In the previous section, we saw how to check whether an equation is dimensionally correct or not. In this section, we will see another application of dimensional analysis, which is: Unit conversion
Let us see an example. We will write it in steps:
1. An engineer measures the length of a wall and notes it down as 3.05 m
• Another engineer measures the length of the wall and notes it down as 10 feet
■ Doing a dimensional analysis will help us to understand why the difference occurred:
2. The first engineer used the SI system
• We know that, in this system, the basic unit for length is 1 m
• So we can write: [L] = [1 m]
3. The second engineer use the Imperial system
• In this system, the basic unit for length is 1 foot
• Let L1 represent the basic unit for length in that system
• From the data book, we have: 1 foot = 0.3048 m
(This information is available online also)
• So we can write: [L1] = [0.3048 m]
4. Let 'x' be the reading obtained by the second engineer
• This implies that, 'x' number of his units will make up the 'length of the wall'
■ The length of the wall is a constant regardless of the system used for measuring it's length
5. So we can write:
3.05 × [L] = x × [L1]
⇒ 3.05 × [1 m] = x × [0.3048 m]
$\mathbf\small{\Rightarrow x = \frac{3.05 \times 1}{0.3048}=10.00}$
• So the reading obtained by the second engineer will be 10 feet
Let us see an example in volume:
1. An engineer calculates the volume of a rectangular prism as 5.67 m3.
• Another engineer calculates the volume of the prism as 200.23 cubic feet
2. Let us do a dimensional analysis and see if the same 200.23 can be obtained:
• We can write:
5.67 × [L3] = x × [L13]
⇒ 5.05 × [1 m3] = x × [0.30483 m3]
$\mathbf\small{\Rightarrow x = \frac{5.67 \times 1}{0.3048^3}=200.23}$
• So the reading obtained by the second engineer will be 200.23 cubic feet
An example in speed:
In an experiment, the speed of a body was obtained as 10 ms-1. Suppose that the measurements were made using another system in which, the basic units are as follows:
• Basic unit for length is km
• Basic unit for time is hours
■ What would be the result of the experiment?
Solution:
1. The dimensions of speed are: $\mathbf\small{[LT^{-1}]}$
• This can be written as: $\mathbf\small{\frac{[L]}{[T]}}$
2. Let 'x' be the new result
• So we can write:
$\mathbf\small{\frac{10[L]}{[T]}=\frac{x[L_1]}{[T_1]}}$
3. L1 is the basic unit for length in the new system. It is given as 1 km
♦ So we get: L1 = 1 km = 1000 m
• T1 is the basic unit for time in the new system. It is given as 1 hour
♦ So we get: T1 = 1 hour = (60 × 60) s
4. Substituting the values, we get:
$\mathbf\small{\frac{10[1\,m]}{[1\,s]}=\frac{x[1000\,m]}{[(60\times 60)s]}}$
$\mathbf\small{\Rightarrow x=\frac{10\times(60\times 60)}{1000}=36}$
• So the result would be 36 km hr-1
Now we will see some solved examples
Solved example 2.37
In an experiment which uses the SI system, the energy was measured as 100 J. Suppose that the measurements were made using another system in which, the basic units are as follows:
• Basic unit for mass is 250 g
• Basic unit for length is 20 cm
• Basic unit for time is half a minute
■ What would be the result of the experiment?
Solution:
1. The dimensions of energy are: $\mathbf\small{[ML^2T^{-2}]}$
• This can be written as: $\mathbf\small{\frac{[M][L^2]}{[T^{2}]}}$
2. Let 'x' be the new result
• So we can write:
$\mathbf\small{\frac{100[M][L^2]}{[T^{2}]}=\frac{x[M_1][L_1^2]}{[T_1^{2}]}}$
3. M1 is the basic unit for mass in the new system. It is given as 250 g
♦ So we get: M1 = 250 g = 0.250 kg
• L1 is the basic unit for length in the new system. It is given as 20 cm
♦ So we get: L1 = 20 cm = 0.20 m
• T1 is the basic unit for time in the new system. It is given as half a minute
♦ So we get: T1 = half a minute = 30 s
4. Substituting the values, we get:
$\mathbf\small{\frac{100[1\,kg][1^2\,m^2]}{[1^2\,s^2]}=\frac{x[0.250\,kg][0.20^2\,m^2]}{[30^2\,s^2]}}$
$\mathbf\small{\Rightarrow x=\frac{100\times 30^2}{0.250\times 0.20^2}=9 \times 10^6}$
• So the result would be 9 × 106 new units.
Solved example 2.38
The value of gravitational constant is G = 6.67 × 10-11 N m2 kg-2 in SI units. Convert it into CGS system of units
Solution:
1. In the CGS system,
• Basic unit for mass is 1 g
• Basic unit for length is 1 cm
• Basic unit for time is 1 s
2. The unit of G is given as N m2 kg-2.
• N is the unit of force. Force has dimensions $\mathbf\small{[MLT^{-2}]}$
• m is the unit of length. Length has dimension [L]
• kg is the unit for mass. Mass has the dimension [M]
3. So the dimensions of G are: $\mathbf\small{\frac{[MLT^{-2}][L^2]}{[M^{2}]}}$
• This can be written as: $\mathbf\small{\frac{[L^3]}{[M][T^{2}]}}$
4. Let 'x' be the new result
• So we can write: $\mathbf\small{\frac{6.67 \times 10^{-11}[L^3]}{[M][T^{2}]}=\frac{x[L_1^3]}{[M_1][T_1^{2}]}}$
5. M1 is the basic unit for mass in the new system. It is 1 g
♦ So we get: M1 = 1 g = 0.001 kg
• L1 is the basic unit for length in the new system. It is 1 cm
♦ So we get: L1 = 1 cm = 0.01 m
• T1 is the basic unit for time in the new system. It is the same 1 s
♦ So we get: T1 = 1 s
6. Substituting the values, we get:
$\mathbf\small{\frac{6.67 \times 10^{-11}[1^3\,m^3]}{[1\,kg][1^{2}\,s^2]}=\frac{x[0.01^3\,m^3]}{[0.001\,kg][1^{2}\,s^2]}}$
$\mathbf\small{\Rightarrow x =\frac{6.67 \times 10^{-11}\times 0.001}{0.01^3}=6.67 \times 10^{-8}}$
Solved example 2.39
A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1 J = 1 kg m2 s-2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude 4.2 α-1 β-2 γ2 in terms of the new units
Solution:
1. Given that: 1 J = 1 kg m2 s-2
• kg is the unit for mass. Mass has the dimension [M]
• m is the unit of length. Length has dimension [L]
• s is the unit for time. Time has the dimension [T]
2. So the dimensions of energy are: $\mathbf\small{\frac{[M][L^2]}{[T^2]}}$
3. Let 'x' be the new result
• So we can write:
$\mathbf\small{\frac{4.2[M][L^2]}{[T^2]}=\frac{x[M_1][L_1^2]}{[T_1^2]}}$
4. M1 is the basic unit for mass in the new system. It is α kg
• L1 is the basic unit for length in the new system. It is β m
• T1 is the basic unit for time in the new system. It is γ s
5. Substituting the values, we get:
$\mathbf\small{\frac{4.2[1\,kg][1^2\;m^2]}{[1^2\,s^2]}=\frac{x[\alpha\,kg][\beta^2\;m^2]}{[\gamma^2\,s^2]}}$
$\mathbf\small{\Rightarrow x=4.2\,\alpha^{-1}\beta^{-2}\gamma^2}$
Solved example 2.40
A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?
Solution:
1. The dimensions of speed are: $\mathbf\small{[LT^{-1}]}$
• This can be written as: $\mathbf\small{\frac{[L]}{[T]}}$
2. When we employ the SI system, the speed of light is 3 × 108 ms-1
• In the 'new system', let the speed of light be 'x'
• So we can write:
$\mathbf\small{\frac{3\times 10^8[L]}{[T]}=\frac{x[L_1]}{[T_1]}}$
3. But 'x' is given to us as 'unity'
• That means, in the new system, the light travels '1 new unit' in one second
4. L1 is the basic unit for length in the new system. We have to find it
5. T1 is the basic unit for time in the new system
• It is clear that: T1 = 1 s
6. Substituting the values, we get:
$\mathbf\small{\frac{3\times 10^8[1\,m]}{[1\,s]}=\frac{1[L_1]}{[1\,s]}}$
$\mathbf\small{\Rightarrow L_1=3\times 10^8\,m}$
• That means, 'the basic unit for length (L1) in the new system' is equal to 3 × 108 m
7. Now we apply the method in reverse:
(The word 'reverse' can be used because, first we calculated L1 using x. Now we are going to calculate x using L1)
(i) Distance in the SI system
= Velocity in the SI system × Time in the SI system
= 3 × 108 ms-1 × [(8 × 60) + 20] s
= 3 × 108 ms-1 × [500] s
= 3 × 500 × 108 m.
(ii) The dimension of distance is: [L]
(iii). Let 'x' be the new result
• So we can write:
3 × 108[L] = x[L1]
(iv) L1 is the basic unit for length in the new system. We obtained it as 3 × 108 m
♦ So we get: L1 = 3 × 108 m
(v) Substituting the values in (iii), we get:
3 × 500 × 108[1 m] = x[3 × 108 m]
⇒ x = 500
• So we can write:
Distance between Sun and Earth = 500 'new units'
Solved example 2.41
The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m2. What is the linear magnification of the projector-screen arrangement.
Solution:
1. Consider the two rectangles in fig.2.26 below:
• The larger rectangle is a 'proportional enlargement' of the smaller rectangle
(Details about proportionality can be seen here)
2. Length l2 of the larger rectangle is obtained by multiplying l1 by a factor 'k'
♦ Where l1 is the length of the smaller rectangle
• Width b2 of the larger rectangle is obtained by multiplying b1 by the same factor 'k'
♦ Where b1 is the width of the smaller rectangle
3. Note that, the same factor 'k' must be used for both length and width. otherwise, there will be distortion
• Thus we get: (l1 × b1) = 1.75 cm2
(l2 × b2) = 1.55 m2 = 1.55 × 104 cm2
⇒ (kl1 × kb1) = 1.55 × 104 cm2
⇒ k2 × (l1 × b1) = 1.55 × 104 cm2.
4. But (l1 × b1) = 1.75 cm2
• So we get: k2(1.75) = 1.55 × 104 cm2
⇒ k2 = (1.55⁄1.75) × 104 = 0.885714286 × 104 = 8857.14
⇒ k = √(8857.14) = 94.11
• So the linear magnification (k) = 94.11
Let us see an example. We will write it in steps:
1. An engineer measures the length of a wall and notes it down as 3.05 m
• Another engineer measures the length of the wall and notes it down as 10 feet
■ Doing a dimensional analysis will help us to understand why the difference occurred:
2. The first engineer used the SI system
• We know that, in this system, the basic unit for length is 1 m
• So we can write: [L] = [1 m]
3. The second engineer use the Imperial system
• In this system, the basic unit for length is 1 foot
• Let L1 represent the basic unit for length in that system
• From the data book, we have: 1 foot = 0.3048 m
(This information is available online also)
• So we can write: [L1] = [0.3048 m]
4. Let 'x' be the reading obtained by the second engineer
• This implies that, 'x' number of his units will make up the 'length of the wall'
■ The length of the wall is a constant regardless of the system used for measuring it's length
5. So we can write:
3.05 × [L] = x × [L1]
⇒ 3.05 × [1 m] = x × [0.3048 m]
$\mathbf\small{\Rightarrow x = \frac{3.05 \times 1}{0.3048}=10.00}$
• So the reading obtained by the second engineer will be 10 feet
Let us see an example in volume:
1. An engineer calculates the volume of a rectangular prism as 5.67 m3.
• Another engineer calculates the volume of the prism as 200.23 cubic feet
2. Let us do a dimensional analysis and see if the same 200.23 can be obtained:
• We can write:
5.67 × [L3] = x × [L13]
⇒ 5.05 × [1 m3] = x × [0.30483 m3]
$\mathbf\small{\Rightarrow x = \frac{5.67 \times 1}{0.3048^3}=200.23}$
• So the reading obtained by the second engineer will be 200.23 cubic feet
An example in speed:
In an experiment, the speed of a body was obtained as 10 ms-1. Suppose that the measurements were made using another system in which, the basic units are as follows:
• Basic unit for length is km
• Basic unit for time is hours
■ What would be the result of the experiment?
Solution:
1. The dimensions of speed are: $\mathbf\small{[LT^{-1}]}$
• This can be written as: $\mathbf\small{\frac{[L]}{[T]}}$
2. Let 'x' be the new result
• So we can write:
$\mathbf\small{\frac{10[L]}{[T]}=\frac{x[L_1]}{[T_1]}}$
3. L1 is the basic unit for length in the new system. It is given as 1 km
♦ So we get: L1 = 1 km = 1000 m
• T1 is the basic unit for time in the new system. It is given as 1 hour
♦ So we get: T1 = 1 hour = (60 × 60) s
4. Substituting the values, we get:
$\mathbf\small{\frac{10[1\,m]}{[1\,s]}=\frac{x[1000\,m]}{[(60\times 60)s]}}$
$\mathbf\small{\Rightarrow x=\frac{10\times(60\times 60)}{1000}=36}$
• So the result would be 36 km hr-1
Now we will see some solved examples
Solved example 2.37
In an experiment which uses the SI system, the energy was measured as 100 J. Suppose that the measurements were made using another system in which, the basic units are as follows:
• Basic unit for mass is 250 g
• Basic unit for length is 20 cm
• Basic unit for time is half a minute
■ What would be the result of the experiment?
Solution:
1. The dimensions of energy are: $\mathbf\small{[ML^2T^{-2}]}$
• This can be written as: $\mathbf\small{\frac{[M][L^2]}{[T^{2}]}}$
2. Let 'x' be the new result
• So we can write:
$\mathbf\small{\frac{100[M][L^2]}{[T^{2}]}=\frac{x[M_1][L_1^2]}{[T_1^{2}]}}$
3. M1 is the basic unit for mass in the new system. It is given as 250 g
♦ So we get: M1 = 250 g = 0.250 kg
• L1 is the basic unit for length in the new system. It is given as 20 cm
♦ So we get: L1 = 20 cm = 0.20 m
• T1 is the basic unit for time in the new system. It is given as half a minute
♦ So we get: T1 = half a minute = 30 s
4. Substituting the values, we get:
$\mathbf\small{\frac{100[1\,kg][1^2\,m^2]}{[1^2\,s^2]}=\frac{x[0.250\,kg][0.20^2\,m^2]}{[30^2\,s^2]}}$
$\mathbf\small{\Rightarrow x=\frac{100\times 30^2}{0.250\times 0.20^2}=9 \times 10^6}$
• So the result would be 9 × 106 new units.
Solved example 2.38
The value of gravitational constant is G = 6.67 × 10-11 N m2 kg-2 in SI units. Convert it into CGS system of units
Solution:
1. In the CGS system,
• Basic unit for mass is 1 g
• Basic unit for length is 1 cm
• Basic unit for time is 1 s
2. The unit of G is given as N m2 kg-2.
• N is the unit of force. Force has dimensions $\mathbf\small{[MLT^{-2}]}$
• m is the unit of length. Length has dimension [L]
• kg is the unit for mass. Mass has the dimension [M]
3. So the dimensions of G are: $\mathbf\small{\frac{[MLT^{-2}][L^2]}{[M^{2}]}}$
• This can be written as: $\mathbf\small{\frac{[L^3]}{[M][T^{2}]}}$
4. Let 'x' be the new result
• So we can write: $\mathbf\small{\frac{6.67 \times 10^{-11}[L^3]}{[M][T^{2}]}=\frac{x[L_1^3]}{[M_1][T_1^{2}]}}$
5. M1 is the basic unit for mass in the new system. It is 1 g
♦ So we get: M1 = 1 g = 0.001 kg
• L1 is the basic unit for length in the new system. It is 1 cm
♦ So we get: L1 = 1 cm = 0.01 m
• T1 is the basic unit for time in the new system. It is the same 1 s
♦ So we get: T1 = 1 s
6. Substituting the values, we get:
$\mathbf\small{\frac{6.67 \times 10^{-11}[1^3\,m^3]}{[1\,kg][1^{2}\,s^2]}=\frac{x[0.01^3\,m^3]}{[0.001\,kg][1^{2}\,s^2]}}$
$\mathbf\small{\Rightarrow x =\frac{6.67 \times 10^{-11}\times 0.001}{0.01^3}=6.67 \times 10^{-8}}$
Solved example 2.39
A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1 J = 1 kg m2 s-2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude 4.2 α-1 β-2 γ2 in terms of the new units
Solution:
1. Given that: 1 J = 1 kg m2 s-2
• kg is the unit for mass. Mass has the dimension [M]
• m is the unit of length. Length has dimension [L]
• s is the unit for time. Time has the dimension [T]
2. So the dimensions of energy are: $\mathbf\small{\frac{[M][L^2]}{[T^2]}}$
3. Let 'x' be the new result
• So we can write:
$\mathbf\small{\frac{4.2[M][L^2]}{[T^2]}=\frac{x[M_1][L_1^2]}{[T_1^2]}}$
4. M1 is the basic unit for mass in the new system. It is α kg
• L1 is the basic unit for length in the new system. It is β m
• T1 is the basic unit for time in the new system. It is γ s
5. Substituting the values, we get:
$\mathbf\small{\frac{4.2[1\,kg][1^2\;m^2]}{[1^2\,s^2]}=\frac{x[\alpha\,kg][\beta^2\;m^2]}{[\gamma^2\,s^2]}}$
$\mathbf\small{\Rightarrow x=4.2\,\alpha^{-1}\beta^{-2}\gamma^2}$
Solved example 2.40
A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?
Solution:
1. The dimensions of speed are: $\mathbf\small{[LT^{-1}]}$
• This can be written as: $\mathbf\small{\frac{[L]}{[T]}}$
2. When we employ the SI system, the speed of light is 3 × 108 ms-1
• In the 'new system', let the speed of light be 'x'
• So we can write:
$\mathbf\small{\frac{3\times 10^8[L]}{[T]}=\frac{x[L_1]}{[T_1]}}$
3. But 'x' is given to us as 'unity'
• That means, in the new system, the light travels '1 new unit' in one second
4. L1 is the basic unit for length in the new system. We have to find it
5. T1 is the basic unit for time in the new system
• It is clear that: T1 = 1 s
6. Substituting the values, we get:
$\mathbf\small{\frac{3\times 10^8[1\,m]}{[1\,s]}=\frac{1[L_1]}{[1\,s]}}$
$\mathbf\small{\Rightarrow L_1=3\times 10^8\,m}$
• That means, 'the basic unit for length (L1) in the new system' is equal to 3 × 108 m
7. Now we apply the method in reverse:
(The word 'reverse' can be used because, first we calculated L1 using x. Now we are going to calculate x using L1)
(i) Distance in the SI system
= Velocity in the SI system × Time in the SI system
= 3 × 108 ms-1 × [(8 × 60) + 20] s
= 3 × 108 ms-1 × [500] s
= 3 × 500 × 108 m.
(ii) The dimension of distance is: [L]
(iii). Let 'x' be the new result
• So we can write:
3 × 108[L] = x[L1]
(iv) L1 is the basic unit for length in the new system. We obtained it as 3 × 108 m
♦ So we get: L1 = 3 × 108 m
(v) Substituting the values in (iii), we get:
3 × 500 × 108[1 m] = x[3 × 108 m]
⇒ x = 500
• So we can write:
Distance between Sun and Earth = 500 'new units'
Solved example 2.41
The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m2. What is the linear magnification of the projector-screen arrangement.
Solution:
1. Consider the two rectangles in fig.2.26 below:
 |
| Fig.2.26 |
(Details about proportionality can be seen here)
2. Length l2 of the larger rectangle is obtained by multiplying l1 by a factor 'k'
♦ Where l1 is the length of the smaller rectangle
• Width b2 of the larger rectangle is obtained by multiplying b1 by the same factor 'k'
♦ Where b1 is the width of the smaller rectangle
3. Note that, the same factor 'k' must be used for both length and width. otherwise, there will be distortion
• Thus we get: (l1 × b1) = 1.75 cm2
(l2 × b2) = 1.55 m2 = 1.55 × 104 cm2
⇒ (kl1 × kb1) = 1.55 × 104 cm2
⇒ k2 × (l1 × b1) = 1.55 × 104 cm2.
4. But (l1 × b1) = 1.75 cm2
• So we get: k2(1.75) = 1.55 × 104 cm2
⇒ k2 = (1.55⁄1.75) × 104 = 0.885714286 × 104 = 8857.14
⇒ k = √(8857.14) = 94.11
• So the linear magnification (k) = 94.11
In the next section, we will see how to deduce relation among physical quantities
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