JEE MAIN - Physics (2022 - 29th July Evening Shift - No. 17)
The root mean square speed of smoke particles of mass $$5 \times 10^{-17} \mathrm{~kg}$$ in their Brownian motion in air at NTP is approximately. [Given $$\mathrm{k}=1.38 \times 10^{-23} \mathrm{JK}^{-1}$$]
$$60 \mathrm{~mm} \mathrm{~s}^{-1}$$
$$12 \mathrm{~mm} \mathrm{~s}^{-1}$$
$$15 \mathrm{~mm} \mathrm{~s}^{-1}$$
$$36 \mathrm{~mm} \mathrm{~s}^{-1}$$
Explanation
At NTP, $T=298 \mathrm{~K}$
$$ \begin{aligned} v_{\mathrm{rms}} &=\sqrt{\frac{3 R T}{M}} \\ &=\sqrt{\frac{3 k N_{A} \times 298}{5 \times 10^{-17} \times N_{A}}} \end{aligned} $$
$\simeq 15 \mathrm{~mm} / \mathrm{s}$
$$ \begin{aligned} v_{\mathrm{rms}} &=\sqrt{\frac{3 R T}{M}} \\ &=\sqrt{\frac{3 k N_{A} \times 298}{5 \times 10^{-17} \times N_{A}}} \end{aligned} $$
$\simeq 15 \mathrm{~mm} / \mathrm{s}$
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