The increase in pressure required to decrease the 400 litre volume of a liquid by 0.001% is (Bulk modulus of the liquid is \(2.1 \times 10^9 \text{ N/m}^2\))
Bulk modulus \(B = -\frac{\Delta P}{\Delta V/V}\). Ignoring the negative sign for magnitude, \(Delta P = B \frac{\Delta V}{V} = (2.1 \times 10^9) \times \left(\frac{0.001}{100}\right) = 2.1 \times 10^4 \text{ Pa} = 21 \text{ kPa}\).
Assertion (A): A raindrop after falling through some height attains a constant velocity.
Reason (R): At constant velocity the viscous drag plus buoyant force is just equal to its weight.
Assertion (A) is true, as falling objects in a fluid reach terminal velocity when resistive forces balance gravity.
Reason (R) is true, stating the force balance condition for constant velocity: Weight = viscous drag + buoyant force. (R) correctly explains (A).
Assertion (A): Water flows faster than honey.
Reason (R): The co-efficient of viscosity of water is less than honey.
Concept: Viscosity and fluid resistance. Viscosity is a measure of a fluid's resistance to flow. A fluid with lower viscosity flows more easily and thus faster. Water has a significantly lower coefficient of viscosity than honey. Both Assertion and Reason are true, and Reason correctly explains Assertion.
Assertion (A): The angle of contact of a liquid decreases with increase in temperature.
Reason (R): With increase in temperature, the surface tension of liquid increases.
Concept: Effect of temperature on liquid properties. Assertion (A) is true; generally, the angle of contact decreases with increasing temperature as intermolecular forces weaken. Reason (R) is false; surface tension of a liquid *decreases* with an increase in temperature, not increases, because the kinetic energy of molecules increases, reducing cohesive forces.
Assertion (A): The shape of a liquid drop is spherical.
Reason (R): The pressure inside the drop is greater than that of outside.
Concept: Surface tension and pressure difference. Assertion (A) is true because surface tension tends to minimize the surface area of a liquid for a given volume, and a sphere has the minimum surface area. Reason (R) is true; due to surface tension, there is an excess pressure inside a spherical liquid drop, given by \(P_{in} - P_{out} = \frac{2T}{R}\). This excess pressure balances the inward pull of surface tension, thus R correctly explains A.
Assertion (A): Surface energy of an oil drop is same whether placed on glass or water surface.
Reason (R): Surface energy is dependent only on the properties of oil.
Concept: Interfacial surface energy. Surface energy (or surface tension) is a property of the interface between two phases. It depends on the intermolecular forces between the molecules of *both* substances forming the interface. Therefore, an oil-glass interface will have different surface energy than an oil-water interface. Both Assertion and Reason are false.
A small iron needle placed slowly on the surface of water floats due to
The needle floats on the water surface because of surface tension, which creates a stretched membrane-like behavior on the surface of the water.
Eight drops of equal radii are falling through air with a steady velocity of \(3\text{ cm/s}\). If the eight drops combine to form a single drop, then its steady velocity will be
Terminal velocity \(v \propto r^2\). Since volume remains constant, \(\frac{4}{3}\pi R^3 = 8 \times \frac{4}{3}\pi r^3 \implies R = 2r\). Thus, the new terminal velocity is \(v' = \left(\frac{R}{r}\right)^2 v = 2^2 \times 3 = 12\text{ cm/s}\).
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 $Z_0 = \frac{4 \times 2.5 \times 10^{-2}}{10^{-3} \times 10^3 \times 10} = 10^{-2} \text{ m} = 1 \text{ cm}$.