Q94 / 180 · NEET 2015 (Phase 2) · Physics

If dimensions of critical velocity, vc of a liquid flowing through a tube are expressed as ηxρyrz, where, η, ρ and 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

  • A
    -1, -1, 1
  • B
    -1, -1, -1
  • C
    1, 1, 1
  • Correct answer
    D
    1, -1, -1

Solution

Dimensions of velocity, coefficient of viscosity and density are,

v=LT-1

η=FAdvdx=MLT-2L2T-1=ML-1T-1

ρ=ML-3

Now, v=ηxρyrz

 ⇒LT-1=ML-1T-1xML-3yLz

Now, by using dimensional homogeneity, we will get,

M0=Mx+y

⇒y=-x

For T, 

x=1

⇒y=-x=-1

For L,

1=-x-3y+z

⇒z=-1

Hence, the correct answer is option: 'D'.

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