$A$ shell having a hole of radius $r$ is dipped in water. It holds the water up to a depth of $h$. Then the value of $r$ is:

$A$ soap bubble in vacuum has a radius of $3 \, cm$ and another soap bubble in vacuum has a radius of $4 \, cm$. If the two bubbles coalesce under isothermal condition,then the radius of the new bubble is ....... $cm$.

When $10^6$ small drops coalesce to make a new larger drop,then the drop:

Formation of bubbles is in Column-$I$ and the pressure difference between them is given in Column-$II$. Match them appropriately.

Column-$I$	Column-$II$
$(a)$ Liquid drop in air	$(i)$ $\frac{4T}{R}$
$(b)$ Bubble of liquid in air	$(ii)$ $\frac{2T}{R}$
	$(iii)$ $\frac{2R}{T}$

What is the pressure inside the drop of mercury of radius $3.00 \; mm$ at room temperature? Surface tension of mercury at that temperature $(20 \; ^{\circ}C)$ is $4.65 \times 10^{-1} \; N m^{-1}$. The atmospheric pressure is $1.01 \times 10^{5} \; Pa$. Also,calculate the excess pressure inside the drop.

The excess pressure inside one soap bubble is four times that of another. What is the ratio of the volume of the first bubble to the second bubble?