Units and Measurements Previous Year Questions — JEE Main Physics
50 past-year Units and Measurements questions from JEE Main (Physics).
Q7 (2025)
Given a charge $q$, current $I$ and permeability of vacuum $\mu_0$ . Which of the following quantity has the dimension of momentum?
- $q I / \mu_0$
- $q \mu_0 I$
- $q^2 \mu_0 I$
- $q \mu_0 / I$
### Related Formula
Let momentum be $P$ ($[P] = [M L T^{-1}]$).
We assume:
$$P = q^x \mu_0^y I^z$$
### Core Logic
Let's find the dimensional formulas of the individual variables:
1. **Charge ($q$):**
$[q] = [A T]$
2. **Current ($I$):**
$[I] = [A]$
3. **Permeability of vacuum ($\mu_0$):**
From Biot-Savart law or force between parallel wires: $F = \frac{\mu_0 I^2 L}{2\pi d}$:
$$[\mu_0] = \frac{[F]}{[I]^2} = \frac{[M L T^{-2}]}{[A]^2} = [M L T^{-2} A^{-2}]$$
### Step 1: Apply Dimensional Homogeneity
Substitute these into our assumed dimensional equation:
$$[M L T^{-1}] = [A T]^x [M L T^{-2} A^{-2}]^y [A]^z$$
$$[M L T^{-1}] = [M]^y [L]^y [T]^{x - 2y} [A]^{x - 2y + z}$$
Comparing exponents on both sides:
- For $[M]$:
$y = 1$
- For $[L]$:
$$y = 1 \quad \text{(consistent)}$$
- For $[T]$:
$$x - 2y = -1 \implies x - 2(1) = -1 \implies x = 1$$
- For $[A]$:
$$x - 2y + z = 0 \implies 1 - 2(1) + z = 0 \implies z = 1$$
Thus, $x = 1, y = 1, z = 1$.
Therefore, the required quantity is:
$$q^1 \mu_0^1 I^1 = q \mu_0 I$$
### Pattern Recognition
Sees: Permeability, charge, and current linked to momentum.
Trap: Deriving the dimensions of $\mu_0$ using complex magnetic formulas. Remember $[\mu_0] = [\text{Force}]/[\text{Current}]^2$ is the quickest way to get its dimensions.
Shortcut: Since $[q] = AT$ and $[I] = A$, $[q \mu_0 I] = [A T] [M L T^{-2} A^{-2}] [A] = [M L T^{-1}]$, which is exactly the dimensions of momentum. Hence, option (2) is correct.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Class 12 Physics: Moving Charges and Magnetism
Q93 (2025)
Match List-I with List-II.
$$ \begin{array}{|l|l|} \hline \textbf{List-I} & \textbf{List-II} \\ \hline \text{(A) Coefficient of viscosity} & \text{(I) } [\mathrm{ML}^0\mathrm{T}^{-3}] \\ \hline \text{(B) Intensity of wave} & \text{(II) } [\mathrm{ML}^{-2}\mathrm{T}^{-2}] \\ \hline \text{(C) Pressure gradient} & \text{(III) } [\mathrm{M}^{-1}\mathrm{LT}^{2}] \\ \hline \text{(D) Compressibility} & \text{(IV) } [\mathrm{ML}^{-1}\mathrm{T}^{-1}] \\ \hline \end{array} $$
Choose the correct answer from the options given below:
- (A)-(I), (B)-(IV), (C)-(III), (D)-(II)
- (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
- (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
- (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
### Related Formula
Definitions of physical quantities:
1. Viscosity: $F = -\eta A \frac{dv}{dx}$
2. Intensity: $I = \frac{\text{Power}}{\text{Area}}$
3. Pressure gradient: $\frac{dP}{dx}$
4. Compressibility: $K = \frac{1}{B} = \frac{\text{Strain}}{\text{Stress}}$
### Core Logic
Let's calculate each dimensional formula:
1. **Coefficient of viscosity ($\eta$):**
$$[\eta] = \frac{[F]}{[A]\left[\frac{dv}{dx}\right]} = \frac{\text{M L T}^{-2}}{\text{L}^2 \left(\frac{\text{L T}^{-1}}{\text{L}}\right)} = \text{M L}^{-1}\text{T}^{-1}$$
Matches **(IV)**.
2. **Intensity of wave ($I$):**
$$[I] = \frac{[\text{Power}]}{[\text{Area}]} = \frac{\text{M L}^2\text{T}^{-3}}{\text{L}^2} = \text{M T}^{-3} = \text{M L}^0\text{T}^{-3}$$
Matches **(I)**.
3. **Pressure gradient ($\frac{dP}{dx}$):**
$$\left[\frac{dP}{dx}\right] = \frac{[\text{Pressure}]}{[\text{Length}]} = \frac{\text{M L}^{-1}\text{T}^{-2}}{\text{L}} = \text{M L}^{-2}\text{T}^{-2}$$
Matches **(II)**.
4. **Compressibility ($K$):**
Compressibility is the inverse of Bulk Modulus:
$$[K] = \frac{1}{[\text{Pressure}]} = \frac{1}{\text{M L}^{-1}\text{T}^{-2}} = \text{M}^{-1}\text{L}\text{T}^2$$
Matches **(III)**.
Thus: (A)-(IV), (B)-(I), (C)-(II), (D)-(III).
### Step 1: Final Conclusion
The correct option is (2).
### Pattern Recognition
Target basic matching terms first: Compressibility is the reciprocal of pressure, giving $[M^{-1}LT^2]$. Wave intensity has units of power per unit area, giving $[MT^{-3}]$. This immediately isolates option (2).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Class 11 Physics: Mechanical Properties of Fluids
Class 11 Physics: Mechanical Properties of Solids
Q2 (2025)
The equation for real gas is given by $\left(P + \frac{a}{V^2}\right)(V - b) = RT$, where $P, V, T$ and $R$ are the pressure, volume, temperature and gas constant, respectively. The dimension of $ab^{-2}$ is equivalent to that of:
- Planck's constant
- Compressibility
- Strain
- Energy density
### Related Formula
By the principle of dimensional homogeneity, terms added or subtracted must have the same dimensions:
$$[P] = \left[\frac{a}{V^2}\right] \implies [a] = [P][V]^2$$
$[V] = [b]$
### Core Logic
Let's find the dimensional formula of the quantities:
- Pressure $P$:
$$[P] = \text{M L}^{-1}\text{T}^{-2}$$
- Volume $V$:
$$[V] = \text{L}^3$$
Substituting these to find $[a]$ and $[b]$:
$$[a] = (\text{M L}^{-1}\text{T}^{-2})(\text{L}^6) = \text{M L}^5\text{T}^{-2}$$
$$[b] = \text{L}^3 \implies [b^{-2}] = \text{L}^{-6}$$
Now, compute the dimensions of $ab^{-2}$:
$$[ab^{-2}] = (\text{M L}^5\text{T}^{-2})(\text{L}^{-6}) = \text{M L}^{-1}\text{T}^{-2}$$
This matches the dimensions of pressure.
Let's evaluate the options:
1. Planck's constant: $[h] = \text{M L}^2\text{T}^{-1}$
2. Compressibility: $[\beta] = \text{M}^{-1}\text{L}\text{T}^2$
3. Strain: dimensionless
4. Energy density (energy per unit volume):
$$\left[\frac{E}{V}\right] = \frac{\text{M L}^2\text{T}^{-2}}{\text{L}^3} = \text{M L}^{-1}\text{T}^{-2}$$
### Step 1: Final Conclusion
Therefore, the dimension of $ab^{-2}$ is equivalent to that of Energy density.
### Pattern Recognition
By writing the relation directly as $[ab^{-2}] = \frac{[a]}{[b]^2}$, and noting $[a] = [P][V]^2$ and $[b] = [V]$, we get $[ab^{-2}] = \frac{[P][V]^2}{[V]^2} = [P]$ (Pressure). Since pressure and energy density have identical dimensions, the answer is immediately Energy density.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Class 11 Physics: Kinetic Theory of Gases
Q166 (2025)
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. Boltzmann constant</td><td style="border: 1px solid #888; padding: 8px;">I. $ML^2T^{-1}$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">B. Coefficient of viscosity</td><td style="border: 1px solid #888; padding: 8px;">II. $MLT^{-3}K^{-1}$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">C. Planck's constant</td><td style="border: 1px solid #888; padding: 8px;">III. $ML^2T^{-2}K^{-1}$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">D. Thermal conductivity</td><td style="border: 1px solid #888; padding: 8px;">IV. $ML^{-1}T^{-1}$</td></tr></tbody></table></div>
Choose the correct answer from the options given below :
- A-III, B-IV, C-I, D-II
- A-II, B-III, C-IV, D-I
- A-III, B-II, C-I, D-IV
- A-III, B-IV, C-II, D-I
### Related Formula
Formulas to find dimensional formulas:
- Boltzmann constant:
$$k_B = \frac{\text{Energy}}{\text{Temperature}}$$
- Coefficient of viscosity:
$$\eta = \frac{F}{A \frac{dv}{dx}}$$
- Planck's constant:
$$h = \frac{E}{\nu}$$
- Thermal conductivity:
$$\frac{dQ}{dt} = K A \frac{dT}{dx} \Rightarrow K = \frac{\text{Heat flow} \cdot \text{thickness}}{\text{Area} \cdot \text{Temperature difference}}$$
### Core Logic
Evaluate each constant individually:
### Step 1: Dimensions of Boltzmann constant ($k_B$)
$$[k_B] = \frac{[ML^2T^{-2}]}{[K]} = [ML^2T^{-2}K^{-1}] \quad \Rightarrow \text{Matches III}$$
### Step 2: Dimensions of Coefficient of viscosity ($\eta$)
$$[\eta] = \frac{[MLT^{-2}]}{[L^2] [T^{-1}]} = [ML^{-1}T^{-1}] \quad \Rightarrow \text{Matches IV}$$
### Step 3: Dimensions of Planck's constant ($h$)
$$[h] = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}] \quad \Rightarrow \text{Matches I}$$
### Step 4: Dimensions of Thermal conductivity ($K$)
$$[K] = \frac{[ML^2T^{-3}] [L]}{[L^2] [K]} = [MLT^{-3}K^{-1}] \quad \Rightarrow \text{Matches II}$$
This sequence yields A-III, B-IV, C-I, D-II, matching Option (1).
### Pattern Recognition
To solve matching sets efficiently, search for the most recognizable dimensions first. Planck's constant $h$ ($ML^2T^{-1}$) and viscosity coefficient $\eta$ ($ML^{-1}T^{-1}$) are highly unique and usually resolve the options instantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Q25 (2025)
A physical quantity C is related to four other quantities p, q, r and s as follows $C = \frac{pq^2}{r^3\sqrt{s}}$ The percentage errors in the measurement of p, q, r and s are $1\% , 2\% , 3\%$ and $2\%$ respectively. The percentage error in the measurement of C will be ________ \%.
### Related Formula
For a physical quantity defined by algebraic powers $C = \frac{p^a q^b}{r^c s^d}$, the maximum fractional error is calculated by summing absolute scaled fractional errors:
$$\frac{\Delta C}{C} = a \frac{\Delta p}{p} + b \frac{\Delta q}{q} + c \frac{\Delta r}{r} + d \frac{\Delta s}{s}$$
Expressed as percentages:
$$\% \text{ error in } C = a(\% \text{ error in } p) + b(\% \text{ error in } q) + c(\% \text{ error in } r) + d(\% \text{ error in } s)$$
### Core Logic
Given expression:
$$C = p^1 q^2 r^{-3} s^{-1/2}$$
Max fractional error equation:
$$\frac{\Delta C}{C} = 1 \left(\frac{\Delta p}{p}\right) + 2 \left(\frac{\Delta q}{q}\right) + 3 \left(\frac{\Delta r}{r}\right) + \frac{1}{2} \left(\frac{\Delta s}{s}\right)$$
### Step 1: Calculate the total percentage error
Substitute the individual percentage errors:
- Error in $p = 1\%$
- Error in $q = 2\%$
- Error in $r = 3\%$
- Error in $s = 2\%$
$$\% \text{ error in } C = 1(1\%) + 2(2\%) + 3(3\%) + \frac{1}{2}(2\%)$$
$$\% \text{ error in } C = 1\% + 4\% + 9\% + 1\% = 15\%$$
The total percentage error in $C$ is $15\%$.
### Pattern Recognition
In error propagation, individual errors always combine constructively to produce the maximum possible uncertainty limit. Hence, negative powers (like division by $r^3$ or $s^{1/2}$) are integrated using positive coefficients during maximum absolute error summation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
