Moment of inertia of a solid about its geometrical axis is given by `1 =2/5 MR^(2)` where M is mass & R is radius. Find out the rate by which its moment of inertia is changing keeping dnsity constant at the moment `R= 1 m`, `M=1 kg` & rate of change of radius w.r.t. time `2 ms^(-1)`

Options

A. `4 kg ms^(-1)`
B. `2 kg m^(2)s^(-1)`
C. `4 kg m^(2) s^(-1)`
D. None of these

Explanation

To find the rate at which the moment of inertia \( I \) of a solid is changing, we start with the formula for the moment of inertia about its geometrical axis:

\[ I = \frac{2}{5} M R^2 \]

Where:
- \( M \) is the mass of the solid,
- \( R \) is the radius of the solid.

Given:
- \( M = 1 \, \text{kg} \)
- \( R = 1 \, \text{m} \)
- The rate of change of radius \( \frac{dR}{dt} = 2 \, \text{m/s} \)

### Step 1: Express Mass in terms of Density and Volume
Since the density \( \rho \) is constant, we can express mass \( M \) in terms of density and volume. The volume \( V \) of a solid sphere is given by:

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

Thus, the mass \( M \) can be expressed as:

\[ M = \rho V = \rho \left(\frac{4}{3} \pi R^3\right) \]

### Step 2: Substitute Mass into the Moment of Inertia Formula
Substituting \( M \) into the moment of inertia formula gives:

\[ I = \frac{2}{5} \left(\rho \frac{4}{3} \pi R^3\right) R^2 = \frac{8}{15} \pi \rho R^5 \]

### Step 3: Differentiate Moment of Inertia with Respect to Time
Now, we differentiate \( I \) with respect to time \( t \):

\[ \frac{dI}{dt} = \frac{8}{15} \pi \rho \frac{d}{dt}(R^5) \]

Using the chain rule, we have:

\[ \frac{d}{dt}(R^5) = 5R^4 \frac{dR}{dt} \]

Thus, we can write:

\[ \frac{dI}{dt} = \frac{8}{15} \pi \rho \cdot 5R^4 \frac{dR}{dt} \]

### Step 4: Substitute Known Values
Now we substitute the known values into the equation:

- \( R = 1 \, \text{m} \)
- \( \frac{dR}{dt} = 2 \, \text{m/s} \)

We also need to find the density \( \rho \) using the mass and volume at \( R = 1 \, \text{m} \):

\[ \rho = \frac{M}{V} = \frac{1 \, \text{kg}}{\frac{4}{3} \pi (1)^3} = \frac{3}{4\pi} \, \text{kg/m}^3 \]

Now substituting \( R = 1 \, \text{m} \) and \( \rho = \frac{3}{4\pi} \):

\[ \frac{dI}{dt} = \frac{8}{15} \pi \left(\frac{3}{4\pi}\right) \cdot 5(1^4)(2) \]

### Step 5: Simplify the Expression
Now simplify the expression:

\[ \frac{dI}{dt} = \frac{8}{15} \cdot \frac{3}{4} \cdot 5 \cdot 2 \]

Calculating this gives:

\[ \frac{dI}{dt} = \frac{8 \cdot 3 \cdot 5 \cdot 2}{15 \cdot 4} = \frac{240}{60} = 4 \, \text{kg m}^2/\text{s} \]

### Final Answer
Thus, the rate at which the moment of inertia is changing is:

\[ \frac{dI}{dt} = 4 \, \text{kg m}^2/\text{s} \]