Exams › JEE Advanced › Physics
A bent tube is submerged in a water stream moving at velocity V = 4 m/s relative to the tube. The closed upper end of the tube is at height h0 = 20 cm above the free surface of the water. There is a small orifice at the closed upper end. Find the height h (in cm) to which water spurts above the free surface. If x = h, find x/10.
- 6
- 8
- 10
- 12
Correct answer: 8
Solution
The bent tube acts as a Pitot tube. The open end faces the oncoming stream, creating a stagnation pressure. By Bernoulli's equation between the free stream and the stagnation point: P_atm + (1/2)*rho*V² = P_stagnation. The stagnation pressure supports a water column of height H above the free surface: P_stagnation = P_atm + rho*g*H. So rho*g*H = (1/2)*rho*V², giving H = V²/(2g) = 16/(2*10) = 0.8 m = 80 cm above the free surface. But the closed end with orifice is at h0 = 20 cm above free surface. The water will spurt through the orifice with velocity determined by the excess pressure. Pressure at top of tube (closed end, height h0): P_top = P_stagnation - rho*g*h0 = P_atm + rho*g*H - rho*g*h0 = P_atm + rho*g*(H - h0). Water spurts out and rises to height h where P_atm + rho*g*(H - h0) = P_atm + rho*g*h (above the orifice exit). So h (above orifice) = H - h0 = 80 - 20 = 60 cm. Total height above free surface = h0 + h_above_orifice = 20 + 60 = 80 cm. So x = 80 cm, x/10 = 8.
Related JEE Advanced Physics questions
- A capillary tube with a square-shaped internal cross-section, each side of length 'a', is inserted vertically into a liquid of density ρ and surface tension σ. If the liquid adheres to the tube walls with a contact angle θ, approximately how high will the liquid rise inside the tube? (Ignore the surface tension effects at the corners of the tube.)
- A hot air balloon with a total mass of 480 kg, including passengers and 1 kg sandbags, is floating steadily at a height of 100 m. The balloon's buoyancy is determined by its fixed volume, V. If N sandbags are removed, the balloon ascends to a stable height of approximately 150 m. The air density changes with altitude according to ρ(h) = ρ₀e⁻ʰ/ʰ₀, where ρ₀ = 1.25 kg/m³ and h₀ = 6000 m. What is the value of N?
- A container accelerates horizontally to the left with acceleration g/2. A uniform rod of length L is partially submerged vertically in a liquid contained in the accelerating container. The density of the liquid is rho and the density of the rod is rho/2. If the length of the rod submerged in the liquid is sqrt(a/b)*L, where a and b are the least positive integers (in lowest terms), find (a + b).
- A vertical U-tube (communicating tube) contains a liquid of density rho and surface tension T. The tube has inner radius r. When the entire system accelerates horizontally with acceleration a, the pressure at point A (at the liquid surface in the left arm, taking the meniscus correction into account) is given by which of the following expressions? (Take atmospheric pressure as P0, and assume the contact angle is 0 deg.)
- A space 2.5 cm wide between two large parallel plane surfaces is filled with oil. The force required to drag a very thin plate of area 0.5 m² at a speed of 0.5 m/s, placed exactly midway between the two surfaces, is 1 N. Find the coefficient of viscosity of the oil.
- A ball of mass 10 kg and density 1 g/cm³ is attached to the base of a container by a spring (spring constant k = 200 N/m). The container is filled with a liquid of density 1.1 g/cm³. The entire system accelerates upward at 2 m/s². Find the elongation (extension) of the spring. (Take g = 10 m/s²)
⚔️ Practice JEE Advanced Physics free + battle 1v1 →