Question

Find the dimensions of
a).Linear momentum
b).Frequency
c).Pressure

Answer 1

Hint: We can find the dimension of a particular quantity by writing its relation with other quantities and breaking it into parts, whose dimensions are known to us. Then we use them to calculate the dimension of the unknown quantity that has been asked in the question.

Complete answer:
a).We know that linear momentum $\left( p \right)$ is the product of mass $\left( m \right)$ of the body and its velocity
$\left( v \right)$. So, if we write this as a formula we get:
$p=m\times v$
Dimension of mass is $\left[ {{M}^{1}}{{L}^{0}}{{T}^{0}} \right]$, while the dimension of velocity is $\left[ {{M}^{0}}{{L}^{1}}{{T}^{-1}} \right]$, the dimension of linear momentum will be:
$\begin{align}
  & \left[ {{M}^{1}}{{L}^{0}}{{T}^{0}} \right]\times \left[ {{M}^{0}}{{L}^{1}}{{T}^{-1}} \right] \\
 & \therefore \left[ {{M}^{1}}{{L}^{1}}{{T}^{-1}} \right] \\
\end{align}$
The dimension of linear momentum is $\left[ {{M}^{1}}{{L}^{1}}{{T}^{-1}} \right]$ and its SI units is $\text{kg m/s}$.
b).We know that frequency $\left( f \right)$ is the inverse of time $\left( t \right)$. So, if we write this as a formula we get:
$f=\dfrac{1}{t}$
Dimension of time is $\left[ {{M}^{0}}{{L}^{0}}{{T}^{1}} \right]$, therefore, the dimension of frequency will be:
$\begin{align}
  & \dfrac{1}{\left[ {{M}^{0}}{{L}^{0}}{{T}^{1}} \right]} \\
 & \therefore \left[ {{M}^{0}}{{L}^{0}}{{T}^{-1}} \right] \\
\end{align}$
The dimension of frequency is $\left[ {{M}^{0}}{{L}^{0}}{{T}^{-1}} \right]$ and its SI units is ${{\text{s}}^{-1}}$.
c).We know that pressure $\left( P \right)$ is defined as force $F$ acting per unit area $\left( A \right)$. So, if we write this as a formula we get:
$P=\dfrac{F}{A}$
Dimension of force can be found out by its formula. We know that force $\left( F \right)$ is the product of mass $\left( m \right)$ of the body and acceleration $\left( a \right)$. So, if we write this as a formula we get:
$F=m\times a$
Dimension of mass is $\left[ {{M}^{1}}{{L}^{0}}{{T}^{0}} \right]$, while the dimension of acceleration is $\left[ {{M}^{0}}{{L}^{1}}{{T}^{-2}} \right]$, the dimension of force will be:
$\begin{align}
  & \left[ {{M}^{1}}{{L}^{0}}{{T}^{0}} \right]\times \left[ {{M}^{0}}{{L}^{1}}{{T}^{-2}} \right] \\
 & \therefore \left[ {{M}^{1}}{{L}^{1}}{{T}^{-2}} \right] \\
\end{align}$
Now, since the dimension of force is $\left[ {{M}^{1}}{{L}^{1}}{{T}^{-2}} \right]$, and the dimension of area is $\left[ {{M}^{0}}{{L}^{2}}{{T}^{0}} \right]$, the dimension of pressure will be:
$\begin{align}
  & \dfrac{\left[ {{M}^{1}}{{L}^{1}}{{T}^{-2}} \right]}{\left[ {{M}^{0}}{{L}^{2}}{{T}^{0}} \right]} \\
 & \therefore \left[ {{M}^{1}}{{L}^{-1}}{{T}^{-2}} \right] \\
\end{align}$
The dimension of pressure is $\left[ {{M}^{1}}{{L}^{-1}}{{T}^{-2}} \right]$ and its SI units is Pascal.

Note:
We need to take care that all the physical quantities must be expressed in their absolute units, i.e., no fractions should be used. One of the most important steps to find the dimension of a particular quantity is to break down into small units and then proceed by considering the dimensions of those units.