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:

The surface tension of a soap solution is $2 \times 10^{-2} \ N/m$. To blow a bubble of radius $1 \ cm$,the work done is:

Three small identical bubbles of water having the same charge on each coalesce to form a bigger bubble. Then the ratio of the potentials on one initial bubble and that on the resultant bigger bubble is:

$A$ soap bubble of radius $1.0 \text{ cm}$ is formed inside another soap bubble of radius $2.0 \text{ cm}$. The radius of another soap bubble which has the same pressure difference as that between the inside of the smaller and outside of the large soap bubble, in metres is:

Under isothermal conditions,two soap bubbles of radii $r_{1}$ and $r_{2}$ coalesce to form a single larger bubble. The radius of the larger bubble is

Let $R_{1}$ and $R_{2}$ be the radii of two mercury drops. $A$ big mercury drop is formed from them under isothermal conditions. The radius of the resultant drop is