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If dimensions of critical velocity vc of a liquid flowing through a tube are expressed as [ηxρyγz] where η, ρ and γ 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.
  • a)
    -1,-1,-1
  • b)
    1,1,1
  • c)
    1,-1,-1
  • d)
    -1,-1,1
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
If dimensions of critical velocity vc of a liquid flowing through a t...
To determine the dimensions of critical velocity (vc) of a liquid flowing through a tube, we need to express vc in terms of the given variables: η (coefficient of viscosity), ρ (density of liquid), and γ (radius of the tube).

The formula for critical velocity (vc) is given by:

vc = (2γ/ρ)^(1/2)

Breaking down the formula, we can determine the dimensions of vc:

1. Breaking down the given variables:
- η (coefficient of viscosity): [η] = ML^(-1)T^(-1)
- ρ (density of liquid): [ρ] = ML^(-3)
- γ (radius of the tube): [γ] = L

2. Substituting the dimensions into the formula for vc:
- vc = (2γ/ρ)^(1/2)
- vc = (2L / (ML^(-3)))^(1/2)
- vc = (2L^(4) / (M))^(1/2)
- vc = (2L^(2) / (M))^(1/2) * L
- vc = (2L^(2) / (M))^(1/2) * L^(1)

From the above expression, we can determine the dimensions of vc as [L^(1)] * [M^(-1/2)].

Therefore, the values of x, y, and z are:
- x = 1 (corresponding to L^(1))
- y = -1/2 (corresponding to M^(-1/2))
- z = -1 (corresponding to L^(-1))

Hence, the correct answer is option 'C': 1, -1/2, -1.
Free Test
Community Answer
If dimensions of critical velocity vc of a liquid flowing through a t...
[vc] = [ηxρyγz] (given) ........(i)
Writing the dimensions of various quantities in eqn. (i), we get
[M0LT−1] = [M L−1T−1]x [M L−3T0]y[M0 LT0]z
= [Mx+yL−x−3y+zT−x]
Applying the principle of homogeneity of dimensions, we get
x + y = 0; −x − 3y + z = 1; −x = −1
On solving, we get x = 1, y = −1, z = −1
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If dimensions of critical velocity vc of a liquid flowing through a tube are expressed as [ηxρyγz] where η, ρ and γ 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.a)-1,-1,-1b)1,1,1c)1,-1,-1d)-1,-1,1Correct answer is option 'C'. Can you explain this answer?
Question Description
If dimensions of critical velocity vc of a liquid flowing through a tube are expressed as [ηxρyγz] where η, ρ and γ 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.a)-1,-1,-1b)1,1,1c)1,-1,-1d)-1,-1,1Correct answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about If dimensions of critical velocity vc of a liquid flowing through a tube are expressed as [ηxρyγz] where η, ρ and γ 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.a)-1,-1,-1b)1,1,1c)1,-1,-1d)-1,-1,1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If dimensions of critical velocity vc of a liquid flowing through a tube are expressed as [ηxρyγz] where η, ρ and γ 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.a)-1,-1,-1b)1,1,1c)1,-1,-1d)-1,-1,1Correct answer is option 'C'. Can you explain this answer?.
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