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The root mean square speed of smoke particles of mass 5 × 10⁻¹⁷ kg in their Brownian motion in air at NTP is approximately. [Given k = 1.38 × 10⁻²³ J K⁻¹]

Asked in JEE Main 29th July 2nd Shift 2022 · Escape speed and Brownian motion

Answer: (3) 15 mm s⁻¹

Step-by-step solution

A smoke particle in Brownian motion shares the thermal energy of the air, so the same formula applies.

vᵣₘₛ=√(3k_BT)/m with T=273 K at NTP.

3k_BT=3×1.38×10⁻²³×273=1.13×10⁻²⁰.

(3k_BT)/m=(1.13×10⁻²⁰)/(5×10⁻¹⁷)=2.26×10⁻⁴.

vᵣₘₛ=√2.26×10⁻⁴=1.5×10⁻² m/s=15 mm/s.

Why the other options are wrong

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