4.
In a laminar flow the velocity of the liquid in contact with the walls of the tube is
A) Zero
B) Maximum
C) In between zero and maximum
D) Equal to critical velocity
Answers
Answer:
Answer:
The dimension of volume is \dfrac{kr^4\dfrac{p}{l}}{\eta}
η
kr
4
l
p
.
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∝r
a
(
l
p
)
b
η
c
V=kr^{a}(\dfrac{p}{l})^{b}\eta^{c}V=kr
a
(
l
p
)
b
η
c
...(I)
Here, k = constant
The dimension formula of volume
V = [L^3T^{-1}]V=[L
3
T
−1
]
The dimension formula of radius
r =[L]r=[L]
The dimension formula of viscosity
\eta=[ML^{-1}T^{-1}]η=[ML
−1
T
−1
]
The dimension formula of pressure gradient
\dfrac{p}{l}=[ML^{-2}T^{-2}]
l
p
=[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
3
T
−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}[L
3
T
−1
]=[L]
a−2b−c
[M]
b+c
[T]
−2b−c
a-2b-c=3a−2b−c=3 ....(II)
-2b-c=-1−2b−c=−1 ....(III)
b+c=0b+c=0 ....(IV)
Put the value of (-2b-c) in equation (II)
a-1=3a−1=3
a=4a=4
From equation (III) and (IV)
b = 1b=1
Put the value of b in equation (IV)
1+c=01+c=0
c= -1c=−1
Put the value of a,b and c in equation (I)
V = kr^4(\dfrac{p}{l})^{1}\eta^{-1}V=kr
4
(
l
p
)
1
η
−1
V=\dfrac{kr^4\dfrac{p}{l}}{\eta}V=
η
kr
4
l
p
Hence, The dimension of volume is \dfrac{kr^4\dfrac{p}{l}}{\eta}
η
kr
4
l
p
.