Exams › JEE Main › Physics
In a siphon, the reservoir's surface area is far larger than the uniform tube's cross-section. The lower (outlet) end of the tube is 6 m below the water surface. Taking water density = 10³ kg/m³ and atmospheric pressure = 10⁵ N/m², what is the maximum height H (measured above the water surface) of the highest point of the siphon for it to keep working?
- 1 m
- 6 m
- 10 m
- 16 m
Correct answer: 10 m
Solution
A siphon works as long as the pressure at its highest point stays positive. The greatest height of the crest above the reservoir surface that atmospheric pressure can sustain is H = P_atm/(rho*g) = 10⁵/(10³ * 10) = 10 m.
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 →