Q 11-09-133JEE MainJEE Main 2020 (3 Sep, Shift 1)Easy
Pressure inside two soap bubbles are $1.01$ and $1.02$ atmosphere, respectively. The ratio of their volumes is:
Answer: (C) $8 : 1$
Excess pressure $\Delta P = \dfrac{4S}{r}$, so $r \propto \dfrac{1}{\Delta P}$. With outside pressure $1$ atm, $\Delta P_1 = 0.01$ atm and $\Delta P_2 = 0.02$ atm, so $r_1 : r_2 = 2 : 1$.
$V_1 : V_2 = 2^{3} : 1 = 8 : 1$.
Solution by Sreeraj P, M.Sc Physics