Question

The dimensions $\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$ corresponds to:
A) Moment of a force or torque.
B) Surface tension.
C) Pressure.
D) Coefficient of viscosity.

Answer 1

Hint: The dimensional formula is the relation between the physical quantity and the fundamental quantities. The dimensional formula also helps us to check whether the formula of the physical quantity is correct or not.

Formula used:
The formula of the moment of force is given by,
$ \Rightarrow \tau = F \times d$
Where the force is F and the perpendicular distance is d.
The formula of the surface tension given by,
$ \Rightarrow \sigma = \dfrac{F}{L}$
Where force is F and length is L.
The formula of the pressure is given by,
$ \Rightarrow P = \dfrac{F}{A}$
Where force is F and area is A.
The formula of force on the liquid surface is given by,
$ \Rightarrow F = \mu \dfrac{{du}}{{dy}}$
Where force is F, the velocity is u and the vertical distance is y, the coefficient of viscosity is given by $\mu $.

Complete step by step solution:
It is given that the dimensional formula is $\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$ and we need to find the physical quantity to which it belongs to, so we will consider each physical quantity and then we will find the correct answer for this problem.
Let us consider torque first.
The formula of the torque is given by,
$ \Rightarrow \tau = F \times d$
Where the force is F and the perpendicular distance is d.
The dimension of force is $kgm{s^{ - 2}}$ and the dimension is in meters (m).
$ \Rightarrow \tau = F \times d$
$ \Rightarrow \tau = kgm{s^{ - 2}} \times m$
The kg is unit of mass, meter is unit of length and second is unit of time.
$ \Rightarrow \tau = kgm{s^{ - 2}} \times m$
$ \Rightarrow \tau = ML{T^{ - 2}} \times L$
$ \Rightarrow \tau = \left[ {M{L^2}{T^{ - 2}}} \right]$.
Let us consider physical quantity surface tension.
The formula of the surface tension given by,
$ \Rightarrow \sigma = \dfrac{F}{L}$
Where force is F and length is L.
The dimensional formula of force is$\left[ {ML{T^{ - 2}}} \right]$ and the dimensional formula of the length is$\left[ L \right]$.
The dimensional formula of surface tension is equal to,
$ \Rightarrow \sigma = \dfrac{F}{L}$
$ \Rightarrow \sigma = \dfrac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ L \right]}}$
$ \Rightarrow \sigma = \left[ {M{L^0}{T^{ - 2}}} \right]$
Let us consider another physical quantity pressure.
The formula of the force is equal to,
$ \Rightarrow P = \dfrac{F}{A}$
Where force is F and area is A.
The dimensional formula of force is equal to $\left[ {ML{T^{ - 2}}} \right]$ and dimensional formula of area is equal to$\left[ {{L^2}} \right]$.
$ \Rightarrow P = \dfrac{F}{A}$
$ \Rightarrow P = \dfrac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ {{L^2}} \right]}}$
$ \Rightarrow P = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$
The dimensional formula of the pressure is equal to$\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$.
Consider another physical quantity coefficient of viscosity.
The formula of force on the liquid surface is given by,
$ \Rightarrow F = \mu \dfrac{{du}}{{dy}}$
Where force is F, the velocity is u and the vertical distance is y, the coefficient of viscosity is given by$\mu $.
$ \Rightarrow F = \mu \dfrac{{du}}{{dy}}$
$ \Rightarrow \mu = \dfrac{{dy}}{{du}} \times F$
$ \Rightarrow \mu = \dfrac{{dy}}{{du}} \times F$
The dimensional formula of force is$\left[ {ML{T^{ - 2}}} \right]$, the dimensional formula of velocity is equal to $\left[ {L{T^{ - 2}}} \right]$ and the dimensional formula of length is$\left[ L \right]$.
The dimensional formula of coefficient of viscosity is equal to,
$ \Rightarrow \mu = \dfrac{{dy}}{{du}} \times F$
$ \Rightarrow \mu = \dfrac{{\left[ L \right]}}{{\left[ {L{T^{ - 2}}} \right]}} \times \left[ {ML{T^{ - 2}}} \right]$
$ \Rightarrow \mu = \dfrac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ {{T^{ - 2}}} \right]}}$
$ \Rightarrow \mu = \left[ {ML{T^0}} \right]$
The dimensional formula of coefficient of viscosity is equal to$\left[ {ML{T^0}} \right]$.
The dimensional formula $\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$ corresponds to pressure.

The correct answer for this problem is option C.

Note: The students are advised to understand and remember the formula of torque, the formula of surface tension, the formula of pressure and formula of the viscous force as it is very useful in solving these types of problems.