Mechanical Properties of Fluids MCQs for NEET — Physics Questions with Answers

When two soap bubbles coalesce, the radius of curvature of the common surface is given by the formula $ \\frac{r_1 r_2}{r_2 - r_1}$. This is derived from the balance of pressures inside and outside the bubbles and the properties of soap films.

The excess pressure inside a soap bubble is given by the formula: $\Delta P = \frac{4T}{r}$, where $T$ is the surface tension and $r$ is the radius of the bubble. This is because a soap bubble has two surfaces (inner and outer), contributing to the factor of 4 in the formula.

When two soap bubbles combine, the volume is conserved. The relationship between the radii of the initial bubbles and the resultant bubble is given by $R^3 = r_1^3 + r_2^3$. Simplifying for radius in terms of the areas, we get $R^2 = r_1^2 + r_2^2$.

$ \triangle p = { 2T \over R } $

$ r_A \gt r_B and p \alpha {1 \over r } $ $ So P_A \lt P_B $ So air will flow from B to A i.e. size ofA will increase

Mass of liquid in capillary tube $ M = \pi R^2 - \rho $ $ M \alpha R^2 \times (1 / R } $ $ M \alpha R $ If radius become double then mass will become twice.

The correct relation for capillary rise is $r = rac{2T ext{ cos } heta}{hdg}$. Here, $r$ is the radius, $T$ is the surface tension, $ heta$ is the contact angle, $h$ is the height of the liquid column, $d$ is the density of the liquid, and $g$ is the acceleration due to gravity. Hence, option o1 is correct.

$ h = { 2T \over rdg} $ $ \therefore h \alpha { 1 \over r }$ $ \therefore r_1 h_1 = r_2 h_2 $