Units and Measurements Previous Year Questions — JEE Main Physics
50 past-year Units and Measurements questions from JEE Main (Physics) — page 6.
Q33 (2024)
A physical quantity $Q$ is found to depend on quantities $a, b, c$ by the relation $Q = \frac{a^4 b^3}{c^2}$. The percentage error in $a, b$ and $c$ are $3\%$, $4\%$ and $5\%$ respectively. Then, the percentage error in $Q$ is:
- $66\%$
- $43\%$
- $34\%$
- $14\%$
### Related Formula
For a quantity $Q = \frac{a^x b^y}{c^z}$, the maximum relative error is:
$$\frac{\Delta Q}{Q} = x \frac{\Delta a}{a} + y \frac{\Delta b}{b} + z \frac{\Delta c}{c}$$
Multiplying by $100$ gives the percentage error equation:
$$\%\text{ error in } Q = x(\%\text{ error in } a) + y(\%\text{ error in } b) + z(\%\text{ error in } c)$$
### Core Logic
Given formula:
$$Q = \frac{a^4 b^3}{c^2}$$
Applying the error propagation rule:
$$\frac{\Delta Q}{Q} \times 100 = 4 \left(\frac{\Delta a}{a} \times 100\right) + 3 \left(\frac{\Delta b}{b} \times 100\right) + 2 \left(\frac{\Delta c}{c} \times 100\right)$$
### Step 1: Substitute the Percentage Errors
Substitute the given values:
* $\%\text{ error in } a = 3\%$
* $\%\text{ error in } b = 4\%$
* $\%\text{ error in } c = 5\%$
$$\%\text{ error in } Q = 4(3\%) + 3(4\%) + 2(5\%)$$
$$\%\text{ error in } Q = 12\% + 12\% + 10\% = 34\%$$
### Pattern Recognition
Error percentages always add up, weighted by their exponents in the mathematical expression, regardless of whether the variable is in the numerator or denominator.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Units and Measurements
Q38 (2024)
Identify the physical quantity that cannot be measured using a spherometer:
- $\text{Radius of curvature of concave surface}$
- $\text{Specific rotation of liquids}$
- $\text{Thickness of thin plates}$
- $\text{Radius of curvature of convex surface}$
### Core Logic
A spherometer is a mechanical instrument designed to measure small vertical displacements to calculate the thickness of thin plates or the radius of curvature of spherical (convex/concave) surfaces.
Specific rotation of liquids is an optical property measured via a polarimeter, completely outside the scope of a spherometer.
### Pattern Recognition
Spherometers operate purely on linear micrometer screw scale geometry, hence restricted strictly to spatial dimensions.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Units and Measurements
Q47 (2024)
Given below are two statements:
Statement (I): Planck's constant and angular momentum have same dimensions.
Statement (II): Linear momentum and moment of force have same dimensions.
In the light of the above statements, choose the correct answer from the options given below:
- $\text{Statement I is true but Statement II is false}$
- $\text{Both Statement I and Statement II are false}$
- $\text{Both Statement I and Statement II are true}$
- $\text{Statement I is false but Statement II is true}$
### Core Logic
Evaluate dimensions step-by-step:
1. **Planck's constant ($h$)**:
$$E = h
u \implies [h] = \frac{[E]}{[
u]} = \frac{\text{ML}^2\text{T}^{-2}}{\text{T}^{-1}} = \text{ML}^2\text{T}^{-1}$$
2. **Angular momentum ($L$)**:
$$L = mvr \implies [L] = \text{M} \cdot \text{LT}^{-1} \cdot \text{L} = \text{ML}^2\text{T}^{-1}$$
Since $[h] = [L]$, Statement I is true.
3. **Linear momentum ($P$)**:
$$P = mv \implies [P] = \text{MLT}^{-1}$$
4. **Moment of force (Torque $\tau$)**:
$$\tau = F r \implies [\tau] = \text{MLT}^{-2} \cdot \text{L} = \text{ML}^2\text{T}^{-2}$$
Since $[P] \neq [\tau]$, Statement II is false.
### Pattern Recognition
Planck's constant can always be paired with angular momentum units (Joule-seconds), while moment of force matches work/energy footprints, not translational momentum.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Units and Measurements
Q38 (2024)
The resistance $R = \frac{V}{I}$ where $V = (200 \pm 5) \mathrm{~V}$ and $I = (20 \pm 0.2) \mathrm{~A}$, the percentage error in the measurement of $R$ is:
- 3.5%
- 7%
- 3%
- 5.5%
### Related Formula
By propagation of maximum relative error in division:
$$R = \frac{V}{I} \implies \frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$$
Percentage error in $R$ is given by:
$$\% \text{ error in } R = \left( \frac{\Delta V}{V} + \frac{\Delta I}{I} \right) \times 100$$
### Core Logic
Given values:
$$V = 200 \mathrm{~V}, \quad \Delta V = 5 \mathrm{~V}$$
$$I = 20 \mathrm{~A}, \quad \Delta I = 0.2 \mathrm{~A}$$
### Step 1: Evaluate Relative Error
$$\frac{\Delta R}{R} = \frac{5}{200} + \frac{0.2}{20}$$
$$\frac{\Delta R}{R} = \frac{5}{200} + \frac{2}{200}$$
$$\frac{\Delta R}{R} = \frac{7}{200}$$
### Step 2: Calculate Percentage Error
$$\% \text{ error in } R = \frac{\Delta R}{R} \times 100 = \frac{7}{200} \times 100 = 3.5\%$$
Thus, the percentage error is $3.5\%$.
### Pattern Recognition
Whenever independent physical quantities are multiplied or divided, their fractional/relative errors always add up. Make sure to keep the base denominator values aligned to make mental calculations quick.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Units and Measurements
Q31 (2024)
If $50$ Vernier divisions are equal to $49$ main scale divisions of a travelling microscope and one smallest reading of main scale is $0.5 \mathrm{~mm}$, the Vernier constant of travelling microscope is:
- $0.1 \mathrm{~mm}$
- $0.1 \mathrm{~cm}$
- $0.01 \mathrm{~cm}$
- $0.01 \mathrm{~mm}$
### Related Formula
$$\text{Vernier Constant (Least Count)} = 1 \mathrm{~MSD} - 1 \mathrm{~VSD}$$
### Core Logic
Given that $50$ Vernier Scale Divisions (VSD) equal $49$ Main Scale Divisions (MSD).
$$50 \mathrm{~VSD} = 49 \mathrm{~MSD}$$
$$1 \mathrm{~VSD} = \frac{49}{50} \mathrm{~MSD}$$
Also, the smallest reading of the main scale ($1 \mathrm{~MSD}$) is $0.5 \mathrm{~mm}$.
### Step 1: Calculate Vernier Constant
$$\text{Vernier Constant} = 1 \mathrm{~MSD} - 1 \mathrm{~VSD}$$
$$= 1 \mathrm{~MSD} - \frac{49}{50} \mathrm{~MSD}$$
$$= \frac{1}{50} \mathrm{~MSD}$$
Substitute the value of $1 \mathrm{~MSD}$:
$$= \frac{1}{50} \times 0.5 \mathrm{~mm} = \frac{0.5}{50} \mathrm{~mm} = \frac{1}{100} \mathrm{~mm}$$
$$= 0.01 \mathrm{~mm}$$
### Pattern Recognition
In Vernier calipers problems where $N \text{ VSD} = (N-1) \text{ MSD}$, the Least Count is always exactly $\frac{1}{N} \text{ MSD}$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Units and Measurements
