Question

If volume of liquid flowing per second through a tube depends on pressure gradient,radius of the tube and coefficient of viscosity.find the dimensions

Answer 1

Volume of liquid flowing depends on Area of Cross section and Velocity of liquid

Answer 2

Answer:

The dimension of volume is $\dfrac{kr^4\dfrac{p}{l}}{\eta}$.

Explanation:

Given that,

If volume of liquid flowing per second through a tube depends on pressure gradient,radius of the tube and coefficient of viscosity.

$V\propto r^{a}(\dfrac{p}{l})^{b}\eta^{c}$

$V=kr^{a}(\dfrac{p}{l})^{b}\eta^{c}$...(I)

Here, k = constant

The dimension formula of volume

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

The dimension formula of radius

$r =[L]$

The dimension formula of viscosity

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

The dimension formula of pressure gradient

$\dfrac{p}{l}=[ML^{-2}T^{-2}]$

Put the dimension formula of all terms in equation (I)

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

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

$a-2b-c=3$....(II)

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

$b+c=0$....(IV)

Put the value of (-2b-c) in equation (II)

$a-1=3$

$a=4$

From equation (III) and (IV)

$b = 1$

Put the value of b in equation (IV)

$1+c=0$

$c= -1$

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

$V = kr^4(\dfrac{p}{l})^{1}\eta^{-1}$

$V=\dfrac{kr^4\dfrac{p}{l}}{\eta}$

Hence, The dimension of volume is $\dfrac{kr^4\dfrac{p}{l}}{\eta}$.