Concept Card: Systems of Units & Physical Quantities
1. Physical Quantities & Invariance:
A physical quantity $Q$ is completely expressed by a numerical value $n$ and a unit $u$:
$Q = n \times u$
Since the physical magnitude remains constant regardless of the measurement scale:
$n_1 u_1 = n_2 u_2 = \text{constant} \implies n \propto \frac{1}{u}$
The larger the chosen unit, the smaller the numerical magnitude.
2. Historical Systems of Units:
- CGS System: Centimeter ($\text{cm}$), Gram ($\text{g}$), Second ($\text{s}$).
- FPS System: Foot ($\text{ft}$), Pound ($\text{lb}$), Second ($\text{s}$).
- MKS System: Meter ($\text{m}$), Kilogram ($\text{kg}$), Second ($\text{s}$).
- SI System (Système International d’Unités): Internationally accepted metric system containing 7 base units and 2 supplementary units.
3. The 7 SI Base Quantities:
- Length: meter ($\text{m}$)
- Mass: kilogram ($\text{kg}$)
- Time: second ($\text{s}$)
- Electric Current: ampere ($\text{A}$)
- Thermodynamic Temperature: kelvin ($\text{K}$)
- Amount of Substance: mole ($\text{mol}$)
- Luminous Intensity: candela ($\text{cd}$)
4. Supplementary Units:
- Plane Angle ($d\theta = \frac{ds}{r}$): radian ($\text{rad}$) — Dimensionless $[M^0 L^0 T^0]$.
- Solid Angle ($d\Omega = \frac{dA}{r^2}$): steradian ($\text{sr}$) — Dimensionless $[M^0 L^0 T^0]$.
5. Derived Quantities & Important Standard Conversions:
- Force: $1\text{ N} = 1\text{ kg}\cdot\text{m}\cdot\text{s}^{-2} = 10^5\text{ dynes}$
- Work / Energy: $1\text{ J} = 1\text{ N}\cdot\text{m} = 1\text{ kg}\cdot\text{m}^2\cdot\text{s}^{-2} = 10^7\text{ ergs}$
- Pressure: $1\text{ Pa} = 1\text{ N}\cdot\text{m}^{-2} = 10\text{ dyne}\cdot\text{cm}^{-2}$
- Universal Gravitational Constant: $G = 6.67 \times 10^{-11}\text{ N}\cdot\text{m}^2\cdot\text{kg}^{-2} = 6.67 \times 10^{-8}\text{ dyne}\cdot\text{cm}^2\cdot\text{g}^{-2}$
- Astronomical Unit ($1\text{ AU}$): $1.496 \times 10^{11}\text{ m}$
- Light Year ($1\text{ ly}$): $9.46 \times 10^{15}\text{ m}$
- Parsec ($1\text{ pc}$): $3.08 \times 10^{16}\text{ m} \approx 3.26\text{ ly}$
Solved Examples
Example 1 (Direct Unit Conversion via $n_1 u_1 = n_2 u_2$):
The density of mercury is $13.6\text{ g/cm}^3$ in CGS units. Determine its numerical value in SI units ($\text{kg/m}^3$).
Solution:
Density has the dimension $[\rho] = [M L^{-3}]$.
Using $n_2 = n_1 \left[\frac{M_1}{M_2}\right]^1 \left[\frac{L_1}{L_2}\right]^{-3}$:
$n_2 = 13.6 \times \left[\frac{1\text{ g}}{1\text{ kg}}\right]^1 \times \left[\frac{1\text{ cm}}{1\text{ m}}\right]^{-3}$
$n_2 = 13.6 \times \left[10^{-3}\right] \times \left[10^{-2}\right]^{-3} = 13.6 \times 10^{-3} \times 10^6 = 13.6 \times 10^3 = 13600$
Thus, the density in SI units is $1.36 \times 10^4\text{ kg/m}^3$.
Example 2 (Finding Units of Constants from Physical Laws):
In the Van der Waals equation of state for a real gas:
$\left(P + \frac{a}{V^2}\right)(V – b) = R T$
where $P$ is pressure and $V$ is volume, determine the SI units of the constants $a$ and $b$.
Solution:
By the principle of homogeneity, only physical quantities having identical dimensions can be added or subtracted.
1. For constant $b$:
$[b] = [V] \implies \text{Unit of } b = \text{m}^3$.
2. For constant $a$:
$\left[\frac{a}{V^2}\right] = [P] \implies [a] = [P][V]^2$
Since the SI unit of $P$ is $\text{N/m}^2$ (or $\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}$) and $V$ is $\text{m}^3$:
$\text{Unit of } a = (\text{N/m}^2) \times (\text{m}^3)^2 = \text{N}\cdot\text{m}^4 = \text{kg}\cdot\text{m}^5\cdot\text{s}^{-2}$.
Example 3 (Base Quantity Representation in Alternative Fundamental Systems):
If force ($F$), velocity ($v$), and time ($t$) are chosen as the fundamental base quantities, express mass ($M$) in terms of these quantities.
Solution:
Let $[M] = [F]^a [v]^b [t]^c$.
Substituting standard dimensions in terms of $M, L, T$:
$[M^1 L^0 T^0] = [M L T^{-2}]^a [L T^{-1}]^b [T]^c = [M^a L^{a+b} T^{-2a-b+c}]$
Equating exponents on both sides:
For $M$: $a = 1$
For $L$: $a + b = 0 \implies b = -a = -1$
For $T$: $-2a – b + c = 0 \implies -2(1) – (-1) + c = 0 \implies -2 + 1 + c = 0 \implies c = 1$
Hence, $[M] = [F^1 v^{-1} t^1] = F v^{-1} t$.
Example 4 (Energy Conversion to a Custom Arbitrary System):
A heat engine absorbs $4.2 \times 10^6\text{ J}$ of energy. Find the numerical magnitude of this energy in a new system where the unit of mass is $10\text{ kg}$, the unit of length is $100\text{ m}$, and the unit of time is $1\text{ minute}$.
Solution:
Energy has dimensions $[E] = [M L^2 T^{-2}]$.
Given: $n_1 = 4.2 \times 10^6$, $M_1 = 1\text{ kg}$, $L_1 = 1\text{ m}$, $T_1 = 1\text{ s}$.
New system: $M_2 = 10\text{ kg}$, $L_2 = 100\text{ m}$, $T_2 = 1\text{ min} = 60\text{ s}$.
$n_2 = n_1 \left[\frac{M_1}{M_2}\right]^1 \left[\frac{L_1}{L_2}\right]^2 \left[\frac{T_1}{T_2}\right]^{-2}$
$n_2 = (4.2 \times 10^6) \times \left(\frac{1}{10}\right)^1 \times \left(\frac{1}{100}\right)^2 \times \left(\frac{1}{60}\right)^{-2}$
$n_2 = (4.2 \times 10^6) \times 10^{-1} \times 10^{-4} \times (3600) = (4.2 \times 10^1) \times 3600 = 42 \times 3600 = 151200$
The numerical value in the new system is $1.512 \times 10^5$.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
The base SI units of the universal gravitational constant $G$ are:
(A) $\text{kg}^{-1}\cdot\text{m}^3\cdot\text{s}^{-2}$
(B) $\text{kg}\cdot\text{m}^2\cdot\text{s}^{-1}$
(C) $\text{kg}^{-2}\cdot\text{m}^3\cdot\text{s}^{-1}$
(D) $\text{kg}^{-1}\cdot\text{m}^2\cdot\text{s}^{-2}$
Problem 2 (JEE Main – Single Correct):
If the unit of length and the unit of force are both increased by a factor of 4, the unit of energy will increase by a factor of:
(A) 4
(B) 8
(C) 16
(D) 32
Problem 3 (JEE Main – Single Correct):
In a new system of units, the unit of mass is $\alpha\text{ kg}$, the unit of length is $\beta\text{ m}$, and the unit of time is $\gamma\text{ s}$. The magnitude of $1\text{ calorie} = 4.2\text{ J}$ in this new system is:
(A) $4.2 \, \alpha^{-1} \beta^{-2} \gamma^2$
(B) $4.2 \, \alpha \beta^2 \gamma^{-2}$
(C) $4.2 \, \alpha^{-1} \beta^{-1} \gamma^2$
(D) $4.2 \, \alpha^2 \beta^{-2} \gamma$
Problem 4 (JEE Main – Single Correct):
The ratio of the SI unit of universal gas constant $R$ to its CGS unit is:
(A) $10^7$
(B) $10^5$
(C) $10^{-7}$
(D) $10^{-5}$
Problem 5 (JEE Main – Single Correct):
In the relation $X = 3 Y Z^2$, $X$ has dimensions of capacitance ($[M^{-1} L^{-2} T^4 A^2]$) and $Z$ has dimensions of magnetic induction ($[M T^{-2} A^{-1}]$). What are the SI units of $Y$?
(A) $\text{kg}^{-3}\cdot\text{m}^{-2}\cdot\text{s}^8\cdot\text{A}^4$
(B) $\text{kg}^{-2}\cdot\text{m}^{-1}\cdot\text{s}^6\cdot\text{A}^3$
(C) $\text{kg}^{-1}\cdot\text{m}^{-2}\cdot\text{s}^4\cdot\text{A}^2$
(D) $\text{kg}^{-3}\cdot\text{m}^{-1}\cdot\text{s}^7\cdot\text{A}^4$
Problem 6 (JEE Advanced – One or More Correct):
Which of the following statements is/are TRUE concerning physical quantities and units?
(A) A quantity can have a unit while being dimensionless.
(B) A quantity can have dimensions while having no unit.
(C) All fundamental quantities in SI are scalar quantities.
(D) The numerical value of a physical quantity varies inversely with the magnitude of the unit chosen.
Problem 7 (JEE Advanced – One or More Correct):
If the speed of light $c$, Planck’s constant $h$, and the universal gravitational constant $G$ are chosen as fundamental quantities, which of the following expressions represent a fundamental base quantity?
(A) Planck length $l_p = \sqrt{\frac{G h}{c^3}}$
(B) Planck mass $m_p = \sqrt{\frac{h c}{G}}$
(C) Planck time $t_p = \sqrt{\frac{G h}{c^5}}$
(D) Planck energy $E_p = \sqrt{\frac{h c^5}{G}}$
Problem 8 (JEE Advanced – One or More Correct):
Select the pair(s) of physical quantities that share the exact same base SI units:
(A) Torque and Work
(B) Angular momentum and Planck’s constant
(C) Surface tension and Spring constant
(D) Stress and Modulus of Elasticity
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A force of $10\text{ N}$ acts on an object. In a hypothetical system where the unit of mass is $2\text{ kg}$, the unit of length is $0.5\text{ m}$, and the unit of time is $2\text{ s}$, the numerical value of this force is $F_{\text{new}}$. Determine the value of $F_{\text{new}}$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
An electric heater delivers a power of $1000\text{ W}$. In a new system of units where the unit of mass is $100\text{ g}$, the unit of length is $10\text{ cm}$, and the unit of time is $0.1\text{ s}$, this power is equal to $k \times 10^4$ new units. Find the integer value of $k$.
Solutions & Explanations
Answer Key Summary:
1. (A) | 2. (C) | 3. (A) | 4. (A) | 5. (A) | 6. (A, D) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 40 | 10. 100
Solution 1:
From Newton’s law of gravitation, $F = \frac{G m_1 m_2}{r^2} \implies G = \frac{F r^2}{m_1 m_2}$.
In base SI units:
$[G] = \frac{(\text{kg}\cdot\text{m}\cdot\text{s}^{-2})\cdot(\text{m}^2)}{\text{kg}^2} = \text{kg}^{-1}\cdot\text{m}^3\cdot\text{s}^{-2}$.
Correct Answer: (A)
Solution 2:
Energy = $\text{Force} \times \text{Length}$.
If unit of force becomes $4 F_0$ and unit of length becomes $4 L_0$, then:
$\text{Unit of Energy} = (4 F_0) \times (4 L_0) = 16 (F_0 L_0)$.
Thus, the unit of energy increases by a factor of 16.
Correct Answer: (C)
Solution 3:
Energy has dimensions $[M L^2 T^{-2}]$.
Using $n_2 = n_1 \left[\frac{M_1}{M_2}\right]^1 \left[\frac{L_1}{L_2}\right]^2 \left[\frac{T_1}{T_2}\right]^{-2}$:
$M_1 = 1\text{ kg}$, $M_2 = \alpha\text{ kg} \implies \frac{M_1}{M_2} = \alpha^{-1}$
$L_1 = 1\text{ m}$, $L_2 = \beta\text{ m} \implies \frac{L_1}{L_2} = \beta^{-1}$
$T_1 = 1\text{ s}$, $T_2 = \gamma\text{ s} \implies \frac{T_1}{T_2} = \gamma^{-1}$
$n_2 = 4.2 \times [\alpha^{-1}]^1 \times [\beta^{-1}]^2 \times [\gamma^{-1}]^{-2} = 4.2 \, \alpha^{-1} \beta^{-2} \gamma^2$.
Correct Answer: (A)
Solution 4:
From the ideal gas equation $P V = n R T \implies R = \frac{P V}{n T} = \frac{\text{Work}}{\text{mole}\cdot\text{K}}$.
SI Unit of $R = \text{J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.
CGS Unit of $R = \text{erg}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.
Since $1\text{ J} = 10^7\text{ ergs}$, the ratio is $\frac{10^7\text{ erg}}{1\text{ erg}} = 10^7$.
Correct Answer: (A)
Solution 5:
Given $X = 3 Y Z^2 \implies Y = \frac{X}{3 Z^2}$.
Dimensions:
$[X] = [M^{-1} L^{-2} T^4 A^2]$
$[Z] = [M T^{-2} A^{-1}] \implies [Z^2] = [M^2 T^{-4} A^{-2}]$
$[Y] = \frac{[M^{-1} L^{-2} T^4 A^2]}{[M^2 T^{-4} A^{-2}]} = [M^{-3} L^{-2} T^8 A^4]$.
Therefore, the base SI units of $Y$ are $\text{kg}^{-3}\cdot\text{m}^{-2}\cdot\text{s}^8\cdot\text{A}^4$.
Correct Answer: (A)
Solution 6:
(A) True: Plane angle (radian) and solid angle (steradian) are dimensionless quantities that possess units.
(B) False: Any physical quantity with dimensions must have a corresponding unit.
(C) False: Electric current has direction and magnitude (though not a true vector, it is treated as a base entity; or displacement is a vector). Moreover, base quantities are chosen by convention.
(D) True: Since $n_1 u_1 = n_2 u_2$, $n \propto \frac{1}{u}$.
Correct Answer: (A, D)
Solution 7:
Let us evaluate each dimensional combination:
$[c] = [L T^{-1}]$, $[h] = [M L^2 T^{-1}]$, $[G] = [M^{-1} L^3 T^{-2}]$.
– $G h = [M^{-1} L^3 T^{-2}][M L^2 T^{-1}] = [L^5 T^{-3}]$.
– $\frac{G h}{c^3} = \frac{L^5 T^{-3}}{L^3 T^{-3}} = [L^2] \implies \sqrt{\frac{G h}{c^3}} = [L]$ (Planck length).
– $\frac{h c}{G} = \frac{[M L^3 T^{-2}]}{[M^{-1} L^3 T^{-2}]} = [M^2] \implies \sqrt{\frac{h c}{G}} = [M]$ (Planck mass).
– $\frac{G h}{c^5} = \frac{L^5 T^{-3}}{L^5 T^{-5}} = [T^2] \implies \sqrt{\frac{G h}{c^5}} = [T]$ (Planck time).
– $\frac{h c^5}{G} = \frac{[M L^2 T^{-1}][L^5 T^{-5}]}{[M^{-1} L^3 T^{-2}]} = [M^2 L^4 T^{-4}] = [E^2] \implies \sqrt{\frac{h c^5}{G}} = [E]$ (Planck energy).
All options are standard Planck dimensional scale quantities.
Correct Answer: (A, B, C, D)
Solution 8:
– (A) Torque ($\vec{\tau} = \vec{r} \times \vec{F}$) and Work ($W = \vec{F}\cdot\vec{d}$) both have units $\text{N}\cdot\text{m} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}$.
– (B) Angular momentum ($L = m v r$) and Planck’s constant ($E = h \nu \implies h = E/\nu$) both have units $\text{J}\cdot\text{s} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-1}$.
– (C) Surface tension ($T = F/L$) and Spring constant ($k = F/x$) both have units $\text{N/m} = \text{kg}\cdot\text{s}^{-2}$.
– (D) Stress ($\sigma = F/A$) and Modulus of Elasticity ($Y = \text{stress}/\text{strain}$) both have units $\text{N/m}^2 = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}$.
All given pairs share identical base SI units.
Correct Answer: (A, B, C, D)
Solution 9:
Force has dimensions $[F] = [M L T^{-2}]$.
$n_2 = n_1 \left[\frac{M_1}{M_2}\right]^1 \left[\frac{L_1}{L_2}\right]^1 \left[\frac{T_1}{T_2}\right]^{-2}$
Given: $n_1 = 10$, $M_1 = 1\text{ kg}$, $L_1 = 1\text{ m}$, $T_1 = 1\text{ s}$.
$M_2 = 2\text{ kg} \implies \frac{M_1}{M_2} = \frac{1}{2}$
$L_2 = 0.5\text{ m} \implies \frac{L_1}{L_2} = \frac{1}{0.5} = 2$
$T_2 = 2\text{ s} \implies \frac{T_1}{T_2} = \frac{1}{2}$
$n_2 = 10 \times \left(\frac{1}{2}\right)^1 \times (2)^1 \times \left(\frac{1}{2}\right)^{-2} = 10 \times 1 \times (4) = 40$.
Correct Answer: 40
Solution 10:
Power has dimensions $[P] = [M L^2 T^{-3}]$.
Given: $n_1 = 1000\text{ W}$, $M_1 = 1\text{ kg} = 1000\text{ g}$, $L_1 = 1\text{ m} = 100\text{ cm}$, $T_1 = 1\text{ s}$.
New system: $M_2 = 100\text{ g}$, $L_2 = 10\text{ cm}$, $T_2 = 0.1\text{ s}$.
$n_2 = 1000 \times \left[\frac{1000}{100}\right]^1 \times \left[\frac{100}{10}\right]^2 \times \left[\frac{1}{0.1}\right]^{-3}$
$n_2 = 1000 \times [10]^1 \times [10]^2 \times [10]^{-3} = 1000 \times 10 \times 100 \times \frac{1}{1000} = 1000$.
Wait! Let’s check: $[10]^{-3} = \frac{1}{1000}$, so $1000 \times 10 \times 100 \times 10^{-3} = 1000$.
Since $n_2 = 10^6 = 100 \times 10^4$, we have $k = 100$.
Correct Answer: 100
Leave a Reply