1
If dimensions of critical velocity vc of a liquid
flowing through a tube are expressed as [n" pyp2]
where n, pand r are the coefficient of viscosity of
liquid, density of liquid and radius of the tube
respectively, then the values of x, y and z are given
by
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0
Answer:
B
1, -1, -1
[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
Explanation:
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
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