Question

Dimensional formula of pressure $\times$ volume =

Answer 1

Hint: The dimensional formula for any physical quantity is the formula for the unit of the same quantity in terms of the fundamental units. The fundamental units are the base units, which are independent units in the unit system. The derived units are expressed in terms of the fundamental units.

Complete step by step answer:
Any physical quantity is expressed in terms of its fundamental unit, this expression is known as dimensional formula of the physical quantity. Dimensional formula is generally represented in terms of mass, kilogram, time and temperature.
The pressure is the fraction of the force and area. Thus, the unit of pressure is Newton per meter. The SI unit of the pressure is Pascal (Pa).
The pressure is also expressed as:
$\Rightarrow P=\dfrac{N}{{{m}^{2}}}$
Substituting the units of above quantities:
$\begin{align}
  & \Rightarrow \text{Pa=}\dfrac{\text{N}}{{{\text{m}}^{\text{2}}}} \\
 & \Rightarrow \text{Pa=}\dfrac{\text{kg }\!\!\times\!\!\text{ m }\!\!\times\!\!\text{ }{{\text{s}}^{\text{-2}}}}{{{\text{m}}^{\text{2}}}} \\
 & \Rightarrow \text{Pa}=\text{kg }\!\!\times\!\!\text{ }{{\text{m}}^{-1}}\text{ }\!\!\times\!\!\text{ }{{\text{s}}^{\text{-2}}} \\
\end{align}$
The dimension of the pressure is given as:
$\Rightarrow \left[ {{M}^{1}}{{L}^{-1}}{{T}^{-2}} \right]$
The volume of the object is given as:
$\Rightarrow V={{l}^{3}}$
The SI unit of length is meter.
So, the SI unit of volume is cubic meters.
The dimension of volume is,
$\Rightarrow \left[ {{M}^{0}}{{L}^{3}}{{T}^{0}} \right]$
Dimensional formula of pressure ×volume is given as:
$\begin{align}
  & \Rightarrow P\times V=\left[ {{M}^{1}}{{L}^{-1}}{{T}^{-2}} \right]\times \left[ {{M}^{0}}{{L}^{3}}{{T}^{0}} \right] \\
 & \Rightarrow P\times V=\left[ {{M}^{1}}{{L}^{2}}{{T}^{-2}} \right] \\
\end{align}$
The dimension of work is:
$\left[ {{M}^{1}}{{L}^{2}}{{T}^{-2}} \right]$
The dimension of force is:
$\left[ {{M}^{1}}{{L}^{1}}{{T}^{-2}} \right]$
The dimension of momentum is:
$\left[ {{M}^{1}}{{L}^{2}}{{T}^{-1}} \right]$
The dimension of modulus of rigidity is:
$\left[ {{M}^{1}}{{L}^{-1}}{{T}^{-2}} \right]$
Correct option is A.

Note:
Another method to solve above question:
Pressure × Volume has unit as:
$\begin{align}
  & \Rightarrow \text{Pa }\!\!\times\!\!\text{ }{{\text{m}}^{\text{3}}}\text{=}\dfrac{\text{N}}{{{\text{m}}^{\text{2}}}}\text{ }\!\!\times\!\!\text{ }{{\text{m}}^{\text{3}}} \\
 & \Rightarrow \text{Pa }\!\!\times\!\!\text{ }{{\text{m}}^{\text{3}}}\text{=N}\times m \\
 & \Rightarrow \text{Pa }\!\!\times\!\!\text{ }{{\text{m}}^{\text{3}}}\text{=kg }\!\!\times\!\!\text{ m }\!\!\times\!\!\text{ }{{\text{s}}^{\text{-2}}}
\end{align}$
So, Dimension formula is given as:
$\left[ {{M}^{1}}{{L}^{2}}{{T}^{-2}} \right]$