Question

Please someone answer both the questions ​

Answer 1

For convenience, Square braces around the dimensional equations are not being used.

4) If Force F, Velocity v and Time T are taken as fundamental quantities. Dimensions of force in dimensional formula of pressure.

F = MLT^-2

V = LT^-1

T = T

We know that,

Pressure = ML^3T^-2

So,

$P = F^a V^b T^c \\ P = (MLT^{-2})^a (LT^{-1}) ^b (T^c) \\ P = M ^{a} L ^{a + b} T ^{ - 2a - b + c} </p><p>$

Comparing with actual dimensional formula,

We get a = 1, b = 2, c = 2

So in new system,

$P = F^1V^2 T^2$

Therefore, Dimensions of Force in dimensional formula of pressure is 1.

5) If the speed of light c, acceleration due to gravity g, pressure p are taken as fundamental quantities then, What is the dimensions of acceleration due to gravity in dimensional formula of Universal gravitational constant

Dimensions of Universal gravitational constant

$G = M^{-1}L^3T^{-2}$

Dimensions of,

Speed of light c = $LT^{-1}$

Acceleration due to gravity g = $LT^{-2}$

Pressure P = $ML^3T^{-2}$

So let,

$G = c^x g^y \rho^z \\ \\ G = (LT^{-1})^x (LT^{-2})^y (ML^3T^{-2})^z \\ \\G = M^z (L^{x+y+3z}) T^{-x-2y-2z}$

Comparing gives,

z = - 1,

x + y - 3 = 3

x + y = 6

Also, - x - 2y +2 = - 2

-x - 2y = - 4

x + 2y = 4

y = - 2, x = 8.

So,

$G = c^8g^-2 \rho^{-1}$

Therefore, The dimensions of Acceleration due to gravity in universal gravitational constant are - 2