The dimensions of (a/b) in the relation P=(a-t2/bx) , where P is the pressure, x is the distance and t is the time, is

Question

The dimensions of (a/b) in the relation P=(a-t2/bx) , where P is the pressure, x is the distance and t is the time, is (A) [M-1L0T-2] (B) [M L0T-2] (C) [M0L0T-2] (D) [MLT-2]

Tardigrade Admin · Accepted Answer

beginarrayl P =( a - t 2/ bx ) P =[ ML -1 T -2] text a will have same unit as t 2 . text So, a =[ T 2] b =[( a - t 2/ px )]=([ T 2]/[ ML -1 T -2 L ])=[ M -1 L 0 T 4] ( a / b )=([ T 2]/[ M -1 L 0 T 4])=[ ML 0 T -2] endarray

Q. The dimensions of $\frac{a}{b}$ in the relation $P = \frac{a - t ^{2}}{b x}$ , where $P$ is the pressure, $x$ is the distance and $t$ is the time, is

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

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

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

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

$P = \frac{a - t ^{2}}{b x} P = [M L^{- 1} T^{- 2}] a will have same unit as t^{2} . So, a = [T^{2}] b = [\frac{a - t ^{2}}{p x}] = \frac{[ T ^{2} ]}{[ M L ^{- 1} T ^{- 2} L ]} = [M^{- 1} L^{0} T^{4}] \frac{a}{b} = \frac{[ T ^{2} ]}{[ M ^{- 1} L ^{0} T ^{4} ]} = [M L^{0} T^{- 2}]$

Q. The dimensions of ba​ in the relation P=bxa−t2​ , where P is the pressure, x is the distance and t is the time, is

[M−1L0T−2]

[ML0T−2]

[M0L0T−2]

[MLT−2]