Chapter 8: Problem 82
Surface tension of the soap solution is (A) \(\frac{m g}{4 \ell \tan \frac{\theta}{2}}\) (B) \(\frac{m g}{2 \ell \tan \frac{\theta}{2}}\) (C) \(\frac{m g}{4 \ell \tan \theta}\) (D) None of these
/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}
Learning Materials
Features
Discover
Chapter 8: Problem 82
Surface tension of the soap solution is (A) \(\frac{m g}{4 \ell \tan \frac{\theta}{2}}\) (B) \(\frac{m g}{2 \ell \tan \frac{\theta}{2}}\) (C) \(\frac{m g}{4 \ell \tan \theta}\) (D) None of these
All the tools & learning materials you need for study success - in one app.
Get started for free
A body of mass \(0.5 \mathrm{~kg}\) is attached to a thread and it just floats in a liquid. The tension in the thread is (A) \(0.5 \mathrm{~kg} \mathrm{wt}\) (B) More than \(0.5 \mathrm{~kg}\) wt (C) Less than \(0.5 \mathrm{~kg} \mathrm{wt}\) (D) Zero
A uniform rod of length \(L\) has a mass per unit length \(\lambda\) and area of cross-section \(A .\) The elongation in the rod is \(l\) due to its own weight if it is suspended from the ceiling of a room. The Young's modulus of the rod is (A) \(\frac{2 \lambda g L^{2}}{A l}\) (B) \(\frac{\lambda g L^{2}}{2 A l}\) (C) \(\frac{2 \lambda g L}{A l}\) (D) \(\frac{\lambda g l^{2}}{A L}\)
One end of a uniform wire of length \(L\) and of weight \(W\) is attached rigidly to a point in the roof, and a weight \(W_{1}\) is suspended from its lower end. If \(S\) is the area of cross-section of the wire, the stress in the wire at a height ( \(3 L / 4\) ) from its lower end is (A) \(W_{1} / S\) (B) \(\left[W_{1}+(W / 4)\right] / S\) (C) \(\left[W_{1}+(3 W / 4)\right] / S\) (D) \(\left[W_{1}+W\right] / S\)
Two opposite forces \(F_{1}=120 \mathrm{~N}\) and \(F_{2}=80 \mathrm{~N}\) act on an elastic plank of modulus of elasticity \(Y=2 \times 10^{11}\) \(\mathrm{N} / \mathrm{m}^{2}\) and length \(\ell=1 \mathrm{~m}\) placed over a smooth horizontal surface. The cross-sectional area of the plank is \(S=0.5 \mathrm{~m}^{2}\). The change in length of the plank is \(x \times\) \(10^{-11} \mathrm{~m}\), then find the value of \(x\).
Which of the following has greatest viscosity? (A) Hydrogen (B) Air (C) Water (D) Ammonia
What do you think about this solution?
We value your feedback to improve our textbook solutions.