Let's solve each question step by step. --- ## **Question 1: Number of Transfer Units (NTU)** Given: - Area, \( A = 54\, m^2 \) - Heat capacity rate, \( C_{min} = 900\, W/K \) - Overall heat transfer coefficient, \( U = 100\, W/m^2K \) **NTU Formula:** \[ NTU = \frac{UA}{C_{min}} \] **Plug in the values:** \[ NTU = \frac{100 \times 54}{900} \] \[ NTU = \frac{5400}{900} = 6 \] **Correct Option:** **6** --- ## **Question 2: Magnitude of Velocity in Polar Coordinates** Given: - \( r = 0.5 \sin\theta \) m - \( \theta = 60^\circ \) - \( \dot{\theta} = 0.5 \) rad/s The magnitude of velocity in polar coordinates is: \[ v = \sqrt{ \left( \dot{r} \right)^2 + \left( r\dot{\theta} \right)^2 } \] First, calculate \( \dot{r} \): \[ r = 0.5 \sin\theta \] Differentiate w.r.t time: \[ \dot{r} = 0.5 \cos\theta \cdot \dot{\theta} \] Substitute \( \theta = 60^\circ = \frac{\pi}{3} \), \( \cos(60^\circ) = 0.5 \), and \( \dot{\theta} = 0.5 \) rad/s: \[ \dot{r} = 0.5 \times 0.5 \times 0.5 = 0.125\, \text{m/s} \] Now, \( r = 0.5 \sin(60^\circ) = 0.5 \times 0.866 = 0.433\, \text{m} \) So, \[ r\dot{\theta} = 0.433 \times 0.5 = 0.2165\, \text{m/s} \] Now, calculate magnitude: \[ v = \sqrt{ (0.125)^2 + (0.2165)^2 } \] \[ v = \sqrt{ 0.015625 + 0.04689 } \] \[ v = \sqrt{ 0.062515 } \approx 0.25\, \text{m/s} \] **Correct Option:** **0.25 m/s** --- ### **Summary of Answers** 1. **6** (Number of transfer units) 2. **0.25 m/s** (Magnitude of velocity)