Question

If pressure 'P', velocity 'V' and time 'T' are taken
as the functional quantities, find the dimensional
formula of force
(A) PV2T2

(B) P2VT2
(C) P2v2T2
(D) PVT​

Answer 1

$\large\underline\mathfrak\red{Answer \: :- }$

(A) PV²T²

$\large\underline\mathfrak\red{Explanation \: :- }$

$⟼ \sf{F = P {}^{ \alpha } V {}^{ \beta } T {}^{ \gamma } }$

or,

$\sf{ [M¹L¹T {}^{ - 2} ]=[M {}^{ \alpha } L {}^{ - \alpha + \beta } T {}^{ - 2 \alpha - \beta + \gamma } ]}$

$\sf{ i.e. \: [M¹L¹T {}^{ - 2} ]=[M {}^{ \alpha } L {}^{ - \alpha + \beta } T {}^{ - 2 \alpha - \beta + \gamma } ]}$

Hence We have,

$\sf{ \alpha = 1} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf{ - \alpha + \beta =1 } \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf{ - 2 \alpha - \beta + \gamma = - 2}$

Solving these we get

$\sf{ \alpha = 1} \\ \sf{ \beta = 2} \\ \sf{ \gamma = 2}$

Hence,

$\sf{P = 1 } \\ \sf{V = 2} \\ \sf{T = 2 }$

$\bf\pink{⟼ PV²T²}$

______________________________________

Learn more :

If (alpha - bit q), alpha and (alpha + Bit a) are zeroes of the polynomial 2x°-16x +15x-2....

https://brainly.in/question/40344241