Q63 / 90 · JEE Main 2021 (27 Aug Shift 2) · Physics

If the R.M.S. speed of oxygen molecules at $0 ° C$ is $160 m s^{- 1}$ . Find the R.M.S. speed of hydrogen molecules at $0 ° C$ .

Solution

Recall the formula of root mean square speed, $v_{R M S} = \sqrt{\frac{3 R T}{M}}$ ,

$\Rightarrow \frac{{(v_{R M S})}_{O_{2}}}{{(v_{R M S})}_{H_{2}}} = \sqrt{\frac{M_{H_{2}}}{M_{O_{2}}}} = \sqrt{\frac{2}{32}} = \frac{1}{4}$

$\Rightarrow {(v_{RMS})}_{H_{2}} = 4 {(v_{RMS})}_{O_{2}} = 4 \times 160 = 640 {m s}^{- 1}$

Hence, the correct answer is option: 'C'.