In the previous section, we saw dimensions of physical quantities. In this section, we will see dimensional analysis and it's applications
Consider the arithmetic operations of addition and subtraction. Let us see some examples involving those operations
Example 1:
• Suppose we have two masses
♦ We can add the 'magnitude of one mass' to the 'magnitude of the other mass'
♦ We can subtract the 'magnitude of the smaller mass' from the 'magnitude of the larger mass'
• Note that, both the quantities involved in addition/subtraction:
♦ have the same unit kg
♦ have the same dimension [M]
Example 2:
• Suppose we have two forces
♦ We can add the 'magnitude of one force' to the 'magnitude of the other force'
♦ We can subtract the 'magnitude of the smaller force' from the 'magnitude of the larger force'
• Note that, both the quantities involved in addition/subtraction:
♦ have the same unit N
♦ have the same dimension $\mathbf\small{\left[MLT^{-2}\right]}$
Example 3:
• Suppose we have two volumes
♦ We can add the 'magnitude of one volume' to the 'magnitude of the other volume'
♦ We can subtract the 'magnitude of the smaller volume' from the 'magnitude of the larger volume'
• Note that, both the quantities involved in addition/subtraction:
♦ have the same unit m3
♦ have the same dimension [L3]
■ It is clear that, addition or subtraction can be carried out only between those quantities which have the same dimensions
• Addition or subtractions like 'volume to mass', 'force to velocity, etc., are not possible
Example 1:
(i) Consider the multiplication between mass and velocity
• We can multiply their magnitudes
• We then specify an appropriate unit for the product
(ii) We know that product of mass and velocity is momentum
• Unit of mass is kg
• Unit of velocity is ms-1.
• So unit of momentum is kg ms-1.
(iii) In the same way, dimensions of momentum can also be calculated:
• Dimension of mass is [M]
• Dimensions of velocity is $\mathbf\small{\left[LT^{-1}\right]}$
• So the dimensions of momentum is $\mathbf\small{\left[MLT^{-1}\right]}$
■ Note that, we are combining dimensions as if they are ordinary algebraic symbols
eg: a × b = ab
Example 2:
(i) Consider the division of force by area
• We can divide their magnitudes
• We then specify an appropriate unit for the quotient
(ii) We know that force divided by area gives pressure
• Unit of force is N
• Unit of area is m2.
• Force is in the numerator and area is in the denominator
• So unit of pressure is N m-2
(iii) In the same way, dimensions of pressure can also be calculated:
• Dimension of force is $\mathbf\small{\left[MLT^{-2}\right]}$
• Dimensions of area is $\mathbf\small{\left[L^2\right]}$
• So the dimensions of pressure is $\mathbf\small{\frac{\left[MLT^{-2}\right]}{\left[L^2\right]}=\left[ML^{-1}T^{-2}\right]}$
■ One [L] in the numerator gets cancelled by the [L] in the denominator. We are combining dimensions as if they are ordinary algebraic symbols
eg: $\mathbf\small{\frac{abc^{-2}}{b^2}=ab^{-1}c^{-2}}$
■ So we can write a conclusion about multiplication/division:
For multiplication/division, the two physical quantities need not have the same dimensions. The dimensions can be combined as if they are ordinary algebraic symbols
1. We have seen that, while doing addition or subtraction, the physical quantities must be of the same dimensions
2. An equation will contain many terms. An example is shown below:
$\mathbf\small{a=bc^2+df^3g}$
3. There are 2 terms on the right side. We are adding those two terms
• This addition will be possible only if they have the same dimensions
4. At a first look, they may appear to have different dimensions
• But on simplification, it is quite possible that, they may end up having the same dimensions
An example:
(i) After simplification, '$\mathbf\small{\left[MLT^{-2}\right]\left[L\right]}$' will become '$\mathbf\small{\left[ML^2T^{-2}\right]}$'
(ii) After simplification, '$\mathbf\small{\left[M\right]\left[LT^{-1}\right]^2}$' will also become '$\mathbf\small{\left[ML^2T^{-2}\right]}$'
• So we can do addition/subtraction on them
5. For doing addition or subtraction, two physical quantities must have the same dimensions. This is called the principle of homogeneity of dimensions.
• This principle can be used for testing the correctness of an equation in the 'initial stage'
6. If the two terms on the right side of the equation in (2) are not of the same dimensions, the equation itself is wrong
• Not only the two terms, the term 'a' on the left side also must have the very same dimensions
7. Let us see an example:
• Test the dimensional consistency or homogeneity of the equation: $\mathbf\small{x=x_0+v_0t+\frac{1}{2}at^2}$
♦ x is the total distance traveled by the object in time t
♦ x0 is the distance already traveled when the stop watch was turned on
♦ v0 is the initial velocity at the instant when the stop watch was turned on
♦ a is the acceleration with which the object traveled
Solution:
(i) There are a total of 4 terms. Let us write the dimensions of each:
♦ [x] = [L]
♦ [x0] = [L]
♦ [v0t] = [LT-1][T] = [L]
♦ [1⁄2at2] = [LT-2][T2] = [L]
(ii) The dimensions of all the three terms on the right side of the equation is [L]
• The dimension of the term on the left side of the equation is also [L]
■ So the given equation is dimensionally correct
8. But it is important to remember that, 'being dimensionally correct' is not the final word
• An equation may be dimensionally correct
♦ But as a whole, it may be wrong
• This is because, dimensional analysis do not take 'factors' into account
♦ While deriving an equation, we may wrongly obtain a factor as (𝛑) instead of (0.5𝛑)
♦ Dimensional analysis cannot detect such mistakes
• We will come across more such shortcomings in the later chapters
9. So we can write a conclusion
The conclusion can be written in two steps:
(i) If an equation fails in a 'dimensional consistency test', then it is guaranteed that, the equation is wrong
(ii) Even if an equation passes in a 'dimensional consistency test', it is not guaranteed that, the equation is correct
Solved example 2.30
• Check whether the following equation is dimensionally correct:
$\mathbf\small{\frac{1}{2}mv^2=mgh}$
♦ m is the mass of the object
♦ v is the velocity of the object
♦ g is the acceleration due to gravity
♦ h is the height
Solution:
1. There are a total of 4 items. Let us write the dimensions of each:
♦ [m] = [M]
♦ [v] = [LT-1]
♦ [g] = [LT-2]
♦ [h] = [L]
2. So the dimensions of the left side are:
[M] × [LT-1]2 = [ML2T-2]
3. Dimensions of the right side are:
[M] × [LT-2] × [L] = [ML2T-2]
4. So both sides have the same dimensions. The equation is dimensionally correct
Solved example 2.31
The time period T of a simple pendulum is given by $\mathbf\small{T=2\pi\sqrt{\frac{l}{g}}}$. Show that this equation is dimensionally correct
Solution:
1. There are 3 items in the equation. let us write the dimensions of each:
♦ [T] = [T]
♦ [l] = [L]
♦ [g] = [LT-2]
2. Dimensions of the left side are: [T]
3. Dimensions of the right side are:
$\mathbf\small{\frac{[L]^{\frac{1}{2}}}{[LT^{-2}]^{\frac{1}{2}}}=\frac{1}{[T^{-1}]}=[T]}$
3. So both sides have the same dimension. The equation is dimensionally correct
Solved example 2.32
Write the dimensions of a and b in the relation $\mathbf\small{P=\frac{b-x^2}{at}}$
P is the power, x is the distance and t is the time
Solution:
1. 'Power' is work done in unit time. So we have: $\mathbf\small{[P]=\frac{[ML^2T^{-2}]}{[T]}=[ML^2T^{-3}]}$
2. [x] = [L]
3. [t] = [T]
4. In the given expression, we see that x2 is being subtracted from b
• So they must be having the same dimensions
• We have: [x2] = [L2]
• Thus we get: [b] = [L2]
5. A 'change in a physical quantity' will have the same dimensions as the 'physical quantity'
• So we get: [b-x2] = [M0L2T0]
6. Substituting the known dimensions in the given expression, we get:
$\mathbf\small{[ML^2T^{-3}]=\frac{[L^2]}{[a][T]}}$
$\mathbf\small{\Rightarrow[a]=\frac{[L^2]}{[ML^2T^{-3}][T]}=[M^{-1}L^0T^{2}]}$
Solved example 2.33
The velocity v of a particle depends upon the time t according to the equation $\mathbf\small{v=a+bt+\frac{c}{d+t}}$. Write the dimensions of a, b, c and d
Solution:
1. The dimension of the quantity on the left side = [v] = [LT-1]
2. On the right side, 3 terms are being added
• Dimensions of each of them must be [LT-1]
3. So we get: [a] = [LT-1]
4. [bt] = [LT-1]
⇒ [b][T] = [LT-1]
⇒ [b] = [LT-2]
5. In the last term, we have: $\mathbf\small{\frac{[c]}{[d+t]}=[LT^{-1}]}$
• We see that, d and t are being added
• So we get: [d] = [T]
6. Sum of two quantities will have the same dimension as the quantities being added
• Thus the last term in the given equation can be expressed dimensionally as: $\mathbf\small{\frac{[c]}{[T]}=[LT^{-1}]}$
$\mathbf\small{\Rightarrow [c]=[M^0LT^0]}$
6. So the final answer is:
• [a] = [M0LT-1]
• [b] = [M0LT-2]
• [c] = [M0LT0]
• [d] = [M0L0T]
Solved example 2.34
In the equation $\mathbf\small{p=\frac{a-t^2}{bx}}$, P is the pressure, x is the distance and t is the time. What is the dimension of $\mathbf\small{\frac{a}{b}}$?
Solution:
1. Pressure is force per unit area. So it's dimensions is given by:
$\mathbf\small{[P]=\frac{[MLT^{-2}]}{[L^2]}=[ML^{-1}T^{-2}]}$
2. We see that, t2 is being subtracted from a. So we get: [a] = [t2] = [T2]
• Also, since 'change in magnitude' does not alter the dimension, we have: [a-t2] = [a] = [t2] = [T2]
3. Substituting all the known dimensions, we get: $\mathbf\small{[ML^{-1}T^{-2}]=\frac{[T^2]}{[b][L]}}$
$\mathbf\small{\Rightarrow[b]=\frac{[T^2]}{[ML^{-1}T^{-2}][L]}=[M^{-1}L^{0}T^{4}]}$
4. So the dimension of $\mathbf\small{\frac{a}{b}}$
= $\mathbf\small{=\left[\frac{a}{b}\right]=\frac{[a]}{[b]}=\frac{[T^2]}{[M^{-1}L^{0}T^{4}]}=[ML^{0}T^{-2}]}$
Solved example 2.35
The SI unit of energy is J = kg m2 s-2; that of speed v is ms-1 and of acceleration a is ms-2. Which of the formulae for kinetic energy (K) given below can you rule out on the basis of dimensional arguments (m stands for the mass of the body) :
(a) $\mathbf\small{K=m^2v^3}$
(b) $\mathbf\small{K=\frac{1}{2}mv^2}$
(c) $\mathbf\small{K=ma}$
(d) $\mathbf\small{K=\frac{3}{16}mv^2}$
(e) $\mathbf\small{K=\frac{1}{2}mv^2+ma}$
Solution:
The unit of energy is given as J = kg m2 s-2
So the dimensions of kinetic energy will be: [ML2T-2]
(a) $\mathbf\small{K=m^2v^3}$
• Dimensions on the left side are: [ML2T-2]
• Dimensions on the right side are: [M]2 × [LT-1]3 = [M2L3T-3]
• The two sides are not equal. So this formula can be ruled out
(b) $\mathbf\small{K=\frac{1}{2}mv^2}$
• Dimensions on the left side are: [ML2T-2]
• Dimensions on the right side are: [M] × [LT-1]2 = [ML2T-2]
• The two sides are equal. So this formula can not be ruled out
(c) $\mathbf\small{K=ma}$
• Dimensions on the left side are: [ML2T-2]
• Dimensions on the right side are: [M] × [LT-2] = [MLT-2]
• The two sides are not equal. So this formula can be ruled out
(d) $\mathbf\small{K=\frac{3}{16}mv^2}$
• Dimensions on the left side are: [ML2T-2]
• Dimensions on the right side are: [M] × [LT-1]2 = [ML2T-2]
• The two sides are equal. So this formula can not be ruled out
(e) $\mathbf\small{K=\frac{1}{2}mv^2+ma}$
• Dimensions on the left side are: [ML2T-2]
• Dimensions on the right side are: [M] × [LT-1]2 + [M] × [LT-2]
= [ML2T-2]+[MLT-2]
• We cannot even add the two terms on the right side.
• So this formula can be ruled out
■ We can write a conclusion. It can be done in 4 steps:
(i) We see that, (a), (c) and (e) can be ruled out
(ii) (b) and (d) cannot be ruled out
• But that doesn't mean that, (b) and (d) are correct
(iii) To find the correct formula, we need to do further investigation. We need to look for the correct definition of kinetic energy
(iv) The advantage of doing dimensional analysis is that, no such further investigation is required to be done for (a), (c) and (e)
Solved example 2.36
A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m0 of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes: $\mathbf\small{m=\frac{m_0}{(1-v^2)^{\frac{1}{2}}}}$
Guess where to put the missing c
Solution:
1. The boy writes the equation as: $\mathbf\small{m=\frac{m_0}{(1-v^2)^{\frac{1}{2}}}}$
2. In the denominator, 'v2' is being subtracted from '1'
IF 'v2' is to be subtracted from 'something',
THEN that 'something' must have the dimensions [LT-1]2
BECAUSE, 'v2' has the dimensions [LT-1]2
3. Here the 'something' is '1'
• '1' is just a number. It has no dimensions
• So the subtraction (1-v2) is not possible
4. If we divide 'v' by 'c', then we get a dimensionless quantity
• Because, 'c' is also a velocity
5. So we must divide 'v2' by 'c2'
• Thus the correct equation is: $\mathbf\small{m=\frac{m_0}{(1-\frac{v^2}{c^2})^{\frac{1}{2}}}}$
Consider the arithmetic operations of addition and subtraction. Let us see some examples involving those operations
Example 1:
• Suppose we have two masses
♦ We can add the 'magnitude of one mass' to the 'magnitude of the other mass'
♦ We can subtract the 'magnitude of the smaller mass' from the 'magnitude of the larger mass'
• Note that, both the quantities involved in addition/subtraction:
♦ have the same unit kg
♦ have the same dimension [M]
Example 2:
• Suppose we have two forces
♦ We can add the 'magnitude of one force' to the 'magnitude of the other force'
♦ We can subtract the 'magnitude of the smaller force' from the 'magnitude of the larger force'
• Note that, both the quantities involved in addition/subtraction:
♦ have the same unit N
♦ have the same dimension $\mathbf\small{\left[MLT^{-2}\right]}$
Example 3:
• Suppose we have two volumes
♦ We can add the 'magnitude of one volume' to the 'magnitude of the other volume'
♦ We can subtract the 'magnitude of the smaller volume' from the 'magnitude of the larger volume'
• Note that, both the quantities involved in addition/subtraction:
♦ have the same unit m3
♦ have the same dimension [L3]
■ It is clear that, addition or subtraction can be carried out only between those quantities which have the same dimensions
• Addition or subtractions like 'volume to mass', 'force to velocity, etc., are not possible
Next we will consider multiplication and division. Let us see some examples involving those operations:
(i) Consider the multiplication between mass and velocity
• We can multiply their magnitudes
• We then specify an appropriate unit for the product
(ii) We know that product of mass and velocity is momentum
• Unit of mass is kg
• Unit of velocity is ms-1.
• So unit of momentum is kg ms-1.
(iii) In the same way, dimensions of momentum can also be calculated:
• Dimension of mass is [M]
• Dimensions of velocity is $\mathbf\small{\left[LT^{-1}\right]}$
• So the dimensions of momentum is $\mathbf\small{\left[MLT^{-1}\right]}$
■ Note that, we are combining dimensions as if they are ordinary algebraic symbols
eg: a × b = ab
Example 2:
(i) Consider the division of force by area
• We can divide their magnitudes
• We then specify an appropriate unit for the quotient
(ii) We know that force divided by area gives pressure
• Unit of force is N
• Unit of area is m2.
• Force is in the numerator and area is in the denominator
• So unit of pressure is N m-2
(iii) In the same way, dimensions of pressure can also be calculated:
• Dimension of force is $\mathbf\small{\left[MLT^{-2}\right]}$
• Dimensions of area is $\mathbf\small{\left[L^2\right]}$
• So the dimensions of pressure is $\mathbf\small{\frac{\left[MLT^{-2}\right]}{\left[L^2\right]}=\left[ML^{-1}T^{-2}\right]}$
■ One [L] in the numerator gets cancelled by the [L] in the denominator. We are combining dimensions as if they are ordinary algebraic symbols
eg: $\mathbf\small{\frac{abc^{-2}}{b^2}=ab^{-1}c^{-2}}$
■ So we can write a conclusion about multiplication/division:
For multiplication/division, the two physical quantities need not have the same dimensions. The dimensions can be combined as if they are ordinary algebraic symbols
Based on the above discussion, we can now learn how to check the dimensional consistency of equations. We will write it in steps:
2. An equation will contain many terms. An example is shown below:
$\mathbf\small{a=bc^2+df^3g}$
3. There are 2 terms on the right side. We are adding those two terms
• This addition will be possible only if they have the same dimensions
4. At a first look, they may appear to have different dimensions
• But on simplification, it is quite possible that, they may end up having the same dimensions
An example:
(i) After simplification, '$\mathbf\small{\left[MLT^{-2}\right]\left[L\right]}$' will become '$\mathbf\small{\left[ML^2T^{-2}\right]}$'
(ii) After simplification, '$\mathbf\small{\left[M\right]\left[LT^{-1}\right]^2}$' will also become '$\mathbf\small{\left[ML^2T^{-2}\right]}$'
• So we can do addition/subtraction on them
5. For doing addition or subtraction, two physical quantities must have the same dimensions. This is called the principle of homogeneity of dimensions.
• This principle can be used for testing the correctness of an equation in the 'initial stage'
6. If the two terms on the right side of the equation in (2) are not of the same dimensions, the equation itself is wrong
• Not only the two terms, the term 'a' on the left side also must have the very same dimensions
7. Let us see an example:
• Test the dimensional consistency or homogeneity of the equation: $\mathbf\small{x=x_0+v_0t+\frac{1}{2}at^2}$
♦ x is the total distance traveled by the object in time t
♦ x0 is the distance already traveled when the stop watch was turned on
♦ v0 is the initial velocity at the instant when the stop watch was turned on
♦ a is the acceleration with which the object traveled
Solution:
(i) There are a total of 4 terms. Let us write the dimensions of each:
♦ [x] = [L]
♦ [x0] = [L]
♦ [v0t] = [LT-1][T] = [L]
♦ [1⁄2at2] = [LT-2][T2] = [L]
(ii) The dimensions of all the three terms on the right side of the equation is [L]
• The dimension of the term on the left side of the equation is also [L]
■ So the given equation is dimensionally correct
8. But it is important to remember that, 'being dimensionally correct' is not the final word
• An equation may be dimensionally correct
♦ But as a whole, it may be wrong
• This is because, dimensional analysis do not take 'factors' into account
♦ While deriving an equation, we may wrongly obtain a factor as (𝛑) instead of (0.5𝛑)
♦ Dimensional analysis cannot detect such mistakes
• We will come across more such shortcomings in the later chapters
9. So we can write a conclusion
The conclusion can be written in two steps:
(i) If an equation fails in a 'dimensional consistency test', then it is guaranteed that, the equation is wrong
(ii) Even if an equation passes in a 'dimensional consistency test', it is not guaranteed that, the equation is correct
Now we will see some solved examples
• Check whether the following equation is dimensionally correct:
$\mathbf\small{\frac{1}{2}mv^2=mgh}$
♦ m is the mass of the object
♦ v is the velocity of the object
♦ g is the acceleration due to gravity
♦ h is the height
Solution:
1. There are a total of 4 items. Let us write the dimensions of each:
♦ [m] = [M]
♦ [v] = [LT-1]
♦ [g] = [LT-2]
♦ [h] = [L]
2. So the dimensions of the left side are:
[M] × [LT-1]2 = [ML2T-2]
3. Dimensions of the right side are:
[M] × [LT-2] × [L] = [ML2T-2]
4. So both sides have the same dimensions. The equation is dimensionally correct
Solved example 2.31
The time period T of a simple pendulum is given by $\mathbf\small{T=2\pi\sqrt{\frac{l}{g}}}$. Show that this equation is dimensionally correct
Solution:
1. There are 3 items in the equation. let us write the dimensions of each:
♦ [T] = [T]
♦ [l] = [L]
♦ [g] = [LT-2]
2. Dimensions of the left side are: [T]
3. Dimensions of the right side are:
$\mathbf\small{\frac{[L]^{\frac{1}{2}}}{[LT^{-2}]^{\frac{1}{2}}}=\frac{1}{[T^{-1}]}=[T]}$
3. So both sides have the same dimension. The equation is dimensionally correct
Solved example 2.32
Write the dimensions of a and b in the relation $\mathbf\small{P=\frac{b-x^2}{at}}$
P is the power, x is the distance and t is the time
Solution:
1. 'Power' is work done in unit time. So we have: $\mathbf\small{[P]=\frac{[ML^2T^{-2}]}{[T]}=[ML^2T^{-3}]}$
2. [x] = [L]
3. [t] = [T]
4. In the given expression, we see that x2 is being subtracted from b
• So they must be having the same dimensions
• We have: [x2] = [L2]
• Thus we get: [b] = [L2]
5. A 'change in a physical quantity' will have the same dimensions as the 'physical quantity'
• So we get: [b-x2] = [M0L2T0]
6. Substituting the known dimensions in the given expression, we get:
$\mathbf\small{[ML^2T^{-3}]=\frac{[L^2]}{[a][T]}}$
$\mathbf\small{\Rightarrow[a]=\frac{[L^2]}{[ML^2T^{-3}][T]}=[M^{-1}L^0T^{2}]}$
Solved example 2.33
The velocity v of a particle depends upon the time t according to the equation $\mathbf\small{v=a+bt+\frac{c}{d+t}}$. Write the dimensions of a, b, c and d
Solution:
1. The dimension of the quantity on the left side = [v] = [LT-1]
2. On the right side, 3 terms are being added
• Dimensions of each of them must be [LT-1]
3. So we get: [a] = [LT-1]
4. [bt] = [LT-1]
⇒ [b][T] = [LT-1]
⇒ [b] = [LT-2]
5. In the last term, we have: $\mathbf\small{\frac{[c]}{[d+t]}=[LT^{-1}]}$
• We see that, d and t are being added
• So we get: [d] = [T]
6. Sum of two quantities will have the same dimension as the quantities being added
• Thus the last term in the given equation can be expressed dimensionally as: $\mathbf\small{\frac{[c]}{[T]}=[LT^{-1}]}$
$\mathbf\small{\Rightarrow [c]=[M^0LT^0]}$
6. So the final answer is:
• [a] = [M0LT-1]
• [b] = [M0LT-2]
• [c] = [M0LT0]
• [d] = [M0L0T]
Solved example 2.34
In the equation $\mathbf\small{p=\frac{a-t^2}{bx}}$, P is the pressure, x is the distance and t is the time. What is the dimension of $\mathbf\small{\frac{a}{b}}$?
Solution:
1. Pressure is force per unit area. So it's dimensions is given by:
$\mathbf\small{[P]=\frac{[MLT^{-2}]}{[L^2]}=[ML^{-1}T^{-2}]}$
2. We see that, t2 is being subtracted from a. So we get: [a] = [t2] = [T2]
• Also, since 'change in magnitude' does not alter the dimension, we have: [a-t2] = [a] = [t2] = [T2]
3. Substituting all the known dimensions, we get: $\mathbf\small{[ML^{-1}T^{-2}]=\frac{[T^2]}{[b][L]}}$
$\mathbf\small{\Rightarrow[b]=\frac{[T^2]}{[ML^{-1}T^{-2}][L]}=[M^{-1}L^{0}T^{4}]}$
4. So the dimension of $\mathbf\small{\frac{a}{b}}$
= $\mathbf\small{=\left[\frac{a}{b}\right]=\frac{[a]}{[b]}=\frac{[T^2]}{[M^{-1}L^{0}T^{4}]}=[ML^{0}T^{-2}]}$
Solved example 2.35
The SI unit of energy is J = kg m2 s-2; that of speed v is ms-1 and of acceleration a is ms-2. Which of the formulae for kinetic energy (K) given below can you rule out on the basis of dimensional arguments (m stands for the mass of the body) :
(a) $\mathbf\small{K=m^2v^3}$
(b) $\mathbf\small{K=\frac{1}{2}mv^2}$
(c) $\mathbf\small{K=ma}$
(d) $\mathbf\small{K=\frac{3}{16}mv^2}$
(e) $\mathbf\small{K=\frac{1}{2}mv^2+ma}$
Solution:
The unit of energy is given as J = kg m2 s-2
So the dimensions of kinetic energy will be: [ML2T-2]
(a) $\mathbf\small{K=m^2v^3}$
• Dimensions on the left side are: [ML2T-2]
• Dimensions on the right side are: [M]2 × [LT-1]3 = [M2L3T-3]
• The two sides are not equal. So this formula can be ruled out
(b) $\mathbf\small{K=\frac{1}{2}mv^2}$
• Dimensions on the left side are: [ML2T-2]
• Dimensions on the right side are: [M] × [LT-1]2 = [ML2T-2]
• The two sides are equal. So this formula can not be ruled out
(c) $\mathbf\small{K=ma}$
• Dimensions on the left side are: [ML2T-2]
• Dimensions on the right side are: [M] × [LT-2] = [MLT-2]
• The two sides are not equal. So this formula can be ruled out
(d) $\mathbf\small{K=\frac{3}{16}mv^2}$
• Dimensions on the left side are: [ML2T-2]
• Dimensions on the right side are: [M] × [LT-1]2 = [ML2T-2]
• The two sides are equal. So this formula can not be ruled out
(e) $\mathbf\small{K=\frac{1}{2}mv^2+ma}$
• Dimensions on the left side are: [ML2T-2]
• Dimensions on the right side are: [M] × [LT-1]2 + [M] × [LT-2]
= [ML2T-2]+[MLT-2]
• We cannot even add the two terms on the right side.
• So this formula can be ruled out
■ We can write a conclusion. It can be done in 4 steps:
(i) We see that, (a), (c) and (e) can be ruled out
(ii) (b) and (d) cannot be ruled out
• But that doesn't mean that, (b) and (d) are correct
(iii) To find the correct formula, we need to do further investigation. We need to look for the correct definition of kinetic energy
(iv) The advantage of doing dimensional analysis is that, no such further investigation is required to be done for (a), (c) and (e)
Solved example 2.36
A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m0 of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes: $\mathbf\small{m=\frac{m_0}{(1-v^2)^{\frac{1}{2}}}}$
Guess where to put the missing c
Solution:
1. The boy writes the equation as: $\mathbf\small{m=\frac{m_0}{(1-v^2)^{\frac{1}{2}}}}$
2. In the denominator, 'v2' is being subtracted from '1'
IF 'v2' is to be subtracted from 'something',
THEN that 'something' must have the dimensions [LT-1]2
BECAUSE, 'v2' has the dimensions [LT-1]2
3. Here the 'something' is '1'
• '1' is just a number. It has no dimensions
• So the subtraction (1-v2) is not possible
4. If we divide 'v' by 'c', then we get a dimensionless quantity
• Because, 'c' is also a velocity
5. So we must divide 'v2' by 'c2'
• Thus the correct equation is: $\mathbf\small{m=\frac{m_0}{(1-\frac{v^2}{c^2})^{\frac{1}{2}}}}$
In the next section, we will see unit conversion
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