NEET Practice Questions (MCQs) with Answers & Solutions

$ Surface energy = Surface tension \times surface ared $ $ E = T \times 2 A $ $ New surface energy F_1 = T \times 2 \left( { A \over 2} \right) $ $ \% decrase in surface energy = { E - E_i \over E } \times 100 $

The ration of the total surface energies before and after the change $ = n^{1 \over 2 } : 1 = 2 ^ {1 \over 3} : 1 $

$ w = 8 \pi R^2 T $

$ { 4 \over 3} \pi R^ 3 $

$ As volume remains constant R^3 = 8000 r^ 3 R = 20 r $ $ {Surface energy of one big drop \over Surface energy of 8000 small drop } = - { 4 \pi R ^2 T \over 8000 4 \pi r^2 -1 } $

$ Tension = { surface energy \over Area } = or T ={ E \over A } $

A liquid that wets a solid completely will have a concave meniscus in a sufficiently long tube. This is because the adhesive forces between the liquid and the solid are stronger than the cohesive forces within the liquid, causing the liquid to climb up the walls of the tube, forming a concave shape.

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 $R^2 = r_1^2 + r_2^2$.