Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

The number of particles crossing a unit area perpendicular to the $x$ -axis in unit time is given by $n = -D\frac{n_2 - n_1}{x_2 - x_1}$ , where $n_1$ and $n_2$ are the number of particles per unit volume for the value of $x$ equal to $x_1$ and $x_2$ respectively. The dimensions of $D$ , known as the diffusion constant, will be:

$[M^0 L T^2]$

$[M^0 L^2 T^{-4}]$

$[M^0 L T^{-3}]$

$[M^0 L^2 T^{-1}]$

Step-by-Step Solution

Find the dimensions of $n$ : $n$ is the number of particles crossing a unit area per unit time. Dimensions of $n = \frac{1}{\text{Area} \times \text{Time}} = \frac{1}{[L^2][T]} = [L^{-2} T^{-1}]$ .
Find the dimensions of $(n_2 - n_1)$ : $n_1$ and $n_2$ are the number of particles per unit volume. Dimensions of $(n_2 - n_1) = \frac{1}{\text{Volume}} = \frac{1}{[L^3]} = [L^{-3}]$ .
Find the dimensions of $(x_2 - x_1)$ : $x_1$ and $x_2$ are positions (distance). Dimensions of $(x_2 - x_1) = [L]$ .
Calculate dimensions of $D$ : Rearranging the given formula, we get $D = -n \frac{x_2 - x_1}{n_2 - n_1}$ . Ignoring the negative sign for dimensions: $[D] = \frac{[n] \times [x_2 - x_1]}{[n_2 - n_1]}$ $[D] = \frac{[L^{-2} T^{-1}] \times [L]}{[L^{-3}]} = \frac{[L^{-1} T^{-1}]}{[L^{-3}]} = [L^2 T^{-1}]$ .

Therefore, the dimensional formula for $D$ is $[M^0 L^2 T^{-1}]$ .

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