Conversion of one system of units to another using dimensions of physical quantities is an important application of dimensional analysis. This method helps in converting units accurately by using the dimensional formula of a quantity and applying proper conversion factors. Solving numerical problems step by step not only improves understanding but also enhances accuracy and speed, which is essential for competitive exams.
Conversion Of a Physical Quantity From One System of Units To Another by Dimensional Analysis
The method of dimensional analysis can be used to obtain the value of a physical quantity in some other system, when its value in one system is given. The measure of a physical quantity is given by
$$ X = n u $$
where u is the size of the unit and n is the numerical value of the physical quantity for the unit chosen. If u₁ and u₂ are units for measurement of the physical quantity in two systems and n₁ and n₂ are the numerical values of the physical quantity for the two units, then
$$ n_1 u_1 = n_2 u_2 $$
If a, b and c are the dimensions of the physical quantity in mass, length and time, then
$$ n_1 [M_1^a L_1^b T_1^c] = n_2 [M_2^a L_2^b T_2^c] $$
Here, M₁, L₁, T₁ and M₂, L₂ and T₂ are the units of mass, length and time in the two systems. Therefore,
$$ n_2 = n_1 \left( \frac{M_1}{M_2} \right)^a \left( \frac{L_1}{L_2} \right)^b \left( \frac{T_1}{T_2} \right)^c $$
The above equation can be used to find the value of the physical quantity in the second or the new system, when its value in the first (or given) system is known.
Numerical Problems with Step-by-Step Solutions
Problem. Prove that $1$ N $= 10^5$ dyne.
Solution. Newton (N) is unit of force in SI and dyne in cgs system.
In given system (SI):
$M_1 = 1$ kg; $L_1 = 1$ m; $T_1 = 1$ s and $n_1 = 1$
In new system (cgs):
$M_2 = 1$ g; $L_2 = 1$ cm; $T_2 = 1$ s and $n_2 = ?$
Now, $n_2 = n_1 \left( \dfrac{M_1}{M_2} \right)^a \left( \dfrac{L_1}{L_2} \right)^b \left( \dfrac{T_1}{T_2} \right)^c$
The dimensional formula of force is $[M L T^{-2}]$.
Therefore, $a = 1$, $b = 1$ and $c = -2$.
Hence, number of dyne in $1$ N is given by
$$n_2 = 1 \times \left( \frac{1 \text{ kg}}{1 \text{ g}} \right)^1 \left( \frac{1 \text{ m}}{1 \text{ cm}} \right)^1 \left( \frac{1 \text{ s}}{1 \text{ s}} \right)^{-2}$$
$$n_2 = \left( \frac{1000 \text{ g}}{1 \text{ g}} \right) \left( \frac{100 \text{ cm}}{1 \text{ cm}} \right) = 10^5$$
$$\therefore \quad 1 \text{ N} = 10^5 \text{ dyne}$$
Problem. Prove that $1$ J $= 10^7$ erg.
Solution. Joule (J) is unit of work in SI and erg in cgs system.
In given system (SI):
$M_1 = 1$ kg; $L_1 = 1$ m; $T_1 = 1$ s and $n_1 = 1$
In new system (cgs):
$M_2 = 1$ g; $L_2 = 1$ cm; $T_2 = 1$ s and $n_2 = ?$
Now, $n_2 = n_1 \left( \dfrac{M_1}{M_2} \right)^a \left( \dfrac{L_1}{L_2} \right)^b \left( \dfrac{T_1}{T_2} \right)^c$
The dimensional formula of work is $[M L^2 T^{-2}]$.
Therefore, $a = 1$, $b = 2$ and $c = -2$.
Hence, number of erg in $1$ J is given by
$$n_2 = 1 \times \left( \frac{1 \text{ kg}}{1 \text{ g}} \right)^1 \left( \frac{1 \text{ m}}{1 \text{ cm}} \right)^2 \left( \frac{1 \text{ s}}{1 \text{ s}} \right)^{-2}$$
$$n_2 = \left( \frac{1000 \text{ g}}{1 \text{ g}} \right) \left( \frac{100 \text{ cm}}{1 \text{ cm}} \right)^2 = 10^7$$
$$\therefore \quad 1 \text{ J} = 10^7 \text{ erg}$$
Problem. Convert the value of the Universal Gravitational Constant ($G$) from the SI system ($6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$) into the CGS system ($\text{cm}^3 \text{ g}^{-1} \text{ s}^{-2}$).
Solution : From Newton’s Law of Gravitation, $F = \dfrac{G m_1 m_2}{r^2}$. Solving for $G$:
$$G = \frac{F r^2}{m_1 m_2}$$
Substituting the dimensions of Force ($MLT^{-2}$), distance ($L$), and mass ($M$):
$$[G] = \frac{[MLT^{-2}][L^2]}{[M][M]} = [M^{-1} L^3 T^{-2}]$$
Here, the exponents are $a = -1, b = 3, c = -2$.
We compare the SI system (System 1) with the CGS system (System 2):
| Physical Quantity | SI System (1) | CGS System (2) |
| Numerical Value | $n_1 = 6.67 \times 10^{-11}$ | $n_2 = ?$ |
| Mass ($M$) | $M_1 = 1 \text{ kg} = 1000 \text{ g}$ | $M_2 = 1 \text{ g}$ |
| Length ($L$) | $L_1 = 1 \text{ m} = 100 \text{ cm}$ | $L_2 = 1 \text{ cm}$ |
| Time ($T$) | $T_1 = 1 \text{ s}$ | $T_2 = 1 \text{ s}$ |
Using the principle that the magnitude remains constant across units ($n_1u_1 = n_2u_2$):
$$n_2 = n_1 \left[ \frac{M_1}{M_2} \right]^a \left[ \frac{L_1}{L_2} \right]^b \left[ \frac{T_1}{T_2} \right]^c$$
Substitute the values and the exponents:
$$n_2 = 6.67 \times 10^{-11} \left[ \frac{1000 \text{ g}}{1 \text{ g}} \right]^{-1} \left[ \frac{100 \text{ cm}}{1 \text{ cm}} \right]^3 \left[ \frac{1 \text{ s}}{1 \text{ s}} \right]^{-2}$$
$$n_2 = 6.67 \times 10^{-11} \times (10^3)^{-1} \times (10^2)^3 \times (1)^{-2}$$
$$n_2 = 6.67 \times 10^{-11} \times 10^{-3} \times 10^6 \times 1$$
$$n_2 = 6.67 \times 10^{-11 – 3 + 6}$$
$$n_2 = 6.67 \times 10^{-8}$$
The value of the Gravitational Constant in the CGS system is:
$$G = 6.67 \times 10^{-8} \text{ cm}^3 \text{ g}^{-1} \text{ s}^{-2}$$
or
$$G = 6.67 \times 10^{-8} \text{ dyne cm}^2/\text{g}^2$$
Problem. The density of a material in cgs system is $8 \mathrm{g cm^{-3}}$. In a system of units, in which unit of length is $5 \mathrm{cm}$ and unit of mass is $20 \mathrm{g}$, what is the density of the material ?
Solution. Let cgs system and the new system be called as the systems 1 and 2 respectively.
If $n_{2}$ is the numerical value of the density of the material in the new system, then
$$n_{2} = n_{1} \left[\frac{M_{1}}{M_{2}}\right]^{a} \left[\frac{L_{1}}{L_{2}}\right]^{b} \left[\frac{T_{1}}{T_{2}}\right]^{c}$$
Now, dimensional formula of density $= [M L^{-3} T^{0}]$
Therefore, $a = 1; b = -3$ and $c = 0$
Hence, $n_{2} = n_{1} \left[\dfrac{M_{1}}{M_{2}}\right]^{1} \left[\dfrac{L_{1}}{L_{2}}\right]^{-3} \left[\dfrac{T_{1}}{T_{2}}\right]^{0}$
$$n_{2} = n_{1} \left[\frac{M_{1}}{M_{2}}\right] \left[\frac{L_{1}}{L_{2}}\right]^{-3}$$
Here, $M_{1} = 1 \mathrm{g}; L_{1} = 1 \mathrm{cm}; n_{1} = 8$ and $M_{2} = 20 \mathrm{g}; L_{2} = 5 \mathrm{cm}$
$$n_{2} = 8 \left[\frac{1 \mathrm{g}}{20 \mathrm{g}}\right] \times \left[\frac{1 \mathrm{cm}}{5 \mathrm{cm}}\right]^{-3} = \frac{8 \times (5)^{3}}{20} = 50$$
Hence, density of the material in new system $= 50$ units
Problem. A new system of units is proposed in which the unit of mass is $\alpha \text{ kg}$, the unit of length is $\beta \text{ m}$, and the unit of time is $\gamma \text{ s}$. How much will $5\text{ J}$ measure in this new system?
Solution : The physical quantity given is energy (Joule). The dimensional formula for energy is: $[\text{Energy}] = [ML^2T^{-2}]$
So, the exponents are: $a = 1, b = 2, c = -2$.
We compare the given system (SI) with the new proposed system:
| Quantity | SI System (n1) | New System (n2) |
| Numerical Value | $n_1 = 5$ | $n_2 = ?$ |
| Mass | $M_1 = 1 \text{ kg}$ | $M_2 = \alpha \text{ kg}$ |
| Length | $L_1 = 1 \text{ m}$ | $L_2 = \beta \text{ m}$ |
| Time | $T_1 = 1 \text{ s}$ | $T_2 = \gamma \text{ s}$ |
Using the principle that the magnitude of a physical quantity remains constant ($n_1u_1 = n_2u_2$):
$$n_2 = n_1 \left[ \frac{M_1}{M_2} \right]^a \left[ \frac{L_1}{L_2} \right]^b \left[ \frac{T_1}{T_2} \right]^c$$
Substitute the dimensions and the values:
$$n_2 = 5 \left[ \frac{1 \text{ kg}}{\alpha \text{ kg}} \right]^1 \left[ \frac{1 \text{ m}}{\beta \text{ m}} \right]^2 \left[ \frac{1 \text{ s}}{\gamma \text{ s}} \right]^{-2}$$
$$n_2 = 5 \times (\alpha^{-1}) \times (\beta^{-2}) \times (\gamma^2)$$
$$n_2 = 5 \times \frac{1}{\alpha} \times \frac{1}{\beta^2} \times \gamma^2$$
$$n_2 = \frac{5\gamma^2}{\alpha\beta^2}$$
The value of $5\text{ J}$ in the new system of units is:
$$n_2 = \frac{5\gamma^2}{\alpha\beta^2}$$
Problem. A calorie is a unit of heat or energy and it equals about $4.2\text{ J}$, where $1\text{ J} = 1\text{ kg m}^2\text{ s}^{-2}$. Suppose we employ a system of units in which the unit of mass equals $\alpha\text{ kg}$, the unit of length equals $\beta\text{ m}$, and the unit of time is $\gamma\text{ s}$. Show that a calorie has a magnitude of $4.2 \alpha^{-1}\beta^{-2}\gamma^{2}$ in terms of these new units.
Solution : The physical quantity is energy (heat). Its dimensional formula is:
$$[\text{Energy}] = [M^1 L^2 T^{-2}]$$
Here, the exponents are $a=1, b=2, c=-2$.
We compare the standard SI system with the new proposed system:
| Quantity | SI System (n1) | New System (n2) |
| Numerical Value | $n_1 = 4.2$ | $n_2 = ?$ |
| Mass | $M_1 = 1\text{ kg}$ | $M_2 = \alpha\text{ kg}$ |
| Length | $L_1 = 1\text{ m}$ | $L_2 = \beta\text{ m}$ |
| Time | $T_1 = 1\text{ s}$ | $T_2 = \gamma\text{ s}$ |
The magnitude of a physical quantity is constant across different units ($n_1u_1 = n_2u_2$). To find the new numerical value, we use:
$$n_2 = n_1 \left[ \frac{M_1}{M_2} \right]^a \left[ \frac{L_1}{L_2} \right]^b \left[ \frac{T_1}{T_2} \right]^c$$
Substituting the values into the formula:
$$n_2 = 4.2 \left[ \frac{1\text{ kg}}{\alpha\text{ kg}} \right]^1 \left[ \frac{1\text{ m}}{\beta\text{ m}} \right]^2 \left[ \frac{1\text{ s}}{\gamma\text{ s}} \right]^{-2}$$
$$n_2 = 4.2 \times \left( \frac{1}{\alpha} \right)^1 \times \left( \frac{1}{\beta} \right)^2 \times \left( \frac{1}{\gamma} \right)^{-2}$$
$$n_2 = 4.2 \times \alpha^{-1} \times \beta^{-2} \times \gamma^{2}$$
The magnitude of a calorie in the new system is:
$$n_2 = 4.2 \alpha^{-1}\beta^{-2}\gamma^{2}$$
Problem. (a) Convert a velocity of $36 \mathrm{kmh^{-1}}$ into $\mathrm{m s^{-1}}$
(b) Convert an acceleration of $45 \mathrm{km min^{-2}}$ into $\mathrm{m s^{-2}}$
Solution. (a) Here, $v = 36 \mathrm{kmh^{-1}}$
Now, $1 \mathrm{km} = 1000 \mathrm{m}$; $1 \mathrm{h} = 60 \times 60 \mathrm{s}$
$$v = 36 \mathrm{kmh^{-1}} = 36 \times (1000 \mathrm{m}) \times (60 \times 60 \mathrm{s})^{-1} $$
$$ v = \frac{36 \times 1000}{60 \times 60} \mathrm{m s^{-1}} = 10 \mathrm{m s^{-1}}$$
(b) Here, $a = 45 \mathrm{km min^{-2}}$
Now, $1 \mathrm{km} = 1000 \mathrm{m}$; $1 \mathrm{min} = 60 \mathrm{s}$
$$a = 45 \mathrm{km min^{-2}} = 45 \times (1000 \mathrm{m}) \times (60 \mathrm{s})^{-2} $$
$$a = \frac{45 \times 1000}{60 \times 60} \mathrm{m s^{-2}} = 12.5 \mathrm{m s^{-2}}$$
Problem. A body has a uniform acceleration of $5 \mathrm{kmh^{-2}}$. Find its value in cgs system.
Solution. Here, $a = 5 \mathrm{kmh^{-2}}$
Now, $1 \mathrm{km} = 10^{5} \mathrm{cm}$; $1 \mathrm{h} = 3600 \mathrm{s}$
$$a = 5 \times (10^{5} \mathrm{cm}) \times (3600 \mathrm{s})^{-2} = \frac{5 \times 10^{5}}{3600 \times 3600} = 0.0386 \mathrm{cm s^{-2}}$$
Problem. Convert a pressure of one atmospheric (=10⁵ N m⁻²) into dyne cm⁻².
Solution. Here, 1 atmospheric pressure = 10⁵ N m⁻²
Now, 1 N = 10⁵ dyne and 1 m = 100 cm
∴ 1 atmospheric pressure $= 10^5 \times (10^5 \text{ dyne}) \times (100 \text{ cm})^{-2} $
1 atmospheric pressure = $ 10^6 \text{ dyne cm}^{-2} $
Problem. In cgs system, the value of Stefan’s constant,
$$\sigma = 5.67 \times 10^{-5} \mathrm{erg \;s^{-1} cm^{-2} \;K^{-4}}$$
Find its value in SI. Given, $1 \mathrm{J} = 10^{7} \mathrm{erg}$
Solution. Here, $\sigma = 5.67 \times 10^{-5} \mathrm{erg \;s^{-1} cm^{-2} \;K^{-4}}$
In SI, unit of work is joule.
Now, $1 \mathrm{erg} = 10^{-7} \mathrm{J}$; $1 \mathrm{cm} = 10^{-2} \mathrm{m}$
Therefore, in SI,
$$\sigma = 5.67 \times 10^{-5} [10^{-7} \mathrm{J] \; s^{-1}} [10^{-2} \mathrm{m}]^{-2} \mathrm{\;K^{-4}}$$
$$\sigma = 5.67 \times 10^{-5} \times 10^{-7} \times 10^{4} \mathrm{J \;s^{-1} \;m^{-2} \;K^{-4}}$$
$$\sigma = 5.67 \times 10^{-8} \mathrm{J s^{-1} m^{-2} K^{-4}}$$
Important Unit Conversions
- Pressure Units
- 1 bar = 10⁶ dyne/cm² = 10⁵ Nm⁻² = 10⁵ pascal
- 76 cm of Hg = 1.013×10⁶ dyne/cm² = 1.013×10⁵ pascal = 1.013 bar
- 1 torricelli (torr) = 1 mm of Hg = 1.333×10³ dyne/cm² = 1.333 millibar
- Velocity Conversion
- 1 kmph = (5/18) ms⁻¹
- Force and Power Conversions
- 1 dyne = 10⁻⁵ N
- 1 H.P = 746 watt
- Energy and Work Units
- 1 kilowatt-hour = 3.6×10⁶ J
- 1 kgwt = g newton
- 1 calorie = 4.2 joule
- 1 electron volt (eV) = 1.602×10⁻¹⁹ joule
- 1 erg = 10⁻⁷ joule
FAQs Very Short Question and Answers based on topic Convert One System of Units To Another by Dimensions of Physical Quantity
What is meant by conversion of units using dimensional analysis?
It is a method of converting a physical quantity from one system of units to another using its dimensional formula and appropriate conversion factors.
Why is dimensional analysis used for unit conversion?
Dimensional analysis ensures accurate conversion by maintaining consistency in fundamental units like mass, length, and time.
What is the basic formula for unit conversion using dimensions?
If a quantity has dimensions $M^a L^b T^c$, then
$$
n_1 u_1 = n_2 u_2
$$
where conversion is done using ratios of fundamental units.
Which physical quantities are used in dimensional analysis?
Fundamental quantities such as mass (M), length (L), time (T), current (A), temperature (K), and others.
Can dimensional analysis be used for all unit conversions?
It is mainly useful for physical quantities that can be expressed in terms of fundamental dimensions. It may not be applicable for dimensionless quantities.
What are the common mistakes in unit conversion?
Using incorrect dimensional formula
Ignoring unit consistency
Wrong power of conversion factors
How does dimensional analysis help in numerical problems?
It simplifies calculations, reduces errors, and provides a systematic approach to solving unit-based problems.
Is dimensional analysis important for competitive exams?
Yes, it is very important for exams like JEE and NEET, as questions on unit conversion and dimensional analysis are frequently asked.
Q9. What is the role of significant figures in unit conversion?
Significant figures ensure that the final answer maintains the correct level of precision after conversion.
How can students improve in this topic?
Regular practice of numerical problems, understanding dimensional formulas, and solving previous year questions can improve accuracy and speed.
Why is pressure measured in multiple units like bar, torr, and Pascal?
Different measurement systems and scientific applications require different units. The SI unit is Pascal, while bar and torr are often used in meteorology and laboratory experiments.
What is the significance of 1 kilowatt-hour in electricity billing?
1 kilowatt-hour (kWh) represents the amount of energy consumed when a device of 1 kW runs for 1 hour. It is the standard unit used in electricity billing.
How does the conversion of kmph to m/s help in solving physics problems?
Many physics formulas use SI units, where velocity is in m/s. Converting kmph to m/s ensures consistency in calculations.
Important Chapter Links
To strengthen your understanding, you should also study Dimensional Analysis and Dimensional Formulae of Physical Quantities and the Principle of Dimensional Homogeneity, which are closely related to unit conversion. It is also helpful to revise Units and Measurements for basic concepts and practice JEE Main Previous Year Questions (PYQs) and IMU CET PYQs to improve problem-solving skills. Exploring these related topics on this website will help you master numerical applications effectively.