]

### Step 2: Calculate the Density
The density \( \rho \) of the material can be expressed as:

\[ \rho = \frac{M}{V} = \frac{M}{\frac{28}{3} \pi R^3} = \frac{3M}{28 \pi R^3} \]

### Step 3: Moment of Inertia of a Spherical Shell
To find the moment of inertia \( I \) of the hollow sphere, we consider it as a collection of infinitesimally thin spherical shells. The moment of inertia \( dI \) of a thin spherical shell of radius \( r \) and thickness \( dr \) is given by:

\[ dI = \frac{2}{3} dm \cdot r^2 \]

Where \( dm \) is the mass of the shell.

### Step 4: Calculate the Mass of the Shell
The mass \( dm \) of the spherical shell can be expressed in terms of density and volume:

\[ dm = \rho \cdot dV \]

The volume \( dV \) of the thin shell is:

\[ dV = 4 \pi r^2 dr \]

Thus,

\[ dm = \rho \cdot 4 \pi r^2 dr = \left(\frac{3M}{28 \pi R^3}\right) \cdot 4 \pi r^2 dr = \frac{12M}{28 R^3} r^2 dr = \frac{3M}{7 R^3} r^2 dr \]

### Step 5: Substitute \( dm \) into \( dI \)
Substituting \( dm \) into the equation for \( dI \):

\[ dI = \frac{2}{3} \left(\frac{3M}{7 R^3} r^2 dr\right) r^2 = \frac{2M}{7 R^3} r^4 dr \]

### Step 6: Integrate to Find Total Moment of Inertia
Now, we integrate \( dI \) from \( R \) to \( 2R \):

\[ I = \int_{R}^{2R} \frac{2M}{7 R^3} r^4 dr \]

Calculating the integral:

\[ I = \frac{2M}{7 R^3} \left[ \frac{r^5}{5} \right]_{R}^{2R} = \frac{2M}{7 R^3} \left( \frac{(2R)^5}{5} - \frac{R^5}{5} \right) \]

\[ = \frac{2M}{7 R^3} \left( \frac{32R^5 - R^5}{5} \right) = \frac{2M}{7 R^3} \left( \frac{31R^5}{5} \right) = \frac{62M R^2}{35} \]

### Final Result
Thus, the moment of inertia of the hollow sphere about a diametric axis is:

\[ I = \frac{62M R^2}{35} \]