Water is entering at pressure 4 × 104 pascal with a velocity of 2m/s in a horizontal pipe with cross-sectional area decreasing from 2 × 10–2 m2 to 0.01 m2 . The pressure at smaller cross-section of pipe  in pascal will be :-323.43.4 × 1043.4 × 105

To find the pressure at the smaller cross-section of the pipe, we can use the principle of conservation of energy.

Step 1: Calculate the initial kinetic energy of the water entering the pipe. The initial kinetic energy (KE1) can be calculated using the formula: KE1 = (1/2) * m * v^2 where m is the mass of the water and v is the velocity of the water. Given that the velocity is 2 m/s, we need to find the mass of the water. We can use the formula: m = density * volume Given that the density of water is approximately 1000 kg/m^3 and the initial cross-sectional area is 2 × 10^(-2) m^2, we can calculate the volume: volume = area * length where length is the length of the pipe. Since the pipe is horizontal, the length does not affect the calculation, so we can assume it to be 1 meter for simplicity. Therefore, the volume is: volume = (2 × 10^(-2) m^2) * 1 m = 2 × 10^(-2) m^3 Now we can calculate the mass: m = (1000 kg/m^3) * (2 × 10^(-2) m^3) = 20 kg Now we can calculate the initial kinetic energy: KE1 = (1/2) * (20 kg) * (2 m/s)^2 = 40 J

Step 2: Calculate the final kinetic energy of the water at the smaller cross-section. The final kinetic energy (KE2) can be calculated using the same formula as before, but with the velocity at the smaller cross-section. Given that the velocity at the smaller cross-section is unknown, we can use the principle of conservation of mass to find it. The principle of conservation of mass states that the mass flow rate is constant in an incompressible fluid. Therefore, we can use the formula: m1 * v1 = m2 * v2 where m1 and v1 are the mass and velocity at the initial cross-section, and m2 and v2 are the mass and velocity at the smaller cross-section. We already know m1 and v1 from the previous calculations, so we can rearrange the equation to solve for v2: v2 = (m1 * v1) / m2 Given that m1 = 20 kg and v1 = 2 m/s, and the final cross-sectional area is 0.01 m^2, we can calculate the mass at the smaller cross-section: m2 = density * volume volume = area * length Assuming the length is still 1 meter, we can calculate the volume: volume = (0.01 m^2) * 1 m = 0.01 m^3 Now we can calculate the mass: m2 = (1000 kg/m^3) * (0.01 m^3) = 10 kg Now we can calculate the velocity at the smaller cross-section: v2 = (20 kg * 2 m/s) / 10 kg = 4 m/s Now we can calculate the final kinetic energy: KE2 = (1/2) * (10 kg) * (4 m/s)^2 = 80 J

Step 3: Calculate the change in kinetic energy. The change in kinetic energy (ΔKE) can be calculated by subtracting the initial kinetic energy from the final kinetic energy: ΔKE = KE2 - KE1 = 80 J - 40 J = 40 J

Step 4: Calculate the change in pressure. The change in pressure (ΔP) can be calculated using the formula: ΔP = (1/2) * ρ * Δv^2 where ρ is the density of the fluid and Δv is the change in velocity. Given that the density of water is 1000 kg/m^3 and the change in velocity is v2 - v1 = 4 m/s - 2 m/s = 2 m/s, we can calculate the change in pressure: ΔP = (1/2) * (1000 kg/m^3) * (2 m/s)^2 = 2000 Pa

Step 5: Calculate the pressure at the smaller cross-section. The pressure at the smaller cross-section can be calculated by adding the change in pressure to the initial pressure: P2 = P1 + ΔP Given that the initial pressure is 4 × 10^4 Pa, we can calculate the pressure at the smaller cross-section: P2 = (4 × 10^4 Pa) + (2000 Pa) = 4.02 × 10^4 Pa

Therefore, the pressure at the smaller cross-section of the pipe is approximately 4.02 × 10^4 Pa.