Suppose mass, velocity time were fundamental physical quantities then find the dimensional formula of pressure.

Question

Suppose mass, velocity time were fundamental physical quantities then find the dimensional formula of pressure. (A) [ M 1 V -1 T -2] (B) [ M 1 V -1 T -3 ] (C) [ M 1 V -2 T -3] (D) [M2 V-2 T-3]

Tardigrade Admin · Accepted Answer

[ P ]= M x V y T z ( MLT -2/ L 2)= M x (( L / T )) y T z M L-1 T-2=Mx L YTz-y ⇒ x=1, y=-1 z-y=-2 ⇒ z=-2+y=-3

Q. Suppose mass, velocity & time were fundamental physical quantities then find the dimensional formula of pressure.

$[M^{1} V^{- 1} T^{- 2}]$

$[M^{1} V^{- 1} T^{- 3}]$

$[M^{1} V^{- 2} T^{- 3}]$

$[M^{2} V^{- 2} T^{- 3}]$

$[P] = M^{x} V^{y} T^{z}$
$\frac{M L T ^{- 2}}{L ^{2}} = M^{x} (\frac{L}{T})^{y} T^{z}$
$M L^{- 1} T^{- 2} = M^{x} L^{Y} T^{z - y}$
$\Rightarrow x = 1, y = - 1$
$z - y = - 2$
$\Rightarrow z = - 2 + y = - 3$

Q. Suppose mass, velocity & time were fundamental physical quantities then find the dimensional formula of pressure.

[M1V−1T−2]

[M1V−1T−3]

[M1V−2T−3]

[M2V−2T−3]

[P]=MxVyTz L2MLT−2​=Mx(TL​)yTz ML−1T−2=MxLYTz−y ⇒x=1,y=−1 z−y=−2 ⇒z=−2+y=−3

$[M^{1} V^{- 1} T^{- 2}]$

$[M^{1} V^{- 1} T^{- 3}]$

$[M^{1} V^{- 2} T^{- 3}]$

$[M^{2} V^{- 2} T^{- 3}]$

$[P] = M^{x} V^{y} T^{z}$
$\frac{M L T ^{- 2}}{L ^{2}} = M^{x} (\frac{L}{T})^{y} T^{z}$
$M L^{- 1} T^{- 2} = M^{x} L^{Y} T^{z - y}$
$\Rightarrow x = 1, y = - 1$
$z - y = - 2$
$\Rightarrow z = - 2 + y = - 3$