A

To find the percentage change in angular velocity of the disc when the temperature decreases by 20 °C, we will use the concept of thermal expansion and its impact on angular velocity.
Step 1: Understand that the angular velocity \( \omega \) of a rotating body is related to its radius \( r \). As the temperature decreases, the disc will contract according to the coefficient of linear expansion.
Step 2: The formula for linear contraction due to a temperature change is given by:
\[ \Delta L = L_0 \cdot \alpha \cdot \Delta T \]
where:
- \( \Delta L \) = change in length
- \( L_0 \) = original length
- \( \alpha \) = coefficient of linear expansion
- \( \Delta T \) = change in temperature
For a disc, the reduction in radius can be correlated with a change in angular velocity.
Step 3: Calculate the change in radius:
Given, \( \alpha = 5 \times 10^{-4} /°C \) and \( \Delta T = -20 °C \), substituting these values we get:
\[ \Delta r = r_0 \cdot (5 \times 10^{-4}) \cdot (-20) = -0.01r_0 \]
Hence, the final radius \( r = r_0 - 0.01r_0 = 0.99r_0 \).
Step 4: The angular velocity \( \omega \) for a solid disc is given by \( \omega = \frac{v}{r} \) where \( v \) is tangential velocity. Since the mass and linear momentum remain constant and as radius decreases, the angular velocity increases according to the conservation of angular momentum \( (I_1 \, \omega_1 = I_2 \, \omega_2) \).
The new angular velocity:
\[ \omega_{new} = \frac{v}{0.99r_0} = \frac{\omega_1 r_0}{0.99r_0} = \frac{\omega_1}{0.99} \]
Step 5: Now we can find the percentage change in angular velocity:
\[ \text{Percentage change} = \left( \frac{\omega_{new} - \omega_1}{\omega_1} \right) \times 100 = \left( \frac{\frac{\omega_1}{0.99} - \omega_1}{\omega_1} \right) \times 100 \]
\[ \text{Percentage change} = \left( \frac{1 - 0.99}{0.99} \right) \times 100 \approx 1.01 \% \]
Therefore, the percentage change in angular velocity of the disc when the temperature decreases by 20 °C is approximately 1%. Hence the correct answer is option A.