The number of particles crossing unit area perpendicular to x-axis in unit time is given by n=-Dn2-n1x2-x1 where n1 and n2 are number of particles per unit volume for the value of x meant to x1 and x2. The dimensions of D are

n=-Dn2-n1X2-X1⇒n=1 L2 T=L-2 T-1 n1=n2=1 L3=L-3 X2=X1=L ∴L-2 T-1=DL-3L D=LL-2 T-1 L-3=L2 T-1 ∴D=M0 L2 T-1

The number of particles crossing unit area perpendicular to x-axis in unit time is given by n=-Dn2-n1x2-x1 where n1 and n2 are number of particles per unit volume for the value of x meant to x1 and x2. The dimensions of D are

n=-Dn2-n1X2-X1⇒n=1 L2 T=L-2 T-1 n1=n2=1 L3=L-3 X2=X1=L ∴L-2 T-1=DL-3L D=LL-2 T-1 L-3=L2 T-1 ∴D=M0 L2 T-1

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The number of particles crossing unit area perpendicular to x-axis in unit time is given by $n = \frac{- D (n_{2} - n_{1})}{x_{2} - x_{1}}$ where n₁ and n₂ are number of particles per unit volume for the value of x meant to x₁ and x₂. The dimensions of D are

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$M^{o} {LT}^{3}$

$M^{o} L^{2} T^{- 4}$

$M^{o} {LT}^{- 2}$

$M^{o} L^{2} T^{- 1}$

answer is D.

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Detailed Solution

$n = \frac{- D (n_{2} - n_{1})}{X_{2} - X_{1}} \Rightarrow n = [\frac{1}{L^{2} T}] = [L^{- 2} T^{- 1}] n_{1} = n_{2} = [\frac{1}{L^{3}}] = [L^{- 3}] X_{2} = X_{1} = [L] ∴ [L^{- 2} T^{- 1}] = \frac{D [L^{- 3}]}{L} D = [\frac{{LL}^{- 2} T^{- 1}}{L^{- 3}}] = [L^{2} T^{- 1}] ∴ D = [M^{0} L^{2} T^{- 1}]$