Question

Turpentine oil is flowing through a tube of length l and radius r. The pressure difference between the two ends of the tube is p. The viscosity of oil is given $\eta = \dfrac{{p({r^2} - {x^2})}}{{4vl}}$, where v is velocity of oil at a distance x from the axis of the tube. The dimensional formula of viscosity is
$\eqalign{
  & A.[{M^0}{L^0}{T^0}] \cr
  & B.[{M^1}{L^{ - 1}}{T^{ - 1}}] \cr
  & C.[M{L^2}{T^{ - 1}}] \cr
  & D.[{M^{ - 1}}{L^{ - 1}}{T^1}] \cr} $

Answer 1

Hint: Expression for viscosity is given. Various terms like pressure, velocity, radius are involved in that expression. We can add or subtract same dimensional physical quantities only. If two physical quantities are equal then they are equal both quantity wise and dimension wise. by writing dimensions of each physical quantity involved right hand side of equation $\eta = \dfrac{{p({r^2} - {x^2})}}{{4vl}}$ and by solving it we get the dimensions of viscosity.

Formula used:
$\eta = \dfrac{{p({r^2} - {x^2})}}{{4vl}}$

Complete answer:
This problem should be solved by a method of dimensions.
first we note all the terms involved in the RHS of $\eta = \dfrac{{p({r^2} - {x^2})}}{{4vl}}$ and then align dimensions of all those terms and then proceed with calculations
Terms involved and their respective dimensions are
pressure = p = $[M{L^{ - 1}}{T^{ - 2}}]$
tube length = l = $[{M^0}{L^1}{T^0}]$
radius = r = $[{M^0}{L^1}{T^0}]$
velocity = v = $[{M^0}{L^1}{T^{ - 1}}]$
Distance from axis = x = $[{M^0}{L^1}{T^0}] $

number 4 is dimensionless

dimension of ${r^2} = {x^2} = [{M^0}{L^2}{T^0}]$ both will be having same dimensions hence subtraction is performed between them
dimension of $p({r^2} - {x^2}) = [{M^1}{L^{ - 1}}{T^{ - 2}}][{M^0}{L^2}{T^0}] = [{M^{(1 + 0)}}{L^{(2 - 1)}}{T^{(0 - 2)}}] = [{M^1}{L^1}{T^{ - 2}}]$ we got it by normal multiplication of two dimensions. powers of the same variables(M or L or T) will be added in doing so
similarly
dimension of $4vl = [{M^0}{L^1}{T^{ - 1}}][{M^0}{L^1}{T^0}] = [{M^{(0 + 0)}}{L^{(1 + 1)}}{T^{(0 - 1)}}] = [{M^0}{L^2}{T^{ - 1}}]$
now final dimension of the equation $\eta = \dfrac{{p({r^2} - {x^2})}}{{4vl}}$ will be$\dfrac{{[{M^1}{L^1}{T^{ - 2}}]}}{{[{M^0}{L^2}{T^{ - 1}}]}} = [{M^{(1 - 0)}}{L^{(1 - 2)}}{T^{( - 2 + 1)}}] = [{M^1}{L^{ - 1}}{T^{ - 1}}]$

Hence answer will be option B

Additional Information:
Same question can be solved by various methods. For example viscous drag when a spherical ball is falling into fluid is given by $F = 6\pi \eta rv$. From this formula also we can get the dimensions of the viscosity.one can simply remember the dimensions of viscosity.

Note:
Units of viscosity can also be obtained from dimensional formula in this case it would be kg/(m-s). if two quantities are equal dimensionally that doesn’t mean they are equal physically too. For example angular momentum and planck's constant have the same dimensions but they are not equal physically i.e both have different applications. same case with the impulse and linear momentum. Some quantities don’t have dimensions but have units for example angle doesn’t have any dimensions but has units(radian).