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Heat and Thermodynamics question

2022 · 29 Jul · Shift 2 · Q60
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  5. /2022 · 29 Jul · Shift 2 · Q60

Heat and Thermodynamics question

2022 · 29 Jul · Shift 2 · Q60

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
The root mean square speed of smoke particles of mass 5×10−17 kg5 \times 10^{-17} \mathrm{~kg}5×10−17 kg in their Brownian motion in air at NTP is approximately. [Given k=1.38×10−23JK−1\mathrm{k}=1.38 \times 10^{-23} \mathrm{JK}^{-1}k=1.38×10−23JK−1]
  1. A
    60 mm s−160 \mathrm{~mm} \mathrm{~s}^{-1}60 mm s−1
  2. B
    12 mm s−112 \mathrm{~mm} \mathrm{~s}^{-1}12 mm s−1
  3. C
    15 mm s−115 \mathrm{~mm} \mathrm{~s}^{-1}15 mm s−1
  4. D
    36 mm s−136 \mathrm{~mm} \mathrm{~s}^{-1}36 mm s−1
View written solutionFree

Correct answer: C

  1. Use the equipartition result for rms speed

For a particle of mass mmm in thermal equilibrium at temperature TTT, the root mean square speed is

vrms=3kTmv_{\text{rms}} = \sqrt{\frac{3kT}{m}}vrms​=m3kT​​

where:

  • k=1.38×10−23 J K−1k = 1.38 \times 10^{-23}\,\text{J K}^{-1}k=1.38×10−23J K−1
  • m=5×10−17 kgm = 5 \times 10^{-17}\,\text{kg}m=5×10−17kg
  • At NTP, take T≈273 KT \approx 273\,\text{K}T≈273K
  1. Substitute the values

vrms=3(1.38×10−23)(273)5×10−17v_{\text{rms}} = \sqrt{\frac{3(1.38 \times 10^{-23})(273)}{5 \times 10^{-17}}}vrms​=5×10−173(1.38×10−23)(273)​​

  1. Calculate the numerator

3×1.38×273=1130.223 \times 1.38 \times 273 = 1130.223×1.38×273=1130.22

So,

3kT=1130.22×10−23=1.13022×10−203kT = 1130.22 \times 10^{-23} = 1.13022 \times 10^{-20}3kT=1130.22×10−23=1.13022×10−20

  1. Divide by mass

1.13022×10−205×10−17=0.226044×10−3=2.26044×10−4\frac{1.13022 \times 10^{-20}}{5 \times 10^{-17}} = 0.226044 \times 10^{-3} = 2.26044 \times 10^{-4}5×10−171.13022×10−20​=0.226044×10−3=2.26044×10−4

  1. Take square root

vrms=2.26044×10−4v_{\text{rms}} = \sqrt{2.26044 \times 10^{-4}}vrms​=2.26044×10−4​

vrms≈1.503×10−2 m s−1v_{\text{rms}} \approx 1.503 \times 10^{-2}\,\text{m s}^{-1}vrms​≈1.503×10−2m s−1

  1. Convert to mm/s

1.503×10−2 m s−1=15.03 mm s−11.503 \times 10^{-2}\,\text{m s}^{-1} = 15.03\,\text{mm s}^{-1}1.503×10−2m s−1=15.03mm s−1

So, approximately,

vrms≈15 mm s−1v_{\text{rms}} \approx 15\,\text{mm s}^{-1}vrms​≈15mm s−1

  1. Check options
  • A: 60 mm s−160\,\text{mm s}^{-1}60mm s−1
  • B: 12 mm s−112\,\text{mm s}^{-1}12mm s−1
  • C: 15 mm s−115\,\text{mm s}^{-1}15mm s−1
  • D: 36 mm s−136\,\text{mm s}^{-1}36mm s−1

Hence, the correct option is C.

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