Let's break down the problem: **Question:** A turbine is working under a head of 60 m, developing a power of 5 MW. What will be the power developed when the head is increased to 120 m (assuming efficiency remains constant)? ### Solution The power developed by a turbine (assuming the same efficiency and discharge) is proportional to the head: \[ P \propto H \] But if the discharge is also proportional to the square root of the head (as in many turbines), the power is proportional to \( H^{3/2} \): \[ P_1 / P_2 = (H_1 / H_2)^{3/2} \] Given: - \( P_1 = 5 \) MW - \( H_1 = 60 \) m - \( H_2 = 120 \) m \[ P_2 = P_1 \times \left(\frac{H_2}{H_1}\right)^{3/2} \] \[ P_2 = 5 \times \left(\frac{120}{60}\right)^{3/2} \] \[ P_2 = 5 \times (2)^{3/2} \] \[ (2)^{3/2} = 2^{1} \times 2^{1/2} = 2 \times 1.414 = 2.828 \] \[ P_2 = 5 \times 2.828 = 14.14 \text{ MW} \] ### **Approximate Answer:** **14 MW** (Option D) ---