If V=√(γ mathrmP/ρ), then dimensions of γ are:

Question

If V=√(&gamma; mathrmP/ρ), then dimensions of &gamma; are: (A) [ mathrmM0 mathrm~L0 mathrm~T0] (B) [ mathrmM0 mathrm~L0 mathrm~T-1] (C) [ mathrmM-1 mathrmL0 mathrm~T0.] (D) [ mathrmM0 mathrmL1 mathrmT0]

Tardigrade Admin · Accepted Answer

beginaligned text mathrmV=√(&gamma; mathrmP/ρ) text or &gamma;=( mathrmV2 ρ/ mathrmP) &[&gamma;]=([ mathrmLT-1]2[ mathrmML-3]/[ mathrmML-1 mathrm~T-2])=[ mathrmM0 mathrm~L0 mathrm~T0] endaligned

Q. If $V = \frac{γ P}{ρ},$ then dimensions of $γ$ are:

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

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

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

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

$V = \frac{γ P}{ρ} or γ = \frac{V ^{2} ρ}{P} [γ] = \frac{[ LT ^{- 1} ] ^{2} [ ML ^{- 3} ]}{[ ML ^{- 1} T ^{- 2} ]} = [M^{0} L^{0} T^{0}]$

Q. If V=ργP​​, then dimensions of γ are:

[M0 L0 T0]

[M0 L0 T−1]

[M−1L0 T0]

[M0L1T0]