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1.Units, Dimensions and Measurement
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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
A${M^0}L{T^2}$
B${M^0}{L^2}{T^{ - 4}}$
C${M^0}L{T^{ - 3}}$
D${M^0}{L^2}{T^{ - 1}}$
Solution
(d) $[n]$ = Number of particles crossing a unit area in unit time = $[{L^{ – 2}}{T^{ – 1}}] $
$\left[ {{n_2}} \right] = \left[ {{n_1}} \right] $ = number of particles per unit volume $= [L^{-3}]$
$ [{x_2}] = [{x_1}]$= positions $D = \frac{{[n]\;\left[ {{x_2} – {x_1}} \right]}}{{\left[ {{n_2} – {n_1}} \right]}} = \frac{{\left[ {{L^{ – 2}}{T^{ – 1}}} \right] \times [L]}}{{[{L^{ – 3}}]}} $ $=\left[ {{L^2}{T^{ – 1}}} \right]$
$\left[ {{n_2}} \right] = \left[ {{n_1}} \right] $ = number of particles per unit volume $= [L^{-3}]$
$ [{x_2}] = [{x_1}]$= positions $D = \frac{{[n]\;\left[ {{x_2} – {x_1}} \right]}}{{\left[ {{n_2} – {n_1}} \right]}} = \frac{{\left[ {{L^{ – 2}}{T^{ – 1}}} \right] \times [L]}}{{[{L^{ – 3}}]}} $ $=\left[ {{L^2}{T^{ – 1}}} \right]$
Standard 11
Physics
