Exams › JEE Main › Physics
Water emerges drop by drop from a dropper of orifice radius r = 5 * 10⁻⁴ m. Taking density of water rho = 10³ kg/m³, g = 10 m/s² and surface tension T = 0.11 N/m, estimate the radius of a drop at the instant it just detaches from the dropper.
- 1.4 * 10⁻³ m
- 3.3 * 10⁻³ m
- 2.0 * 10⁻³ m
- 4.1 * 10⁻³ m
Correct answer: 1.4 * 10⁻³ m
Solution
Just before detachment, the surface tension acting along the circle of contact (length 2*pi*r) supports the drop's weight. Equating the maximum upward surface-tension force to the drop's weight gives the detachment radius R.
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 →