The number of particles crossing per unit area perpendicular to Z axis per unit time is given by N=-D ((N2-N1)/(Z2-Z1)), where N2 and N1 are the number of particles per unit volume at Z2 and Z1 respectively. What is the dimensional formula for D ?

Question

The number of particles crossing per unit area perpendicular to Z axis per unit time is given by N=-D ((N2-N1)/(Z2-Z1)), where N2 and N1 are the number of particles per unit volume at Z2 and Z1 respectively. What is the dimensional formula for D ? (A) [ M 0 L -1 T 2] (B) [ M 0 L -1 T -1] (C) [M0 L2 T-1] (D) [M0 L2 T2]

Tardigrade Admin · Accepted Answer

N=-D ((N2-N1)/(Z2-Z1)) Dimensionally, D=(N(Z2-Z1)/(N2-N1)) Given, N2, N1 arrow Number of particles per unit volume. N2, N1 arrow (N/V) ⇒ [L-3] Z2-Z1 arrow[L] N arrow ( text Number of particles / text Area ⋅(T)) N arrow[L-2 T-1] So, D=(L-2 T-1 × L/L-3) ⇒ [ L 2 T -1]

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

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

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

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

Q. The number of particles crossing per unit area perpendicular to Z axis per unit time is given by N=−D(Z2​−Z1​)(N2​−N1​)​, where N2​ and N1​ are the number of particles per unit volume at Z2​ and Z1​ respectively. What is the dimensional formula for D ?

[M0L−1T2]

[M0L−1T−1]

[M0L2T−1]

[M0L2T2]

N=−D(Z2​−Z1​)(N2​−N1​)​ Dimensionally, D=(N2​−N1​)N(Z2​−Z1​)​ Given, N2​,N1​→ Number of particles per unit volume. N2​,N1​→VN​⇒[L−3] Z2​−Z1​→[L] N→ Area ⋅(T) Number of particles ​ N→[L−2T−1] So, D=L−3L−2T−1×L​ ⇒[L2T−1]

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

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

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

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