Concept Card: Master Review & Strategy Sheet for Units, Errors & Dimensions
1. Dimensional Analysis Master Relations:
The dimensions of any physical quantity $Q$ are represented as $[Q] = [M^a L^b T^c A^d K^e \text{mol}^f \text{cd}^g]$.
- Fundamental Electromagnetic Dimensionless Ratios:
- Fine-structure constant: $\alpha = \frac{e^2}{4\pi \varepsilon_0 \hbar c} \approx \frac{1}{137} \implies [M^0 L^0 T^0]$
- Speed of light: $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \implies [L T^{-1}]$
- Impedance of free space: $Z_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}} \approx 377\,\Omega \implies [M L^2 T^{-3} A^{-2}]$
- Planck System of Natural Units:
- Planck Length: $l_p = \sqrt{\frac{G \hbar}{c^3}} \implies [L]$
- Planck Mass: $m_p = \sqrt{\frac{\hbar c}{G}} \implies [M]$
- Planck Time: $t_p = \sqrt{\frac{G \hbar}{c^5}} \implies [T]$
2. Error Propagation & Experimental Strategy:
- Power Product Rule: If $Z = \frac{A^p B^q}{C^r}$, the maximum fractional error is:
$\frac{\Delta Z}{Z} = p\left(\frac{\Delta A}{A}\right) + q\left(\frac{\Delta B}{B}\right) + r\left(\frac{\Delta C}{C}\right)$
Strategy: The physical variable with the highest exponent contributes the largest uncertainty. It must be measured with the highest precision instrument. - Linear Combinations: If $Z = a X \pm b Y$, absolute errors always add:
$\Delta Z = a \Delta X + b \Delta Y$ - General Differential Form: For any non-linear function $Z = f(x_1, x_2, \dots)$:
$\Delta Z_{\text{max}} = \sum_{i} \left|\frac{\partial f}{\partial x_i}\right| \Delta x_i$
3. Measuring Instruments Reading Formulation:
- Vernier Calipers:
$\text{LC} = 1\text{ MSD} – 1\text{ VSD} = \frac{1\text{ MSD}}{n}$ (when $n\text{ VSD} = (n-1)\text{ MSD}$)
$\text{True Reading} = \text{MSR} + (\text{VSR} \times \text{LC}) – \text{Zero Error}$ - Screw Gauge:
$\text{Pitch} = \frac{\text{Distance moved on main scale}}{\text{Number of full rotations}}, \quad \text{LC} = \frac{\text{Pitch}}{\text{Total Circular Divisions}}$
$\text{True Reading} = \text{LSR} + (\text{CSR} \times \text{LC}) – \text{Zero Error}$
Solved Examples
Example 1 (Error Analysis in Stokes’ Viscosity Experiment):
A spherical ball of radius $r = (2.00 \pm 0.02)\text{ mm}$ falls with terminal speed $v = (10.0 \pm 0.2)\text{ cm/s}$ through a viscous liquid of density $\sigma = 1.20\text{ g/cm}^3$ (treated as exact). The density of the ball is $\rho = (7.80 \pm 0.06)\text{ g/cm}^3$. Using Stokes’ Law for terminal speed:
$v = \frac{2}{9}\frac{r^2 (\rho – \sigma) g}{\eta}$
find the maximum percentage error in the determination of the coefficient of viscosity $\eta$.
Solution:
1. Express $\eta$ in terms of measured quantities:
$\eta = \frac{2 g}{9} \cdot \frac{r^2 (\rho – \sigma)}{v}$
2. Fractional error formula:
$\frac{\Delta \eta}{\eta} = 2\left(\frac{\Delta r}{r}\right) + \frac{\Delta (\rho – \sigma)}{\rho – \sigma} + \frac{\Delta v}{v}$
Since $\sigma$ has negligible error, $\Delta(\rho – \sigma) = \Delta \rho$.
3. Calculate individual fractional uncertainties:
$\frac{\Delta r}{r} = \frac{0.02\text{ mm}}{2.00\text{ mm}} = 0.01 = 1.0\%$
$\frac{\Delta v}{v} = \frac{0.2\text{ cm/s}}{10.0\text{ cm/s}} = 0.02 = 2.0\%$
$\rho – \sigma = 7.80 – 1.20 = 6.60\text{ g/cm}^3 \implies \frac{\Delta \rho}{\rho – \sigma} = \frac{0.06}{6.60} = \frac{1}{110} \approx 0.91\%$
4. Total percentage error in $\eta$:
$\%\eta = 2(1.0\%) + 0.91\% + 2.0\% = 2.0\% + 0.91\% + 2.0\% = 4.91\% \approx 4.9\%$.
Example 2 (Synthesizing Dimensions in an Unconventional System):
If Force ($F$), Energy ($E$), and Velocity ($V$) are chosen as the fundamental base quantities, determine the dimensional formula of the Universal Gravitational Constant $G$.
Solution:
1. Relate standard base quantities to $F, E, V$:
– Energy $E = F \times d \implies [L] = [E][F^{-1}] = E F^{-1}$.
– Velocity $V = \frac{L}{T} \implies [T] = \frac{[L]}{[V]} = \frac{E F^{-1}}{V} = E F^{-1} V^{-1}$.
– Force $F = M \frac{V}{T} \implies [M] = \frac{F [T]}{[V]} = \frac{F (E F^{-1} V^{-1})}{V} = E V^{-2}$.
2. Express Universal Gravitational Constant $G$:
Dimensions of $G$ in $M, L, T$: $[G] = [M^{-1} L^3 T^{-2}]$.
Substitute the expressions in terms of $E, F, V$:
$[G] = [M]^{-1} [L]^3 [T]^{-2} = (E V^{-2})^{-1} \times (E F^{-1})^3 \times (E F^{-1} V^{-1})^{-2}$
$[G] = (E^{-1} V^2) \times (E^3 F^{-3}) \times (E^{-2} F^2 V^2)$
Group powers of $E, F, V$:
– Power of $E$: $-1 + 3 – 2 = 0$
– Power of $F$: $-3 + 2 = -1$
– Power of $V$: $2 + 2 = 4$
$[G] = [E^0 F^{-1} V^4] = F^{-1} V^4$.
Example 3 (Integrated Instrument Analysis: Calipers and Micrometer):
The thickness of a glass slab is measured using a Vernier Calipers ($1\text{ MSD} = 1\text{ mm}$, $10\text{ VSD} = 9\text{ MSD}$, zero error $= 0$) yielding $\text{MSR} = 15\text{ mm}$ and $\text{VSR} = 6$. The diameter of a small lead shot embedded in it is measured using a Screw Gauge ($\text{pitch} = 0.5\text{ mm}$, $50\text{ CSD}$, zero error $= +0.03\text{ mm}$) giving $\text{LSR} = 2.0\text{ mm}$ and $\text{CSR} = 24$. Find the ratio of the thickness of the slab to the diameter of the lead shot.
Solution:
1. Slab thickness (Vernier Calipers):
$\text{LC}_V = \frac{1\text{ mm}}{10} = 0.1\text{ mm}$
$\text{Thickness } t = \text{MSR} + (\text{VSR} \times \text{LC}_V) = 15\text{ mm} + (6 \times 0.1\text{ mm}) = 15.6\text{ mm}$.
2. Lead shot diameter (Screw Gauge):
$\text{LC}_S = \frac{0.5\text{ mm}}{50} = 0.01\text{ mm}$
$\text{Observed diameter} = \text{LSR} + (\text{CSR} \times \text{LC}_S) = 2.0\text{ mm} + (24 \times 0.01\text{ mm}) = 2.24\text{ mm}$
$\text{True diameter } d = \text{Observed} – (\text{Zero Error}) = 2.24\text{ mm} – (+0.03\text{ mm}) = 2.21\text{ mm}$.
3. Ratio of thickness to diameter:
$\text{Ratio} = \frac{t}{d} = \frac{15.6\text{ mm}}{2.21\text{ mm}} \approx 7.06$.
Example 4 (Calculus Error Propagation in Resonance Tube Experiment):
In a resonance tube experiment to determine the speed of sound $v$ in air, the first resonance occurs at liquid column length $l_1 = (17.5 \pm 0.1)\text{ cm}$ and the second resonance occurs at $l_2 = (52.5 \pm 0.1)\text{ cm}$ using a tuning fork of frequency $\nu = (500 \pm 1)\text{ Hz}$. Determine the speed of sound and its maximum percentage error.
Solution:
1. Formula for speed of sound eliminating end correction:
$v = 2\nu (l_2 – l_1)$
$l_2 – l_1 = 52.5\text{ cm} – 17.5\text{ cm} = 35.0\text{ cm} = 0.35\text{ m}$
$v = 2 \times (500\text{ s}^{-1}) \times (0.35\text{ m}) = 350\text{ m/s}$.
2. Absolute error in length difference:
$\Delta(l_2 – l_1) = \Delta l_2 + \Delta l_1 = 0.1\text{ cm} + 0.1\text{ cm} = 0.2\text{ cm}$.
3. Fractional error in speed of sound:
$\frac{\Delta v}{v} = \frac{\Delta \nu}{\nu} + \frac{\Delta (l_2 – l_1)}{l_2 – l_1} = \frac{1}{500} + \frac{0.2\text{ cm}}{35.0\text{ cm}} = 0.002 + 0.00571 = 0.00771$
4. Percentage error:
$\%v = 0.00771 \times 100\% \approx 0.77\%$.
Absolute error: $\Delta v = 350 \times 0.00771 \approx 2.7\text{ m/s} \implies v = (350.0 \pm 2.7)\text{ m/s}$.
Worksheet: Unit Test (10 Questions)
Problem 1 (JEE Main – Single Correct):
The momentum of an object is measured with an error of $2\%$. If the mass of the object has an uncertainty of $1\%$, the maximum percentage error in the calculated kinetic energy $E = \frac{p^2}{2m}$ is:
(A) $3\%$
(B) $5\%$
(C) $4\%$
(D) $6\%$
Problem 2 (JEE Main – Single Correct):
Magnetic dipole moment per unit volume is termed Magnetization ($M$). The base SI dimensional formula of Magnetization is:
(A) $[M^0 L^{-1} T^0 A^1]$
(B) $[M^0 L^2 T^0 A^1]$
(C) $[M^1 L^{-2} T^1 A^{-1}]$
(D) $[M^0 L^{-2} T^0 A^1]$
Problem 3 (JEE Main – Single Correct):
In Searle’s apparatus to find Young’s modulus $Y = \frac{4 M g L}{\pi d^2 l}$, the length $L$ is measured with a meter scale, diameter $d$ with a screw gauge, load mass $M$ with a physical balance, and extension $l$ with a micrometer. Which of the following quantities should be measured with the greatest accuracy to minimize the error in $Y$?
(A) Load mass $M$
(B) Wire length $L$
(C) Wire diameter $d$
(D) Wire elongation $l$
Problem 4 (JEE Main – Single Correct):
A screw gauge has a pitch of $0.5\text{ mm}$ and $50$ divisions on its circular scale. When the anvils touch, the circular scale zero is $4$ divisions above the reference line. When a sheet is inserted, the main scale shows $3\text{ divisions}$ ($1.5\text{ mm}$) and the circular scale reads $32$. The true thickness of the sheet is:
(A) $1.78\text{ mm}$
(B) $1.86\text{ mm}$
(C) $1.82\text{ mm}$
(D) $1.74\text{ mm}$
Problem 5 (JEE Main – Single Correct):
The energy density (energy per unit volume) of an electrostatic field is given by $u = k \cdot \varepsilon_0^a E^b$, where $E$ is electric field intensity and $\varepsilon_0$ is permittivity. The values of $a$ and $b$ are:
(A) $a = 1, b = 2$
(B) $a = 1, b = 1$
(C) $a = 2, b = 1$
(D) $a = -1, b = 2$
Problem 6 (JEE Advanced – One or More Correct):
Which of the following expressions represent a pure dimensionless quantity ($[M^0 L^0 T^0]$)?
(A) $\frac{e^2}{4\pi \varepsilon_0 \hbar c}$ (Fine structure constant)
(B) $\frac{R}{N_A k_B}$ (Ratio of Gas constant to product of Avogadro number and Boltzmann constant)
(C) $\frac{B^2}{2\mu_0 P}$ (Ratio of magnetic energy density to static pressure)
(D) $\frac{G M}{r c^2}$ (Gravitational potential parameter)
Problem 7 (JEE Advanced – One or More Correct):
A student determines the density of a wire $\rho = \frac{4 m}{\pi D^2 L}$ by measuring mass $m = (0.300 \pm 0.003)\text{ g}$, diameter $D = (0.500 \pm 0.005)\text{ mm}$, and length $L = (10.00 \pm 0.05)\text{ cm}$. Which of the following statements is/are correct?
(A) The fractional error in mass $m$ is $1\%$.
(B) The fractional error in diameter $D$ is $1\%$.
(C) The measurement of diameter contributes $2\%$ to the overall percentage error in density.
(D) The total percentage error in density is $3.5\%$.
Problem 8 (JEE Advanced – One or More Correct):
Let $[X]$ represent the dimension of quantity $X$. Which of the following statements is/are TRUE?
(A) $\left[\frac{L}{C}\right]$ has dimensions of resistance squared ($[R^2]$).
(B) $\left[\sqrt{\frac{L}{C}}\right]$ has dimensions of resistance ($[R]$).
(C) $[R C]$ has dimensions of time ($[T]$).
(D) $\left[\frac{1}{\sqrt{L C}}\right]$ has dimensions of angular frequency ($[T^{-1}]$).
Problem 9 (JEE Main / Advanced – Numerical Value Type):
In a resonance tube experiment, the first two resonances are obtained at lengths $l_1 = 18.0\text{ cm}$ and $l_2 = 54.0\text{ cm}$ using a tuning fork of frequency $480\text{ Hz}$. Calculate the speed of sound in air in $\text{m/s}$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A physical quantity $Z$ depends on four independent experimental observables $A, B, C, D$ according to $Z = \frac{A^3 B^{1/2}}{C^2 D^3}$. If the percentage errors in $A, B, C, D$ are $1\%, 4\%, 2\%, 1\%$ respectively, the maximum percentage error in $Z$ is $k\%$. Find the integer value of $k$.
Solutions & Explanations
Answer Key Summary:
1. (B) | 2. (A) | 3. (C) | 4. (B) | 5. (A) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 345.6 or 346 | 10. 12
Solution 1:
$E = \frac{p^2}{2m} \implies \frac{\Delta E}{E} = 2\left(\frac{\Delta p}{p}\right) + \frac{\Delta m}{m}$.
$\%E = 2(2\%) + 1\% = 4\% + 1\% = 5\%$.
Correct Answer: (B)
Solution 2:
Magnetization $M = \frac{\text{Magnetic Dipole Moment}}{\text{Volume}} = \frac{I \cdot A}{V}$.
$[M] = \frac{[A][L^2]}{[L^3]} = [A L^{-1}] = [M^0 L^{-1} T^0 A^1]$.
Units: $\text{Ampere/meter}$.
Correct Answer: (A)
Solution 3:
$Y = \frac{4 M g L}{\pi d^2 l} \implies \frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \frac{\Delta L}{L} + 2\left(\frac{\Delta d}{d}\right) + \frac{\Delta l}{l}$.
The diameter $d$ appears with power $2$. Furthermore, $d$ is a very small quantity (fractions of a millimeter), meaning any absolute error produces a large relative error that is doubled in the formula. Thus, $d$ requires the greatest measurement precision.
Correct Answer: (C)
Solution 4:
$\text{LC} = \frac{0.5\text{ mm}}{50} = 0.01\text{ mm}$.
The circular zero is $4$ divisions above reference line $\implies$ Negative Zero Error $= -4 \times 0.01\text{ mm} = -0.04\text{ mm}$.
Observed reading $= \text{LSR} + (\text{CSR} \times \text{LC}) = 1.5\text{ mm} + (32 \times 0.01\text{ mm}) = 1.82\text{ mm}$.
True thickness $= \text{Observed} – (\text{Zero Error}) = 1.82\text{ mm} – (-0.04\text{ mm}) = 1.86\text{ mm}$.
Correct Answer: (B)
Solution 5:
Energy density $[u] = \frac{\text{Energy}}{\text{Volume}} = [M L^{-1} T^{-2}]$.
Permittivity $[\varepsilon_0] = [M^{-1} L^{-3} T^4 A^2]$.
Electric field $[E] = [M L T^{-3} A^{-1}]$.
$[u] = [\varepsilon_0]^a [E]^b \implies [M L^{-1} T^{-2}] = [M^{-1} L^{-3} T^4 A^2]^a [M L T^{-3} A^{-1}]^b = [M^{b – a} L^{-3a + b} T^{4a – 3b} A^{2a – b}]$.
– From $A$: $2a – b = 0 \implies b = 2a$.
– From $M$: $b – a = 1 \implies 2a – a = 1 \implies a = 1$.
Then $b = 2(1) = 2$.
Thus $u = \frac{1}{2} \varepsilon_0 E^2 \implies a = 1, b = 2$.
Correct Answer: (A)
Solution 6:
– (A) True: Fine structure constant is pure number $\approx 1/137$.
– (B) True: $R = N_A k_B \implies \frac{R}{N_A k_B} = 1$ (exact number).
– (C) True: $\frac{B^2}{2\mu_0}$ and $P$ both have units $\text{N/m}^2$ (pressure/energy density), so their ratio is dimensionless.
– (D) True: $[G M / r] = [v^2] \implies \frac{G M}{r c^2} = \frac{v^2}{c^2}$ is a ratio of squared speeds, dimensionless.
All are dimensionless.
Correct Answer: (A, B, C, D)
Solution 7:
– $\frac{\Delta m}{m} = \frac{0.003}{0.300} = 0.01 = 1\%$. (A is True)
– $\frac{\Delta D}{D} = \frac{0.005}{0.500} = 0.01 = 1\%$. (B is True)
– In $\rho = \frac{4 m}{\pi D^2 L}$, exponent of $D$ is $2$, contributing $2 \times 1\% = 2\%$. (C is True)
– $\frac{\Delta L}{L} = \frac{0.05}{10.00} = 0.005 = 0.5\%$.
Total percentage error $= 1\% + 2(1\%) + 0.5\% = 3.5\%$. (D is True)
Correct Answer: (A, B, C, D)
Solution 8:
– Since $[L/R] = [T]$ and $[R C] = [T]$, $\frac{[L/R]}{[R C]} = 1 \implies \left[\frac{L}{C R^2}\right] = 1 \implies \left[\frac{L}{C}\right] = [R^2]$. (A is True)
– $\left[\sqrt{\frac{L}{C}}\right] = [R]$. (B is True)
– $[R C] = [T]$. (C is True)
– $[\omega] = \left[\frac{1}{\sqrt{L C}}\right] = [T^{-1}]$. (D is True)
All are True.
Correct Answer: (A, B, C, D)
Solution 9:
$v = 2\nu (l_2 – l_1)$.
$l_2 – l_1 = 54.0\text{ cm} – 18.0\text{ cm} = 36.0\text{ cm} = 0.36\text{ m}$.
$v = 2 \times 480 \times 0.36 = 960 \times 0.36 = 345.6\text{ m/s} \approx 346\text{ m/s}$.
Correct Answer: 345.6 (or 346)
Solution 10:
$Z = \frac{A^3 B^{1/2}}{C^2 D^3}$.
$\%Z = 3(\%A) + \frac{1}{2}(\%B) + 2(\%C) + 3(\%D)$
$\%Z = 3(1\%) + \frac{1}{2}(4\%) + 2(2\%) + 3(1\%) = 3\% + 2\% + 4\% + 3\% = 12\%$.
$k = 12$.
Correct Answer: 12