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?

$A$ big drop is formed by coalescing $1000$ small identical drops of water. If $E_1$ be the total surface energy of $1000$ small drops of water and $E_2$ be the surface energy of the single big drop of water,then the ratio $E_1 : E_2$ is $x : 1$,where $x = . . . . . . $.

When a big drop of water is formed from $n$ small drops of water, the energy loss is $3E$, where $E$ is the energy of the bigger drop. If $R$ is the radius of the bigger drop and $r$ is the radius of the smaller drop, then the number of smaller drops $(n)$ is:

$A$ soap bubble, blown by a mechanical pump at the mouth of a tube, increases in volume with time at a constant rate. The graph that correctly depicts the time dependence of pressure inside the bubble is given by

Two soap bubbles, $A$ and $B$, have radii in the ratio $3 : 2$. The ratio of the excess pressure inside bubble $A$ to that of bubble $B$ is:

Under isothermal conditions,two soap bubbles of radii $r_1$ and $r_2$ coalesce to form a single soap bubble of radius $R$. The radius of the new bubble is