ANS:B - 5π/2

To calculate the force exerted by the water on the right angle bend in the pipe, we need to consider the change in momentum of the water as it flows through the bend. Given:

• Flow velocity (VVV) = 1 m/s
• Internal diameter of the pipe (DDD) = 10 cm = 0.1 m

First, calculate the cross-sectional area (AAA) of the pipe: A=πD24=π(0.1)24=π⋅0.014=π400 m2A = \frac{\pi D^2}{4} = \frac{\pi (0.1)^2}{4} = \frac{\pi \cdot 0.01}{4} = \frac{\pi}{400} \text{ m}^2A=4πD2​=4π(0.1)2​=4π⋅0.01​=400π​ m2 Next, calculate the mass flow rate (m˙\dot{m}m˙) of water: m˙=ρ⋅V⋅A\dot{m} = \rho \cdot V \cdot Am˙=ρ⋅V⋅A where ρ\rhoρ is the density of water. For simplicity, assume ρ≈1000\rho \approx 1000ρ≈1000 kg/m³ (density of water). m˙=1000⋅1⋅π400=1000π400=5π2 kg/s\dot{m} = 1000 \cdot 1 \cdot \frac{\pi}{400} = \frac{1000 \pi}{400} = \frac{5 \pi}{2} \text{ kg/s}m˙=1000⋅1⋅400π​=4001000π​=25π​ kg/s Now, calculate the change in momentum (Δp\Delta pΔp) of the water as it negotiates the bend. The force exerted on the bend is equal to the rate of change of momentum: F=m˙⋅VF = \dot{m} \cdot VF=m˙⋅V Substitute the values: F=5π2⋅1=5π2 NF = \frac{5 \pi}{2} \cdot 1 = \frac{5 \pi}{2} \text{ N}F=25π​⋅1=25π​ N Therefore, the force in Newtons exerted on the bend by water is 5π2\frac{5 \pi}{2}25π​ N. So, the correct answer is: 5π2\frac{5 \pi}{2}25π​ This represents the force exerted by the water on the bend due to its change in momentum as it flows through the pipe.