Units and Measurements Previous Year Questions — JEE Main Physics
50 past-year Units and Measurements questions from JEE Main (Physics) — page 9.
Q44 (2026)
Match the LIST-I with LIST-II
<div style="overflow-x: auto; margin: 1rem 0;"><table style="width: 100%; border-collapse: collapse; text-align: left; font-size: 14px;"><thead><tr><th style="border: 1px solid #888; padding: 8px;">List-I</th><th style="border: 1px solid #888; padding: 8px;">List-II</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">(A) Spring constant</td><td style="border: 1px solid #888; padding: 8px;">(I) $ML^{2}T^{-2}K^{-1}$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">(B) Thermal conductivity</td><td style="border: 1px solid #888; padding: 8px;">(II) $ML^{0}T^{-2}$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">(C) Boltzmann constant</td><td style="border: 1px solid #888; padding: 8px;">(III) $ML^{2}T^{-3}A^{-2}$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">(D) Inductive reactance</td><td style="border: 1px solid #888; padding: 8px;">(IV) $MLT^{-3}K^{-1}$</td></tr></tbody></table></div>
Choose the correct answer from the options given below:
- \text{A-II, B-I, C-IV, D-III}
- \text{A-I, B-IV, C-II, D-III}
- \text{A-III, B-II, C-IV, D-I}
- \text{A-II, B-IV, C-I, D-III}
### Related Formula
$$F = Kx, \quad \frac{dQ}{dt} = -KA\frac{dT}{dx}, \quad E = k_B T, \quad X_L = \omega L$$
### Core Logic
(A) Spring constant $K$: $[ML^0 T^{-2}]$ -> (II)
(B) Thermal conductivity $K$: $[MLT^{-3} K^{-1}]$ -> (IV)
(C) Boltzmann constant $k_B$: $[ML^2 T^{-2} K^{-1}]$ -> (I)
(D) Inductive reactance $X_L$: $[ML^2 T^{-3} A^{-2}]$ -> (III)
Therefore, matching is A-II, B-IV, C-I, D-III.
### Pattern Recognition
Sees: Match List dimensional formulas.
Shortcut: Derive or recall key physical constants dimensions (Spring constant, Thermal conductivity, Boltzmann constant, Inductive reactance).
Check: Matches option (4). ✓
### Chapter Mix
Class 11 Physics: Units and Measurements
Q26 (2026)
If $\in$, E and t represent the free space permittivity, electric field and time respectively, then the unit of $\frac{\in E}{t}$ will be :
- $\mathrm{Am}$
- $\mathrm{Am}^2$
- $\mathrm{A/m}^2$
- $\mathrm{A/m}$
### Related Formula
$$E = \frac{1}{4\pi\in} \cdot \frac{q}{r^2}$$
$$\left[\frac{\in E}{t}\right] = \left[\frac{q}{t \cdot r^2}\right]$$
### Core Logic
Substituting electric field equation into the expression gives:
$$\frac{\in E}{t} = \frac{\in}{t} \cdot \frac{1}{4\pi\in} \frac{q}{r^2} = \frac{q}{4\pi t r^2}$$
Substituting dimensional formulas for current ($I = q/t \rightarrow \mathrm{A}$) and area ($r^2 \rightarrow \mathrm{m}^2$):
$$\left[\frac{\in E}{t}\right] = \frac{\mathrm{A} \cdot \mathrm{T}}{\mathrm{T} \cdot \mathrm{L}^2} = \mathrm{A} \mathrm{L}^{-2} = \mathrm{A/m}^2$$
### Step 1: Final Conclusion
Hence, the unit of $\frac{\in E}{t}$ is $\mathrm{A/m}^2$.
### Pattern Recognition
Sees: $\in E / t$ product.
Shortcut: Permittivity times Electric Field is Displacement Field $D = \in E$, which has units of Charge per unit Area ($C/m^2$). Dividing by time yields $C/(s \cdot m^2) = A/m^2$ directly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Class 12 Physics: Electrostatics
Q29 (2026)
Four persons measure the length of a rod as 20.00 cm, 19.75 cm, 17.01 cm and 18.25 cm. The relative error in the measurement of average length of the rod is :
- $0.24$
- $0.18$
- $0.06$
- $0.08$
### Related Formula
$$l_{\text{mean}} = \frac{\sum l_{i}}{n}$$
$$\Delta l_{\text{mean}} = \frac{\sum |\Delta l_{i}|}{n}$$
$$\text{Relative Error} = \frac{\Delta l_{\text{mean}}}{l_{\text{mean}}}$$
### Step 1: Calculate Mean Value
$$l_{\text{mean}} = \frac{20.00 + 19.75 + 17.01 + 18.25}{4}$$
$$l_{\text{mean}} = \frac{75.01}{4} = 18.7525 \approx 18.75 \text{ cm}$$
### Step 2: Calculate Mean Absolute Error
Deviations from the mean:
$|\Delta l_{1}| = |20.00 - 18.75| = 1.25$
$|\Delta l_{2}| = |19.75 - 18.75| = 1.00$
$|\Delta l_{3}| = |17.01 - 18.75| = 1.74$
$|\Delta l_{4}| = |18.25 - 18.75| = 0.50$
$$\Delta l_{\text{mean}} = \frac{1.25 + 1.00 + 1.74 + 0.50}{4}$$
$$\Delta l_{\text{mean}} = \frac{4.49}{4} = 1.1225 \approx 1.12 \text{ cm}$$
### Step 3: Calculate Relative Error
$$\text{Relative Error} = \frac{\Delta l_{\text{mean}}}{l_{\text{mean}}}$$
$$\text{Relative Error} = \frac{1.12}{18.75} = 0.05973 \approx 0.06$$
### Pattern Recognition
Sees: "relative error" + "multiple readings" → first find mean, then find absolute differences from mean, average those differences, and finally divide by the mean.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Q44 (2026)
In a screw gauge, the zero of the circular scale lies 3 divisions above the horizontal pitch line when their metallic studs are brought in contact. Using this instrument thickness of a sheet is measured. If pitch scale reading is 1 mm and the circular scale reading is 51 then the correct thickness of the sheet is ____ mm.
[Assume least count is 0.01 mm]
- $1.50$
- $1.48$
- $1.54$
- $1.51$
### Related Formula
$$\text{Zero Error} = \text{Division} \times \text{Least Count}$$
$$\text{True Reading} = \text{Measured Reading} - \text{Zero Error}$$
### Core Logic
Since the zero of the circular scale lies *above* the horizontal reference line when the studs are in contact, the screw gauge has a negative zero error. This means the instrument fundamentally "reads" a value less than the actual value, so the error must be added to the raw reading.
### Step 1: Evaluate Zero Error
$$\text{Zero error } e = -3 \times \text{LC} = -3 \times 0.01 \text{ mm} = -0.03 \text{ mm}$$
### Step 2: Calculate Reading
$$\text{Measured Reading} = \text{Pitch Scale Reading} + (\text{Circular Scale Reading} \times \text{LC})$$
$$\text{Measured Reading} = 1 \text{ mm} + (51 \times 0.01 \text{ mm})$$
$$\text{Measured Reading} = 1.51 \text{ mm}$$
### Step 3: Apply Zero Correction
$$\text{Correct Thickness} = \text{Measured Reading} - e$$
$$\text{Correct Thickness} = 1.51 - (-0.03) = 1.54 \text{ mm}$$
### Pattern Recognition
Sees: "zero lies above reference line" → Negative zero error. True value = Measured + |Error|. If it lies below, positive error.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Q48 (2026)
A ball of radius r and density $\rho$ dropped through a viscous liquid of density $\sigma$ and viscosity $\eta$ attains its terminal velocity at time t, given by $t = A \rho^{a} r^{b} \eta^{c} \sigma^{d}$ , where A is a constant and a, b c and d are integers. The value of $\frac{b + c}{a + d}$ is ____.
### Related Formula
Time dimension: $[T] = \mathrm{T}$
Density dimension: $[\rho] = [\sigma] = \mathrm{ML^{-3}}$
Radius dimension: $[r] = \mathrm{L}$
Viscosity dimension: $[\eta] = \mathrm{ML^{-1}T^{-1}}$
### Core Logic
Given dimensional equation:
$$T = [\rho]^a [r]^b [\eta]^c [\sigma]^d$$
$$T = (\mathrm{ML^{-3}})^a (\mathrm{L})^b (\mathrm{ML^{-1}T^{-1}})^c (\mathrm{ML^{-3}})^d$$
Expand the bases:
$$T^1 = \mathrm{M}^{a+c+d} \mathrm{L}^{-3a+b-c-3d} \mathrm{T}^{-c}$$
### Step 1: Compare Exponents
For Time ($T$):
$$-c = 1 \implies c = -1$$
For Mass ($M$):
$$a + c + d = 0 \implies a - 1 + d = 0 \implies a + d = 1$$
For Length ($L$):
$$-3a + b - c - 3d = 0$$
$$b - c - 3(a + d) = 0$$
Substitute $c = -1$ and $a + d = 1$:
$$b - (-1) - 3(1) = 0$$
$$b + 1 - 3 = 0 \implies b = 2$$
### Step 2: Final Calculation
We need the value of $\frac{b+c}{a+d}$:
Numerator $b+c = 2 + (-1) = 1$
Denominator $a+d = 1$
$$\frac{b+c}{a+d} = \frac{1}{1} = 1$$
### Pattern Recognition
Whenever variables are lumped in a product string $X = y^a z^b$, equating dimensions on both sides produces a solvable linear system. The sum groups $(a+d)$ can sometimes be substituted directly without fully isolating $a$ or $d$ individually.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Class 11 Physics: Mechanical Properties of Fluids
