A dimensionally consistent relation for the volume V of a liquid of coefficient of viscosity 'η' flowing per second, through a tube of radius r and length l and having a pressure difference P across its ends, is
(1) V= πPr^4/ 8ηl
(2V= πη/8Pr^4
(3) V= 8Pη/πr^4
(4) V= πPη/8r^4
Answers
Answer:
V= rate of flow =
second
volume
=[M
o
L
3
T
−1
]
P= Pressare =[M
1
L
−1
T
−2
]
r= radius =[M
0
L
1
T
0
]
η= coefficient of viscosity =[M
1
L
−1
T
1
]
L= length =[M
0
L
1
T
o
]
Taking (i) option,
V=
8ηL
πP
r
4
[M
0
L
3
T
−1
]=
[M
1
L
−1
T
−1
][M
0
L
1
T
0
]
[M
1
L
−1
T
−2
][M
0
L
1
T
0
]
4
[M
0
L
3
T
−1
=
[M
1
L
−1
T
−1
][M
0
L
1
T
0
]
[M
1
L
−1
T
−2
][M
0
L
4
T
0
]
[M
0
L
3
T
−1
]=
[M
1
L
0
T
−1
]
[M
2
L
3
T
−2
]
[M
0
L
3
T
−1
=[M
0
L
3
T
−1
]
Hence proved
Answer:
[v]=[LT
−1
]
η=
A
dx
dv
F
⇒[η]=
[L
2
][T
−1
]
[MLT
−2
]
=[M
1
L
−1
T
−1
]
[ρ]=[ML
−3
]
⇒[LT
−1
]=[M
1
L
−1
T
−1
]
x
[ML
−3
]
y
[L]
z
Equating the exponents of M, L and T on both LHS and RHS$$
⇒M
0
=M
(x+y)
⇒y=−x
For T,
−1=−x
x = 1
y=−x=−1
For L
1=−x−3y+z→−1+3+z
1=2+z→z=−1
⇒z=−1