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Q. The dimensional formula of Reynold’s number is same as

UPSEEUPSEE 2019

Solution:

As, the formula of Reynold’s number
$R_{e} = \frac{V_{c}\rho D}{\eta} $
So, the dimension of
$ R_{e} = \frac{\left[M^{0}LT^{-1}\right]\left[ML^{-3}T^{0}\right]\left[M^{0}LT^{0}\right]}{\left[ML^{-1}T^{-1}\right]} $
$\Rightarrow \left[R_{e}\right] = \left[M^{0}L^{0}T^{0}\right]$
Hence, it is a dimensionless quantity.
Whereas dimension of
(i) Coefficient of viscosity $= [M^0L^{-1} T^{-1} ]$
(ii) Coefficient of friction $ = [M^0 L^0 T^0 ]$
(iii) Universal gravitational constant $ = [M^{-1} L^{3} T^{-2} ]$
(iv) Velocity of light $= [M^0 LT^{-1} ] $
Hence, the dimensional formula of Reynold’s number is same as coefficient of friction.