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NEET 2015 (AIPMT) · Applications of Dimensional Analysis

Applications of Dimensional Analysis question from NEET 2015 (AIPMT)

Q71NEET 2015 · AIPMTMedium

Applications of Dimensional Analysis

If dimensions of critical velocity vvc of a liquid flowing through a tube are expressed as [ηxρyrz][\eta^{x}\rho^{y}r^{z}] where η,ρ\eta, \rho 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

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Answer: (c)

  1. [η]=ML−1T−1[\eta] = ML^{-1}T^{-1}, [ρ]=ML−3[\rho] = ML^{-3}, [r]=L[r] = L, [vc]=LT−1[v_c] = LT^{-1}.
  2. M: x+y=0x + y = 0; T: −x=−1-x = -1; L: −x−3y+z=1-x - 3y + z = 1.
  3. x=1x = 1, y=−1y = -1, z=1+1−3=−1z = 1 + 1 - 3 = -1 (so vc∝ηρrv_c \propto \dfrac{\eta}{\rho r}).
  4. So the answer is (c) 1, −1-1, −1-1.
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