SATHEE: Physics And Measurement Question 193

Physics And Measurement Question 193

Question: Number of particles is given by $n = - D \frac{n_{2} - n_{1}}{x^{2} - x_{1}}$ crossing a unit area perpendicular to X-axis in unit time, where $n_{1}^{}$ and $n_{2}^{}$ are number of particles per unit volume for the value of $x_{}^{}$ meant to $x_{2}^{}$ and $x_{1}^{}$ . Find dimensions of D called as diffusion constant

Options:

A) $M_{0}^{} L T_{}^{2}$

B) $M_{}^{0} L_{}^{2} T_{}^{- 4}$

C) $M_{}^{0} L_{}^{} T_{}^{- 3}$

D) $M_{}^{0} L_{}^{2} T_{}^{- 1}$

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Answer:

Correct Answer: D

Solution:

[d] [n]= Number of particles crossing a unit area in unit time = $[L^{- 2} T^{- 1}]$

$[n_{2}] = [n_{1}]$ =number of particles per unit valume= $[L^{- 3}]$

$[x_{2}] = [x_{1}]$ = positions

$∴ D = \frac{[n] [x_{2} - x_{1}]}{[n_{2} - n_{1}]} = \frac{[L^{- 2} T^{- 1}] \times [L]}{[L^{- 3}]} = [L^{2} T^{- 1}]$

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Physics And Measurement Question 193

Question: Number of particles is given by n=−Dn2−n1x2−x1 crossing a unit area perpendicular to X-axis in unit time, where n1 and n2 are number of particles per unit volume for the value of x meant to x2 and x1 . Find dimensions of D called as diffusion constant

Options:

Answer:

Solution:

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