Q57 / 75 · JEE Main 2025 (2 Apr Shift 1) · Physics
$\gamma_{\mathrm{A}}$ is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. $\gamma_B$ is the specific heat ratio of polyatomic gas B having 3 translational, 3 rotational degrees of freedom and 1 vibrational mode. If $\frac{\gamma_{\mathrm{A}}}{\gamma_{\mathrm{B}}}=\left(1+\frac{1}{\mathrm{n}}\right)$, then the value of $n$ is _______.
Correct answer: 3
Solution
$\begin{aligned} & \frac{\gamma_A}{\gamma_B}=\frac{\mathrm{f}_{\mathrm{A}}+2}{\mathrm{f}_{\mathrm{A}}} \times \frac{\mathrm{f}_{\mathrm{B}}}{\mathrm{f}_{\mathrm{B}}+2} \\ & =\frac{3+2}{3} \times \frac{(6+2)}{(6+2)+2} \\ & =\frac{5}{3} \times \frac{8}{10}=\frac{40}{30} \\ & \because \frac{40}{30}=1+\frac{1}{\mathrm{n}} \\ & \Rightarrow \frac{40}{30}-1=\frac{1}{\mathrm{n}}\end{aligned}$
$\Rightarrow \mathrm{n}=3$
$\Rightarrow \mathrm{n}=3$
Hence, the correct answer is 3.