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 $\mathrm{k}=1.38 \times 10^{-23} \mathrm{JK}^{-1}$]

(C) $15 \mathrm{~mm} \mathrm{~s}^{-1}$

(A) $60 \mathrm{~mm} \mathrm{~s}^{-1}$

(B) $12 \mathrm{~mm} \mathrm{~s}^{-1}$

(D) $36 \mathrm{~mm} \mathrm{~s}^{-1}$

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Heat & Thermodynamics

The root mean square speed of smoke particles of mass

5 \times 1 0^{- 17} kg

in their Brownian motion in air at NTP is approximately. [Given

k = 1.38 \times 1 0^{- 23} JK^{- 1}

]

(A) $60 mm s^{- 1}$
(B) $12 mm s^{- 1}$
(C) $15 mm s^{- 1}$ ✓ Correct
(D) $36 mm s^{- 1}$

Answer: (C)

Solution

The root mean square speed is given by:

v_{r m s} = \frac{3 k T}{m}

At NTP, T ≈ 293 K. Substituting the values:

v_{r m s} = \frac{3 \times 1.38 \times 1 0 ^{- 23} \times 293}{5 \times 1 0 ^{- 17}}

Calculating:

v_{r m s} = \frac{1.213 \times 1 0 ^{- 20}}{5 \times 1 0 ^{- 17}} = 2.426 \times 1 0^{- 4} \approx 0.015 m/s = 15 mm/s

The answer is (C) 15 mm/s.

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Single correct option

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