A rigid spherical body is spinning around an axis without any external torque. Due to temperature its volume increases by `3%`. Then percentage change in its angular speed is:

Options

A. `-2%`
B. `-1%`
C. `-3%`
D. `1%`

Answer

To solve the problem step by step, we will use the principle of conservation of angular momentum and the relationship between the volume of the sphere and its radius.

### Step 1: Understand the Conservation of Angular Momentum
Since there is no external torque acting on the spherical body, the angular momentum of the system is conserved. Therefore, we can write:
\[ L_1 = L_2 \]
Where \( L_1 \) is the initial angular momentum and \( L_2 \) is the final angular momentum.

### Step 2: Express Angular Momentum in Terms of Moment of Inertia and Angular Speed
The angular momentum \( L \) can be expressed as:
\[ L = I \omega \]
Where \( I \) is the moment of inertia and \( \omega \) is the angular speed. For a solid sphere, the moment of inertia about its centroidal axis is given by:
\[ I = \frac{2}{5} m r^2 \]
Thus, we can write:
\[ I_1 \omega_1 = I_2 \omega_2 \]

### Step 3: Substitute the Moment of Inertia
Substituting the moment of inertia into the conservation equation, we get:
\[ \frac{2}{5} m r_1^2 \omega_1 = \frac{2}{5} m r_2^2 \omega_2 \]
The terms \( \frac{2}{5} m \) cancel out, leading to:
\[ r_1^2 \omega_1 = r_2^2 \omega_2 \]

### Step 4: Relate the Radii Using Volume Change
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Given that the volume increases by 3%, we can express this as:
\[ V_2 = V_1 (1 + 0.03) = V_1 \cdot \frac{103}{100} \]
Thus:
\[ \frac{4}{3} \pi r_2^3 = \frac{4}{3} \pi r_1^3 \cdot \frac{103}{100} \]
Cancelling \( \frac{4}{3} \pi \) gives:
\[ r_2^3 = r_1^3 \cdot \frac{103}{100} \]

### Step 5: Find the Relationship Between Radii
Taking the cube root of both sides, we find:
\[ \frac{r_2}{r_1} = \left(\frac{103}{100}\right)^{1/3} \]

### Step 6: Substitute Back into the Angular Momentum Equation
Substituting \( r_2 \) in terms of \( r_1 \) into the angular momentum equation:
\[ \omega_1 = \frac{r_2^2}{r_1^2} \omega_2 \]
This gives:
\[ \frac{\omega_1}{\omega_2} = \left(\frac{r_2}{r_1}\right)^2 = \left(\left(\frac{103}{100}\right)^{1/3}\right)^2 = \left(\frac{103}{100}\right)^{2/3} \]

### Step 7: Calculate the Percentage Change in Angular Speed
The percentage change in angular speed is given by:
\[ \text{Percentage Change} = \frac{\omega_2 - \omega_1}{\omega_1} \times 100 \]
This can also be expressed as:
\[ \text{Percentage Change} = \left(\frac{\omega_2}{\omega_1} - 1\right) \times 100 \]
Substituting the expression we derived:
\[ \frac{\omega_2}{\omega_1} = \left(\frac{100}{103}\right)^{2/3} \]
Calculating this gives:
\[ \text{Percentage Change} = \left(\left(\frac{100}{103}\right)^{2/3} - 1\right) \times 100 \approx -1.95\% \]

### Final Step: Round the Result
Rounding this result gives approximately -2%. Therefore, the final answer is:
\[ \text{Percentage change in angular speed} \approx -2\% \]