Step-by-step Solution:

1. Understand the Physical Situation: This problem involves the Doppler effect. The sound is emitted from a moving source (the police car), reflects off a stationary object (the building), and is then detected by a moving observer (the driver in the same car). This can be analyzed in two separate steps.

2. List Given Values and Convert Units:

   • Frequency of the siren (source), $f_s = 8 kHz = 8000 Hz$.
   • Velocity of the police car (source and observer), $v_{car} = 36 km/hr$.
   • Speed of sound in air, v = 320 m/s.

   First, convert the car's velocity to m/s: $v_{car} = 36 \frac{km}{hr} \times \frac{1000 m}{1 km} \times \frac{1 hr}{3600 s} = 10 \ m/s$

3. Step 1: Frequency Received by the Building The police car is the source moving towards a stationary listener (the building). The general formula for the Doppler effect for frequency received by a listener ($f_L$) from a source ($f_S$) is: $f_L = f_S \left( \frac{v \pm v_L}{v \mp v_S} \right)$ where $v_L$ is the listener's velocity and $v_S$ is the source's velocity. The top signs are used for approach, and the bottom signs for recession.

   In this step:

   • Source (car) is moving towards the listener (building), so we use $-v_S$ in the denominator. $v_S = v_{car} = 10 m/s$.
   • Listener (building) is stationary, so $v_L = 0$.

   The frequency received by the building, let's call it $f_b$, is: $f_b = f_s \left( \frac{v}{v - v_{car}} \right)$ $f_b = 8000 \left( \frac{320}{320 - 10} \right) = 8000 \left( \frac{320}{310} \right) Hz$

4. Step 2: Frequency of Reflected Sound Heard by the Driver The building now acts as a stationary source, emitting sound waves with the frequency it received, $f_b$. The car driver is the observer, moving towards this stationary source.

   In this step:

   • Source (building) is stationary, so $v_S = 0$.
   • Observer (driver) is moving towards the source, so we use $+v_L$ in the numerator. $v_L = v_{car} = 10 m/s$.
   • The source frequency is $f_b$.

   The final frequency heard by the driver, $f_{driver}$, is: $f_{driver} = f_b \left( \frac{v + v_{car}}{v} \right)$ Substituting the expression for $f_b$ from Step 1: $f_{driver} = \left[ 8000 \left( \frac{320}{310} \right) \right] \left( \frac{320 + 10}{320} \right)$ $f_{driver} = 8000 \left( \frac{320}{310} \right) \left( \frac{330}{320} \right)$ The 320 terms cancel out: $f_{driver} = 8000 \left( \frac{330}{310} \right) = 8000 \left( \frac{33}{31} \right)$

5. Calculate the Final Frequency: $f_{driver} \approx 8000 \times 1.064516 \approx 8516.13 \ Hz$ Converting to kHz: $f_{driver} \approx 8.516 \ kHz$

6. Alternative Direct Formula: For a case where a source moves towards a stationary reflector, the frequency of the reflected sound heard by the observer (who is at the source) is given directly by: $f_{heard} = f_s \left( \frac{v + v_{obs}}{v - v_{src}} \right)$ Here, the observer and the source are the same car, so $v_{obs} = v_{src} = v_{car} = 10 m/s$. $f_{heard} = 8000 \left( \frac{320 + 10}{320 - 10} \right) = 8000 \left( \frac{330}{310} \right) = 8000 \left( \frac{33}{31} \right) \approx 8516.13 \ Hz \approx 8.52 \ kHz$

7. Compare with Options: The calculated frequency is approximately 8.52 kHz. Looking at the options: A: 8.50 kHz B: 8.25 kHz C: 7.75 kHz D: 7.50 kHz

   The value 8.52 kHz is closest to 8.50 kHz. The small difference is likely due to rounding in the problem's given values or the options. Therefore, option A is the correct answer.