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 the 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=1L2T=L−2T−1n1=n2=1L3=L−3⇒n1=n2=1L3=L−3X2=X1=L ∴L−2T−1=DL−3LD=LL−2T−1L−3=L2T−1 ∴D=M0L2T−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 the 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=1L2T=L−2T−1n1=n2=1L3=L−3⇒n1=n2=1L3=L−3X2=X1=L ∴L−2T−1=DL−3LD=LL−2T−1L−3=L2T−1 ∴D=M0L2T−1

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

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$M^{\circ} {LT}^{- 2}$

$M^{\circ} L^{3}$

$M^{\circ} L^{2} T^{- 4}$

$M^{\circ} L^{2} T^{- 1}$

answer is D.

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Detailed Solution

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