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 :
A long capillary is dipped in a beaker containing water. Water rises in capillary upto some height \(h\). Match the statements in list-I with most appropriate effects on water level mentioned in list-II:
**List-I**
(A) Soap solution is added to water
(B) Arrangement taken in a freely falling lift
(C) In a lift accelerating uniformly upward
(D) Arrangement is taken in a lift accelerating uniformly downward
**List-II**
(p) \(h\) decreases
(q) \(h\) increases
(r) \(h\) remains same
(s) water will rise upto complete height of capillary
(t) water level in capillary goes below the outside level
Soap reduces surface tension, so \(h\) decreases (A-p). In a free fall, effective gravity \(g_{eff} = 0\), so water rises to full height (B-s). Upward acceleration increases \(g_{eff}\) hence \(h\) decreases (C-p). Downward acceleration decreases \(g_{eff}\) hence \(h\) increases (D-q).
The velocity of a small ball of mass \(M\) and density \(d\), when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is \(\frac{d}{2}\), then the viscous force acting on the ball will be
When the ball reaches terminal velocity, net force is zero: \(F_v + F_B = Mg\). The buoyant force is \(F_B = V \rho_{\text{glycerine}} g = V \left(\frac{d}{2}\right) g = \frac{Mg}{2}\). Thus, the viscous force is \(F_v = Mg - \frac{Mg}{2} = \frac{Mg}{2}\).
If a soap bubble expands, the pressure inside the bubble :
(2022)
The excess pressure inside a soap bubble is given by $\Delta P = \frac{4T}{R}$. As the bubble expands, its radius $R$ increases. Therefore, the excess pressure (and total pressure inside) decreases.
A soap bubble, having radius of $1 \text{ mm}$, is blown from a detergent solution having a surface tension of $2.5 \times 10^{-2} \text{ N/m}$. The pressure inside the bubble equals at a point $Z_0$ below the free surface of water in a container. Taking $g = 10 \text{ m/s}^2$, density of water $= 10^3 \text{ kg/m}^3$, the value of $Z_0$ is :
(2019)
Pressure inside the bubble is $P = P_0 + \frac{4T}{r}$. Pressure at depth $Z_0$ is $P = P_0 + \rho g Z_0$. Equating them, $\rho g Z_0 = \frac{4T}{r}$. Solving gives $Z_0 = \frac{4 \times 2.5 \times 10^{-2}}{10^{-3} \times 10^3 \times 10} = 10^{-2} \text{ m} = 1 \text{ cm}$.
A rectangular film of liquid is extended from $(4 \text{ cm} \times 2 \text{ cm})$ to $(5 \text{ cm} \times 4 \text{ cm})$. If the work done is $3 \times 10^{-4} \text{ J}$, the value of the surface tension of the liquid is:
(2016 – II)
Work done in stretching a liquid film is $W = T \times 2\Delta A$ (since it has two surfaces). The change in area is $\Delta A = (5 \times 4) - (4 \times 2) = 12 \text{ cm}^2 = 12 \times 10^{-4} \text{ m}^2$. Thus, $$T = \frac{3 \times 10^{-4}}{2 \times 12 \times 10^{-4}} = 0.125 \text{ Nm}^{-1}$$.
A liquid does not wet the solid surface if angle of contact is:
(2020-Covid)
For a liquid to not wet a solid surface, it must form an obtuse angle of contact. Therefore, the angle of contact must be greater than $90^\circ$.
A capillary tube of radius $r$ is immersed in water and water rises in it to a height $h$. The mass of the water in the capillary is $5\text{ g}$. Another capillary tube of radius $2r$ is immersed in water. The mass of water that will rise in this tube is:
(2020)
Mass of water risen in capillary $m = \pi r^2 h \rho$. Since $h \propto \frac{1}{r}$, we get $m \propto r$. Thus, $m_2 = m_1 \left(\frac{r_2}{r_1}\right) = 5 \times \left(\frac{2r}{r}\right) = 10\text{ g}$.
Three liquids of densities $\rho_1$, $\rho_2$ and $\rho_3$ (with $\rho_1 > \rho_2 > \rho_3$), having the same value of surface tension $T$, rise to the same height in three identical capillaries. The angles of contact $\theta_1$, $\theta_2$ and $\theta_3$ obey:
(2016 – II)
Capillary rise $h = \frac{2T \cos\theta}{r \rho g}$. Since $h, T, r, g$ are constant, $\cos\theta \propto \rho$. Since $\rho_1 > \rho_2 > \rho_3$, $\cos\theta_1 > \cos\theta_2 > \cos\theta_3$. For acute angles, this means $$0 \le \theta_1 < \theta_2 < \theta_3 < \frac{\pi}{2}$$.