Exams › JEE Main › Physics
A liquid of density ρ is coming out of a hose pipe of radius a with a horizontal speed v and hits a mesh. 50% of the liquid passes through the mesh unaffected. 25% loses all of its momentum and 25% comes back with the same speed. The resultant pressure on the mesh will be:
- 3/4 ρv²
- 1/4 ρv²
- 1/2 ρv²
- ρv²
Correct answer: 3/4 ρv²
Solution
Mass rate mdot = rho*pi*a^2*v. The 25% that stops contributes 0.25*mdot*v and the 25% that rebounds contributes 0.25*mdot*2v; total force = 0.75*mdot*v. Pressure = Force/(pi*a^2) = (3/4)*rho*v^2.
Related JEE Main Physics questions
- A body is in motion inside a liquid, and the viscous resistive force on it is directly proportional to its speed. What are the dimensions of the proportionality constant?
- A capillary tube has its inner surface lined with wax and is then placed in water. Relative to a clean, unwaxed capillary, how do the contact angle (θ) and the height (h) to which water rises change?
- A charged, isolated spherical soap bubble of radius r has internal pressure equal to atmospheric pressure. If the charge on the bubble is given by Xπ√(2Tε), what is the value of X?
- Two capillary tubes, one of length L and radius R and the other of length 2L and radius 2R, are joined one after the other in series. If the flow rate through a single capillary is X = πPR⁴ / 8ηL, then the combined flow rate through the series arrangement is:
- A liquid will fail to wet the surface of a solid when the angle of contact is
- A gold sphere of a given size falls through a viscous liquid and attains a terminal speed of 0.2 m/s. If the gold has density 19.5 kg/m³ and the liquid has density 1.5 kg/m³, what terminal speed will a silver sphere of the same size have in the same liquid, given that silver has density 10.5 kg/m³?
⚔️ Practice JEE Main Physics free + battle 1v1 →