Question

Derive by the method of dimensions,an expression for volume of a liquid flowing out per second through a narrow pipe. Assume that the rate of flow of liquid depend on coefficient of viscosity of liquid the radius of pipe and length of pipe and pressure across pipe

Answer 1

Your answer is here. Please rank me the branliest. I took the constant as π/8

Answer 2

Answer:

The dimension formula is $V=k\dfrac{R^4(\dfrac{P}{L})^1}{\eta^{1}}$.

Explanation:

Given that,

$V\propto R^a(\dfrac{P}{L})^b(\eta)^c$....(I)

The dimension formula is

$V=[L]^3[T]^{-1}$

$R=[L]$

$\dfrac{P}{L}=[M][T]^{-2}[L]^{-2}$

$\eta=[M][L]^{-1}[T]^{-1}$

Now, put the dimension formula in equation (I)

$[L]^3[T]^{-1}=[L]^a([M][T]^{-2}[L]^{-2})^b([M][L]^{-1}[T]^{-1})^c$

On compare the powers

a-2b-c=3....(I)

b+c=0.....(II)

-2b-c=-1.....(III)

From equation (II)

b= -c

Now, Put the value of b in equation (III)

2c-c=-1

c=-1

Now, the value of b is

b= 1

Put the value of c and b in equation (I)

a-2+1=3

a=4

Now,

$V=kR^4(\dfrac{P}{L})^1(\eta)^{-1}$

$V=k\dfrac{R^4(\dfrac{P}{L})^1}{\eta^{1}}$

Hence, The dimension formula is $V=k\dfrac{R^4(\dfrac{P}{L})^1}{\eta^{1}}$.