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Physics Question on Units and measurement

The dimension of Pa\frac{P}{a} in the equation p=b−t2axp=\frac{b-{{t}^{2}}}{ax} where p is pressure, xx is distance and t is time are

A

[MLT−2][ML{{T}^{-2}}]

B

[MT−2][M{{T}^{-2}}]

C

[ML3T−2][M{{L}^{3}}{{T}^{-2}}]

D

[LT−3][L{{T}^{-3}}]

Answer

[MT−2][M{{T}^{-2}}]

Explanation

Solution

p=b−t2axp=\frac{b-{{t}^{2}}}{ax} or pax=b−t2pax=b-{{t}^{2}} [pax]=[b]=[T2][pax]=[b]=[{{T}^{2}}] [a]=[T2][p][x][a]=\frac{[{{T}^{2}}]}{[p][x]} =[T2][ML−1T−2][L]=\frac{[{{T}^{2}}]}{[M{{L}^{-1}}{{T}^{-2}}][L]} [a]=[M−1T4][a]=[{{M}^{-1}}{{T}^{4}}] [ba]=[T2][M−1T4]\left[ \frac{b}{a} \right]=\frac{[{{T}^{2}}]}{[{{M}^{-1}}{{T}^{4}}]} =[MT−2]=[M{{T}^{-2}}]