Question

The fundamental unit which has the same power in the dimension formula of the surface tension and viscosity is
(A). Mass
(B). Length
(C). Time
(D). None of the above

Answer 1

Hint: Here we first start with start with surface tension $\dfrac{F}{l}$ whose dimension is $\left[ {M{T^{ - 2}}} \right]$ then we find the dimension of viscosity using formula $F = - \eta \dfrac{{Au}}{x}$. So dimension for viscosity comes out to be $\left[ {M{L^{ - 1}}{T^{ - 1}}} \right]$ so from here we can see mass have the same power in both the dimensional formulas.

Complete step-by-step solution -
We know that viscosity is the internal resistance that comes into the picture when the fluid is in motion. Viscosity opposes the relative motion between surfaces. And Surface tension also a type of force is the tendency of liquid surfaces to shrink into the minimum surface area possible.it.
So surface tensions is surface force per unit length that is $\dfrac{F}{l}$ and it’s the dimension is same as that of the force per unit length that is $\left[ {M{T^{ - 2}}} \right]$
We know the expression for viscous force is
$F = - \eta \dfrac{{Au}}{x}$------------------------------------- (1)
Here, $\eta $ = coefficient of viscosity
$x$= distance of layer from a fixed layer and its dimension is $\left[ L \right]$
$F$= Viscous force and its dimension is $\left[ {ML{T^{ - 2}}} \right]$
$A$= Area of the layer and its dimension is $\left[ {{L^2}} \right]$
$u$= Velocity of the layer from a fixed layer and its dimension is $\left[ {L{T^{ - 1}}} \right]$

Now writing equation (1) in other way we get
$\eta = - \dfrac{{xF}}{{Au}}$---------------------------------------- (2)
Now to calculate the dimension of coefficient of viscosity we use equation (2) and substitute the dimensions of $F$,$A$, $u$, and $x$ in equation (2).
$\eta = \dfrac{{\left[ L \right]\left[ {ML{T^{ - 2}}} \right]}}{{\left[ {{L^2}} \right]\left[ {L{T^{ - 1}}} \right]}} = \left[ {M{L^{ - 1}}{T^{ - 1}}} \right]$
So we can see that the power of mass in both surface tension $\left[ {M{T^{ - 2}}} \right]$ and viscosity $\left[ {M{L^{ - 1}}{T^{ - 1}}} \right]$ is same that is one. Therefore the correct option is A.

Note: For these types of questions we need to first write the dimensions of the given values in the question if need calculate the dimensions. Also, take care of the power. Then compare the dimensional formula and find which among the mass, length, and time have the same power.