A uniform solid brass sphere is rotating with angular speed `omega_0` about a diameter. If its temperature is now increased by `100^@C`, what will be its new angular speed. (given`alpha_B=2.0xx10^-5per^@C`)

Options

A. `1.1omega_(0)`
B. `1.01omega_(0)`
C. `0.996omega_(0)`
D. `0.842omega_(0)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the new radius after temperature increase
When the temperature of the brass sphere is increased, it expands. The new radius \( r' \) can be calculated using the formula:
\[ r' = r_0 (1 + \alpha \Delta T) \]
where:
- \( r_0 \) is the initial radius,
- \( \alpha \) is the coefficient of linear expansion (given as \( 2 \times 10^{-5} \, \text{per} \, ^\circ C \)),
- \( \Delta T \) is the change in temperature (given as \( 100 \, ^\circ C \)).

Substituting the values:
\[ r' = r_0 \left(1 + 2 \times 10^{-5} \times 100\right) = r_0 \left(1 + 2 \times 10^{-3}\right) = r_0 (1 + 0.002) \]

### Step 2: Apply the conservation of angular momentum
According to the conservation of angular momentum:
\[ I_1 \omega_1 = I_2 \omega_2 \]
where:
- \( I_1 \) is the moment of inertia before heating,
- \( I_2 \) is the moment of inertia after heating,
- \( \omega_1 \) is the initial angular speed (\( \omega_0 \)),
- \( \omega_2 \) is the new angular speed.

The moment of inertia \( I \) for a solid sphere is given by:
\[ I = \frac{2}{5} m r^2 \]

Thus, we have:
\[ \frac{2}{5} m r_0^2 \omega_0 = \frac{2}{5} m (r')^2 \omega_2 \]

### Step 3: Simplify the equation
We can cancel \( \frac{2}{5} m \) from both sides:
\[ r_0^2 \omega_0 = (r')^2 \omega_2 \]

### Step 4: Substitute for \( r' \)
Substituting \( r' \):
\[ r_0^2 \omega_0 = \left(r_0 (1 + 0.002)\right)^2 \omega_2 \]
\[ r_0^2 \omega_0 = r_0^2 (1 + 0.002)^2 \omega_2 \]

### Step 5: Expand \( (1 + 0.002)^2 \)
Using the binomial expansion for small values:
\[ (1 + x)^2 \approx 1 + 2x \quad \text{for small } x \]
So,
\[ (1 + 0.002)^2 \approx 1 + 2 \times 0.002 = 1 + 0.004 \]

### Step 6: Substitute back into the equation
Now substituting back:
\[ r_0^2 \omega_0 = r_0^2 (1 + 0.004) \omega_2 \]
Cancelling \( r_0^2 \) from both sides:
\[ \omega_0 = (1 + 0.004) \omega_2 \]

### Step 7: Solve for \( \omega_2 \)
Rearranging gives:
\[ \omega_2 = \frac{\omega_0}{1 + 0.004} = \frac{\omega_0}{1.004} \]

### Step 8: Approximate \( \omega_2 \)
Calculating \( \omega_2 \):
\[ \omega_2 \approx \omega_0 (1 - 0.004) = \omega_0 (0.996) \]

Thus, the new angular speed \( \omega_2 \) is approximately:
\[ \omega_2 \approx 0.996 \omega_0 \]

### Final Answer
The new angular speed of the brass sphere after the temperature increase is:
\[ \omega_2 = 0.996 \omega_0 \]