Pure water rises through a height $h$ in a capillary tube of internal radius $r$. Surface tension of water is $T$. The pressure difference between the water level in the container and the lowest point of the concave meniscus is

$A$ spherical soap bubble inside an air chamber at pressure $P_0 = 10^5 \ Pa$ has a certain radius $R$ such that the excess pressure inside the bubble is $\Delta P = 144 \ Pa$. Now,the chamber pressure is reduced to $8P_0 / 27$ so that the bubble radius and its excess pressure change. In this process,all temperatures remain unchanged. Assume air to be an ideal gas and the excess pressure $\Delta P$ in both cases to be much smaller than the chamber pressure. The new excess pressure $\Delta P'$ in $Pa$ is:

The radii of two soap bubbles are $r_1$ and $r_2$. In isothermal condition,they combine with each other to form a single bubble. The radius of the resultant bubble is

Two soap bubbles of radii $2 \, cm$ and $4 \, cm$ join to form a double bubble in air. The radius of curvature of the interface is .......... $cm$.

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

When two soap bubbles of radii $a$ and $b$ $(b > a)$ coalesce,the radius of curvature of the common surface is: