n=-D[((n2-n1)/(x2-x1))] , where n is number of particles in unit area perpendicular to x-axis in unit time, n1, n2 are number of particles per unit volume, x2 and x1 are distance of two points from any point, then the dimension of D is

Question

n=-D[((n2-n1)/(x2-x1))] , where n is number of particles in unit area perpendicular to x-axis in unit time, n1, n2 are number of particles per unit volume, x2 and x1 are distance of two points from any point, then the dimension of D is (A) [L2T-1] (B) [L-2T-1] (C) [L-2T1] (D) [L3T-1]

Tardigrade Admin · Accepted Answer

[n]=(1/[Area × Time])=(1/[L2T]) =[L-2T-1] [x2]=[x1]=[L] and [n2-n1]=(1/Volume)=(1/[L3])=[L-3]; ⇒ [D]=[(L-2T-1 × L/L-3)] =[L2T-1]

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

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

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

$[L^{3} T^{- 1}]$

$[n] = \frac{1}{[ A re a \times T im e ]} = \frac{1}{[ L ^{2} T ]}$
$= [L^{- 2} T^{- 1}]$
$[x_{2}] = [x_{1}] = [L]$ and $[n_{2} - n_{1}] = \frac{1}{V o l u m e} = \frac{1}{[ L ^{3} ]} = [L^{- 3}];$
$\Rightarrow [D] = [\frac{L ^{- 2} T ^{- 1} \times L}{L ^{- 3}}]$
$= [L^{2} T^{- 1}]$

Q. n=−D[(x2​−x1​)(n2​−n1​)​] , where n is number of particles in unit area perpendicular to x-axis in unit time, n1​, n2​ are number of particles per unit volume, x2​ and x1​ are distance of two points from any point, then the dimension of D is

[L2T−1]

[L−2T−1]

[L−2T1]

[L3T−1]

[n]=[Area×Time]1​=[L2T]1​ =[L−2T−1] [x2​]=[x1​]=[L] and [n2​−n1​]=Volume1​=[L3]1​=[L−3]; ⇒[D]=[L−3L−2T−1×L​] =[L2T−1]

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

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

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

$[L^{3} T^{- 1}]$

$[n] = \frac{1}{[ A re a \times T im e ]} = \frac{1}{[ L ^{2} T ]}$
$= [L^{- 2} T^{- 1}]$
$[x_{2}] = [x_{1}] = [L]$ and $[n_{2} - n_{1}] = \frac{1}{V o l u m e} = \frac{1}{[ L ^{3} ]} = [L^{- 3}];$
$\Rightarrow [D] = [\frac{L ^{- 2} T ^{- 1} \times L}{L ^{- 3}}]$
$= [L^{2} T^{- 1}]$