Two gases A and B are filled at the same pressure in separate cylinders with movable pistons of radius $r_A$ and $r_B$, respectively. On supplying an equal amount of heat to both gases reversibly at constant pressure, the pistons of A and B are displaced by $16$ cm and $9$ cm, respectively. If the change in their internal energy is the same, then the ratio $\dfrac{r_A}{r_B}$ is

Solution & Explanation

\textbf{Answer:} (d) $\dfrac{3}{4}$ \textbf{Solution:} By the first law, $Q=\Delta U+W$. Both gases receive equal heat $Q$ and have equal change in internal energy $\Delta U$, so the work done at constant pressure must be equal: $$W_A=W_B.$$ The work done in displacing a piston of area $A=\pi r^2$ by a distance $d$ at constant pressure $P$ is $$W=P\,\Delta V=P\,(\pi r^2)\,d.$$ Since the pressures are equal, $W_A=W_B$ gives $$r_A^2\,d_A=r_B^2\,d_B\quad\Rightarrow\quad r_A^2(16)=r_B^2(9).$$ Therefore $$\frac{r_A^2}{r_B^2}=\frac{9}{16}\quad\Rightarrow\quad \frac{r_A}{r_B}=\frac{3}{4}.$$ This matches option (d).