Units and Measurements Previous Year Questions — JEE Main Physics
50 past-year Units and Measurements questions from JEE Main (Physics) — page 2.
Q10 (2025)
A quantity $Q$ is formulated as $X^{-2}Y^{\frac{3}{2}}Z^{- rac{2}{5}}$. $X, Y$ and $Z$ are independent parameters which have fractional errors of $0.1, 0.2$ and $0.5$, respectively in measurement. The maximum fractional error of $Q$ is:
- $0.1$
- $0.8$
- $0.7$
- $0.6$
### Related Formula
For a quantity $Q = X^a Y^b Z^c$, the maximum fractional error is:
$$\frac{\Delta Q}{Q} = |a| \frac{\Delta X}{X} + |b| \frac{\Delta Y}{Y} + |c| \frac{\Delta Z}{Z}$$
where,
$\frac{\Delta X}{X}, \frac{\Delta Y}{Y}, \frac{\Delta Z}{Z}$ are fractional errors of individual variables
### Core Logic
Given formula: $Q = X^{-2} Y^{3/2} Z^{-2/5}$.
Identify the absolute exponents:
- $|a| = |-2| = 2$
- $|b| = \left|\frac{3}{2}\right| = \frac{3}{2}$
- $|c| = \left|-\frac{2}{5}\right| = \frac{2}{5}$
Now write the error expression:
$$\frac{\Delta Q}{Q} = 2 \frac{\Delta X}{X} + \frac{3}{2} \frac{\Delta Y}{Y} + \frac{2}{5} \frac{\Delta Z}{Z}$$
Substitute the given values:
- $\frac{\Delta X}{X} = 0.1$
- $\frac{\Delta Y}{Y} = 0.2$
- $\frac{\Delta Z}{Z} = 0.5$
### Step 1: Compute Maximum Fractional Error
Calculate term by term:
$$\frac{\Delta Q}{Q} = 2 (0.1) + \frac{3}{2} (0.2) + \frac{2}{5} (0.5)$$
$$\frac{\Delta Q}{Q} = 0.2 + 0.3 + 0.2 = 0.7$$
### Pattern Recognition
Sees: Exponential algebraic relation for errors.
Trap: Exponents are negative, but maximum error is cumulative. Always take the *absolute* value of exponents when summing errors! ✓
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Q16 (2025)
Match List-I with List-II.
$$\begin{array}{|l|l|l|l|} \hline \textbf{List-I} & & \textbf{List-II} & \\ \hline \text{(A)} & \text{Young's Modulus} & \text{(I)} & \mathrm{ML^{-1}T^{-1}} \\ \text{(B)} & \text{Torque} & \text{(II)} & \mathrm{ML^{-1}T^{-2}} \\ \text{(C)} & \text{Coefficient of Viscosity} & \text{(III)} & \mathrm{M^{-1}L^3T^{-2}} \\ \text{(D)} & \text{Gravitational Constant} & \text{(IV)} & \mathrm{ML^2T^{-2}} \\ \hline \end{array}$$
Choose the correct answer from the options given below:
- $\text{(A)-(I), (B)-(III), (C)-(II), (D)-(IV)}$
- $\text{(A)-(II), (B)-(I), (C)-(IV), (D)-(III)}$
- $\text{(A)-(IV), (B)-(II), (C)-(III), (D)-(I)}$
- $\text{(A)-(II), (B)-(IV), (C)-(I), (D)-(III)}$
### Related Formula
$$\text{Young's Modulus: } Y = \frac{F/A}{\Delta \ell / \ell}$$
$$\text{Torque: } \tau = F \cdot r$$
$$\text{Viscosity Force: } F = \eta A \frac{dv}{dx}$$
$$\text{Gravitational Force: } F = \frac{G m_1 m_2}{r^2}$$
### Core Logic
Evaluating dimensions component-by-component:
- **(A) Young's Modulus**:
$$[Y] = \frac{[F]}{[A]} = \frac{\mathrm{MLT^{-2}}}{\mathrm{L^2}} = \mathrm{ML^{-1}T^{-2}} \quad \rightarrow \text{(II)}$$
- **(B) Torque**:
$$[\tau] = [F][r] = (\mathrm{MLT^{-2}})(\mathrm{L}) = \mathrm{ML^2T^{-2}} \quad \rightarrow \text{(IV)}$$
- **(C) Coefficient of Viscosity**:
$$[\eta] = \frac{[F]}{[A][dv/dx]} = \frac{\mathrm{MLT^{-2}}}{(\mathrm{L^2})(\mathrm{T^{-1}})} = \mathrm{ML^{-1}T^{-1}} \quad \rightarrow \text{(I)}$$
- **(D) Gravitational Constant**:
$$[G] = \frac{[F][r^2]}{[m_1][m_2]} = \frac{(\mathrm{MLT^{-2}})(\mathrm{L^2})}{\mathrm{M^2}} = \mathrm{M^{-1}L^3T^{-2}} \quad \rightarrow \text{(III)}$$
Matching path yields: (A)-(II), (B)-(IV), (C)-(I), (D)-(III).
### Pattern Recognition
Torque and energy share the identical dimensional formula $\mathrm{ML^2T^{-2}}$. Modulus and pressure share $\mathrm{ML^{-1}T^{-2}}$. Spotting these matching associations cuts solving time significantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Q23 (2025)
A physical quantity $Q$ is related to four observables $a, b, c, d$ as follows: $Q = \frac{ab^4}{cd}$ where, $a = (60 \pm 3)\mathrm{~Pa}$ ; $b = (20 \pm 0.1)\mathrm{~m}$ ; $c = (40 \pm 0.2)\mathrm{~Nsm}^{-2}$ and $d = (50 \pm 0.1)\mathrm{~m}$ , then the percentage error in $Q$ is $\frac{x}{1000}$ , where $x = $ ______.
### Related Formula
$$\frac{\Delta Q}{Q} = \frac{\Delta a}{a} + 4\frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{\Delta d}{d}$$
### Core Logic
Write down fractional errors from the raw text configurations:
- $\frac{\Delta a}{a} = \frac{3}{60} = 0.05$
- $\frac{\Delta b}{b} = \frac{0.1}{20} = 0.005$
- $\frac{\Delta c}{c} = \frac{0.2}{40} = 0.005$
- $\frac{\Delta d}{d} = \frac{0.1}{50} = 0.002$
Compute the total fractional error expression:
$$\frac{\Delta Q}{Q} = [0.05 + 4(0.005) + 0.005 + 0.002]$$
$$\frac{\Delta Q}{Q} = 0.05 + 0.02 + 0.005 + 0.002 = 0.077$$
Percentage error expression configuration:
$$\% \text{ Error} = \frac{\Delta Q}{Q} \times 100 = 7.7 \%$$
Given that percentage error equals $\frac{x}{1000}$:
$$\frac{x}{1000} = 7.7 \implies x = 7700$$
### Pattern Recognition
Powers scale up error contributions via direct multiplication multipliers. The term $b^4$ contributes exactly 4 times its basic fraction error component to the compilation step.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Q21 (2025)
A tiny metallic rectangular sheet has length and breadth of $5 \mathrm{~mm}$ and $2.5 \mathrm{~mm}$ , respectively. Using a specially designed screw gauge which has pitch of $0.75 \mathrm{~mm}$ and 15 divisions in the circular scale, you are asked to find the area of the sheet. In this measurement, the maximum fractional error will be $\frac{\mathrm{x}}{100}$ where $\mathrm{x}$ is ________.
### Core Logic
First, find the least count of the measurement tool:
$$\text{Least Count} = \frac{\text{Pitch}}{\text{Number of circular scale divisions}} = \frac{0.75 \mathrm{~mm}}{15} = 0.05 \mathrm{~mm}$$
{{SOL_IMG1}}
The area of the rectangular metallic sheet is calculated as:
$$\mathrm{A} = \mathrm{L} \cdot \mathrm{W}$$
Expressing the absolute error via fractional configuration parts:
$$\frac{\mathrm{dA}}{\mathrm{A}} = \frac{\mathrm{dL}}{\mathrm{L}} + \frac{\mathrm{dW}}{\mathrm{W}}$$
Substituting the instrument limits ($\mathrm{dL} = \mathrm{dW} = 0.05 \mathrm{~mm}$):
$$\frac{\mathrm{dA}}{\mathrm{A}} = \frac{0.05}{5} + \frac{0.05}{2.5} = \frac{1}{100} + \frac{2}{100} = \frac{3}{100}$$
### Step 1: Final Value Match
Comparing this to the target format $\frac{\mathrm{x}}{100}$ gives:
$\mathrm{x} = 3$
### Pattern Recognition
The absolute measurement uncertainty matches the instrument's least count value directly. Sum up individual fractional errors to compute the total area uncertainty parameter.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Q23 (2025)
In a measurement, it is asked to find modulus of elasticity per unit torque applied on the system. The measured quantity has dimension of $\left[\mathrm{M}^{\mathrm{a}}\mathrm{L}^{\mathrm{b}}\mathrm{T}^{\mathrm{c}}\right]$ . If $b = 3$ , the value of $c$ is
### Core Logic
Let's find the dimensional formula for the ratio of Modulus of Elasticity to Torque:
$$\text{Target Dimensions} = \frac{[\text{Modulus of Elasticity}]}{[\text{Torque}]}$$
$$\text{Target Dimensions} = \frac{[\mathrm{M L}^{-1} \mathrm{T}^{-2}]}{[\mathrm{M L}^2 \mathrm{T}^{-2}]} = [\mathrm{M}^0 \mathrm{L}^{-3} \mathrm{T}^0]$$
### Step 1: Exponent Matching
Comparing this output to the target layout formula $[\mathrm{M}^{\mathrm{a}} \mathrm{L}^{\mathrm{b}} \mathrm{T}^{\mathrm{c}}]$:
$\mathrm{c} = 0$
### Pattern Recognition
Both dimensions share identical time dependence factors ($\mathrm{T}^{-2}$), meaning they cancel out completely. This leaves the time exponent value as exactly zero.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
