Two unequal soap bubbles are formed one on each side of a tube closed in the middle by a tap. What happens when the tap is opened to put the two bubbles in communication ?
The radii of the two columns in U tube are r1 and r2 (r1 > r2). When a liquid of density \[\rho\] (angle of contact is 0º) is filled in it, the level difference of liquid in two arms is h. The surface tension of liquid is (g = acceleration due to gravity):
If force of cohesion of liquid is three times the force of adhesion between liquid and glass tube molecules, then the shape of meniscus of liquid in this glass tube is:
The terminal velocity of a copper ball of radius 2.0 mm falling through a tank of oil at 20°C is 6.5 cms–1. Calculate the viscosity of the oil at 20°C. (Density of oil is \[1.7\times 10^{3}kgm^{-3}\] and density of copper is \[8.9\times 10^{3}kgm^{-3}\]) :
A lead sphere of mass m falls in viscous liquid with terminal velocity v0. Another lead sphere of mass M falls through the same viscous liquid with terminal velocity 4v0. the ratio M/m is :
Two objects A & B of equal density and radius rA = 1 mm and rB = 2 mm are moving in same medium then find the ratio of their terminal velocity \( \frac{v_{B}}{v_{A}}\) in the medium.
The terminal velocity v of a spherical object moving through a viscous medium is directly proportional to the square of its radius, expressed as v
\(r^2 \), when the density and the medium remain constant.
\( \frac{v_B}{v_A} = \left(\frac{r_B}{r_A}\right)^2 = \left(\frac{2}{1}\right)^2 = 4 \)
Thus, the required ratio \( \frac{v_B}{v_A} \) is 4 (Option 3).
The viscous drag acting on a metal sphere of diameter \(1\text{ mm}\), falling through a fluid of viscosity \(0.8\text{ Pa s}\) with a velocity of \(2\text{ m s}^{-1}\) is equal to
According to Stokes' law, \(F = 6\pi \eta r v\). Here \(r = 0.5 \times 10^{-3}\text{ m}\), \(\eta = 0.8\text{ Pa s}\), and \(v = 2\text{ m s}^{-1}\). Calculating, we get \( F = 6 \times 3.14 \times 0.8 \times 0.5 \times 10^{-3} \times 2 \approx 15 \times 10^{-3}\text{ N}\).
If surface tension of a liquid is \(T\), then excess force required to lift a ring of radius \(R\) from surface of this liquid is:
A ring has two free surfaces in contact with the liquid (inner and outer boundaries). Therefore, the total contact length is \(L = 2(2\pi R) = 4\pi R\), and the excess force required is \(F = T \times 4\pi R\).
Given below are two statements:
Assertion (A): When two soap bubbles of different radii are brought into contact, the common interface of contact bulges into the bubble of larger radii.
Reason (R): Pressure inside a soap bubble of a lesser radius is more than pressure inside the soap bubble of a larger radius.
The excess pressure inside a soap bubble is \(P = \frac{4T}{R}\). Thus, pressure is greater inside the smaller bubble (lesser radius). When in contact, the common interface bulges towards the larger bubble due to this pressure difference. Both are true and Reason explains Assertion.
An object of mass m and density \(\rho\) is falling in a viscous liquid of density \(\frac{\rho}{4}\). When it attains terminal velocity, then viscous force acting on it will be:
At terminal velocity, the upward forces balance the downward force: \(F_v + F_b = mg\). Since buoyant force is \(F_b = V \rho_l g = \left(\frac{m}{\rho}\right)\left(\frac{\rho}{4}\right)g = \frac{mg}{4}\), we get \(F_v = mg - \frac{mg}{4} = \frac{3mg}{4}\).