Physics > Unit and Dimensions > 9.0 Dimensional equation
Unit and Dimensions
1.0 Introduction
2.0 Physical quantity
3.0 SI units
3.1 Definition of standard units
3.2 System of units
3.3 Rules for writing units
3.4 Characteristics of a standard unit
3.5 Advantages of SI
4.0 SI prefixes
5.0 Conversion of units
6.0 Important practical units
7.0 Dimensions
8.0 Dimensional formula
9.0 Dimensional equation
10.0 List of dimensional formula
11.0 Application of dimensional analysis
11.1 To check the dimensional consistency of equations
11.2 To deduce relation among the physical quantities
11.3 To convert one system of unit into another system of unit
12.0 Limitations of dimensional analysis
9.1 Solved examples
3.2 System of units
3.3 Rules for writing units
3.4 Characteristics of a standard unit
3.5 Advantages of SI
11.2 To deduce relation among the physical quantities
11.3 To convert one system of unit into another system of unit
Question: The dimensions of $\frac{a}{b}$ in the equation $P = \frac{{a - {t^2}}}{{bx}}$ where $P$ is pressure, $x$ is distance and $t$ is time.
Solution: The equation is given as,
$$P = \frac{{a - {t^2}}}{{bx}}$$
As $a$ and $t^2$ are subtracted. So, $a$ should have the same dimension as that of $t^2$.
Therefore, $$\left[ a \right] = \left[ {{T^2}} \right]$$
Also, $$\left[ x \right] = \left[ L \right]$$
We known the dimension of pressure is, $$\left[ P \right] = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$$So, $$\left[ {M{L^{ - 1}}{T^{ - 2}}} \right] = \frac{{\left[ {{T^2}} \right]}}{{\left[ b \right]\left[ L \right]}}$$$$\left[ b \right] = \frac{{\left[ {{T^2}} \right]}}{{\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]\left[ L \right]}}$$$$\left[ b \right] = \left[ {{M^{ - 1}}{L^0}{T^4}} \right]$$.
So, dimension of $\frac{a}{b}$ is, $$\left[ {\frac{a}{b}} \right] = \left[ {M{L^0}{T^{ - 2}}} \right]$$
Question: The time of oscillation of a simple pendulum depends on the length of the string $(l)$, mass of the bob $(m)$ and acceleration due to gravity $(g)$. Derive the formula.
Solution: From the question, we can write,
$$t \propto {l^a}{m^b}{g^c}$$
To remove the proportional sign we will add a proportionality constant $(k)$.
$$t = k{l^a}{m^b}{g^c} \quad ...(i)$$
Writing dimension for all the quantities as,
$$\left[ t \right] = \left[ T \right]$$$$\left[ l \right] = \left[ L \right]$$$$\left[ m \right] = \left[ M \right]$$$$\left[ g \right] = \left[ {L{T^{ - 2}}} \right]$$
Substituting all the values in equation $(i)$ we get,
$$\left[ T \right] = k{\left[ L \right]^a}{\left[ M \right]^b}{\left[ {L{T^{ - 2}}} \right]^c}$$$$\left[ {{T^1}} \right] = k\left[ {{L^{a + c}}{M^b}{T^{ - 2c}}} \right]$$Solving the above equation for $a,b,c$ we get,
$$ - 2c = 1$$$$c = - \frac{1}{2}$$$$b = 0$$and $$a + c = 0$$$$a = - c$$So, $$a = \frac{1}{2}$$Substituing the values of $a,b,c$ in equation $(i)$ we get,
$$t = k{\left( l \right)^{\frac{1}{2}}}{\left( m \right)^0}{\left( g \right)^{ - \frac{1}{2}}}$$$$t = k\sqrt {\frac{l}{g}} $$
