RRB Technician Grade 1 Model Paper 1

Concept:Excess pressure inside a soap bubble is inversely proportional to its radius: $P = \frac{4T}{R}$.Since $T$ is the same for both bubbles, the bubble with smaller radius has larger excess pressure.Explanation:Pressure outside each bubble is atmospheric pressure, $P_o = 1$ atm.For the first bubble, inside pressure is $P_1' = 1.01$ atm.So excess pressure: $P' = P_i' - P_o = 1.01 - 1 = 0.01$ atm.For the second bubble, inside pressure is $P_1'' = 1.02$ atm.So excess pressure: $P'' = P_i'' - P_o = 1.02 - 1 = 0.02$ atm.Using $P = \frac{4T}{R}$, we get $P \propto \frac{1}{R}$.Let the radii be $R_1$ and $R_2$. Then $\frac{P'}{P''} = \frac{R_2}{R_1}$.So $\frac{0.01}{0.02} = \frac{R_2}{R_1}$, which gives $\frac{R_1}{R_2} = \frac{0.02}{0.01} = 2$.Volume of a spherical bubble is $V = \frac{4}{3}\pi R^3$, so $V \propto R^3$.Therefore, $\frac{V_1}{V_2} = \left(\frac{R_1}{R_2}\right)^3 = (2)^3 = 8$.Thus, the ratio of their volumes is $8 : 1$.Answer:$8 : 1$ (Option C).