Exams › JEE Main › Physics
A light straight wire AB of length 10 cm is free to slide along a vertical frame. A soap film is held between the frame and the wire. What load W must be hung from the wire so that it stays in equilibrium? Ignore friction. Surface tension of the soap solution is 25 dyne/cm and g = 10 m/s².
- 5 * 10⁻³ N
- 2.5 * 10⁻³ N
- 5 * 10⁻⁴ N
- 10 * 10⁻³ N
Correct answer: 5 * 10⁻³ N
Solution
The soap film pulls the wire up with force from both of its surfaces: F = 2*T*L. Convert units: T = 25 dyne/cm = 25*10⁻³ N / 10⁻² m = 2.5*10⁻² N/m? Better keep CGS: T = 25 dyne/cm, L = 10 cm, so F = 2*25*10 = 500 dyne = 500*10⁻⁵ N = 5*10⁻³ N. For equilibrium W = F = 5*10⁻³ N.
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 →