The translational degrees of freedom ($f_{t}$) and rotational degrees of freedom $\left(\mathrm{f}_{\mathrm{r}}\right)$ of $\mathrm{CH}_{4}$ molecule are :

(B) $\mathrm{f}_{\mathrm{t}}=3$ and $\mathrm{f}_{\mathrm{r}}=3$

(A) $\mathrm{f}_{\mathrm{t}}=2$ and $\mathrm{f}_{\mathrm{r}}=2$

(C) $f_{t}=3$ and $f_{r}=2$

(D) $f_{t}=2$ and $f_{r}=3$

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Heat & Thermodynamics

The translational degrees of freedom (

f_{t}

) and rotational degrees of freedom

(f_{r})

CH_{4}

molecule are :

(A) $f_{t} = 2$ and $f_{r} = 2$
(B) $f_{t} = 3$ and $f_{r} = 3$ ✓ Correct
(C) $f_{t} = 3$ and $f_{r} = 2$
(D) $f_{t} = 2$ and $f_{r} = 3$

Answer: (B)

Solution

_{4}

is a non-linear, polyatomic (pentatomic) molecule. It has 3 translational degrees of freedom (motion of the centre of mass along the three axes) and, being non-linear, 3 rotational degrees of freedom (rotation about three independent axes). So

f_{t} = 3

and

f_{r} = 3

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Single correct option

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