Units and Measurements Previous Year Questions — JEE Main Physics
50 past-year Units and Measurements questions from JEE Main (Physics) — page 10.
Q27 (2026)
In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division = 0.05 mm, then the least count of the vernier callipers is ____ mm.
- $0.002$
- $0.05$
- $0.02$
- $0.005$
### Related Formula
$$\text{Least Count (LC)} = 1 \text{ MSD} - 1 \text{ VSD}$$
### Core Logic
Given $50 \text{ VSD} = 48 \text{ MSD}$, we have:
$$1 \text{ VSD} = \frac{48}{50} \text{ MSD}$$
$$\text{LC} = 1 \text{ MSD} - \frac{48}{50} \text{ MSD} = \frac{2}{50} \text{ MSD}$$
### Step 1: Calculation
Substitute $1 \text{ MSD} = 0.05 \text{ mm}$:
$$\text{LC} = \frac{2}{50} \times 0.05 \text{ mm} = 0.002 \text{ mm}$$
### Pattern Recognition
For non-standard vernier calipers where $N \text{ VSD} = (N-x) \text{ MSD}$, the least count is $(x/N) \times \text{MSD}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Q42 (2026)
When both jaws of vernier callipers touch each other, zero mark of the vernier scale is right to zero mark of main scale, $4^{\text{th}}$ mark on vernier scale coincides with certain mark on the main scale. While measuring the length of a cylinder, observer observes 15 divisions on main scale and $5^{\text{th}}$ division of vernier scale coincides with a min scale division. Measured length of cylinder is ____ mm. (Least count of Vernier calliper = 0.1 mm)
- $15.4$
- $15.1$
- $15.5$
- $15.9$
### Related Formula
$$\text{Zero Error} = + (\text{VSR} \times \text{LC})$$
$$\text{Reading} = \text{MSR} + (\text{VSR} \times \text{LC}) - \text{Zero Error}$$
### Core Logic
Since the zero of the vernier scale is to the right of the main scale zero, there is a positive zero error. We must subtract this zero error from the observed reading.
### Step 1: Calculate Zero Error
Zero error $= +(4 \times 0.1 \mathrm{~mm}) = +0.4 \mathrm{~mm}$
### Step 2: Calculate Final Reading
Observed Reading $= 15 \mathrm{~mm} + (5 \times 0.1 \mathrm{~mm}) = 15.5 \mathrm{~mm}$
True Reading $= \text{Observed Reading} - \text{Zero Error}$
$= 15.5 \mathrm{~mm} - 0.4 \mathrm{~mm}$
$= 15.1 \mathrm{~mm}$
### Pattern Recognition
Vernier zero right of main zero $\rightarrow$ positive error $\rightarrow$ subtract it. Vernier zero left of main zero $\rightarrow$ negative error $\rightarrow$ add it.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Q32 (2026)
In an experiment, a set of reading are obtained $-1.24\mathrm{~mm}$ , $1.25\mathrm{~mm}$ , $1.23\mathrm{~mm}$ , $1.21\mathrm{~mm}$ . The expected least count of the instrument used in recording these readings is ____ mm.
- $0.01$
- $0.001$
- $0.1$
- $0.05$
### Core Logic
The given readings are $1.24 \text{ mm}$, $1.25 \text{ mm}$, $1.23 \text{ mm}$, $1.21 \text{ mm}$.
Each of these readings is recorded up to two decimal places in millimeters.
### Step 1: Determine Least Count
The least count of an instrument represents the smallest value that it can measure accurately, which is typically indicated by the decimal precision of the recorded data.
Since the smallest variation possible between readings written to two decimal places is $0.01 \text{ mm}$ (e.g., the difference between $1.24$ and $1.23$), the least count must be $0.01 \text{ mm}$.
### Pattern Recognition
Look at the decimal places. If a measurement is $X.YZ$, the least count is generally the place value of the last significant digit, i.e., $0.01$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Q44 (2026)
The time period of a simple harmonic oscillator is $T = 2\pi \sqrt{\frac{k}{m}}$ . The measured value of mass (m) of the object is $10 \text{ g}$ with an accuracy of $10 \text{ mg}$, and time for 50 oscillations of the spring is found to be $60 \text{ s}$ using a watch of $2 \text{ s}$ resolution. Percentage error in determination of spring constant(k) is ____%.
- $3.43$
- $3.35$
- $7.60$
- $6.76$
### Related Formula
$$T = 2\pi \sqrt{\frac{m}{k}}$$
*(Note: The question states $T = 2\pi \sqrt{k/m}$, which is dimensionally incorrect. Assuming it meant $T = 2\pi \sqrt{m/k}$ or just working with the error propagation mathematically.)*
Regardless, rearranging for $k$ gives $k \propto \frac{m}{T^2}$ or $k \propto mT^2$.
Using standard error analysis logic for multiplication/division:
$$\frac{\Delta k}{k} = \frac{\Delta m}{m} + 2\frac{\Delta T}{T}$$
### Step 1: Extract Given Data
$m = 10 \text{ g} = 10 \times 10^{-3} \text{ kg}$
$\Delta m = 10 \text{ mg} = 10 \times 10^{-3} \text{ g}$
Total time for $N=50$ oscillations is $t = 60 \text{ s}$.
Resolution $\Delta t = 2 \text{ s}$.
Time period $T = \frac{t}{N} = \frac{60}{50} = 1.2 \text{ s}$.
Error in Time period $\Delta T = \frac{\Delta t}{N} = \frac{2}{50} \text{ s}$.
### Step 2: Calculating Fractional Errors
$$\frac{\Delta T}{T} = \frac{2/50}{60/50} = \frac{2}{60}$$
Wait, the solution uses $\frac{2}{50 \times 1.2}$.
Let's apply it properly:
$$\frac{\Delta T}{T} = \frac{\Delta t / N}{t / N} = \frac{\Delta t}{t} = \frac{2}{60}$$
$$\frac{\Delta m}{m} = \frac{10 \times 10^{-3} \text{ g}}{10 \text{ g}} = 10^{-3} = 0.001$$
### Step 3: Calculating Percentage Error in K
$$\frac{\Delta k}{k} = 2 \left( \frac{2}{60} \right) + 0.001$$
$$\frac{\Delta k}{k} = \frac{4}{60} + 0.001 = 0.0666... + 0.001 = 0.0676$$
Percentage error:
$$\%\text{Error} = \frac{\Delta k}{k} \times 100 = 0.0676 \times 100 = 6.76\%$$
### Pattern Recognition
For total time $t = NT$, the relative error $\frac{\Delta T}{T}$ is exactly equal to $\frac{\Delta t}{t}$. Don't overcomplicate by dividing by $N$ separately for both, it cancels out.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Class 11 Physics: Oscillations
Q45 (2026)
Match 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. Coefficient of viscosity</td><td style="border: 1px solid #888; padding: 8px;">I. $[ML^{-1}T^{-2}]$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">B. Surface tension</td><td style="border: 1px solid #888; padding: 8px;">II. $[ML^{2}T^{-2}]$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">C. Pressure</td><td style="border: 1px solid #888; padding: 8px;">III. $[ML^{0}T^{-2}]$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">D. Surface energy</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 :
- $\text{A-I, B-II, C-IV, D-III}$
- $\text{A-IV, B-III, C-I, D-II}$
- $\text{A-I, B-III, C-II, D-IV}$
- $\text{A-IV, B-I, C-II, D-III}$
### Core Logic
(A) Coefficient of viscosity ($\eta$):
$$F = \eta A \frac{dv}{dx} \Rightarrow \eta = \frac{F}{A \frac{dv}{dx}}$$
$$[\eta] = \frac{[MLT^{-2}]}{[L^2][T^{-1}]} = [ML^{-1}T^{-1}] \Rightarrow \text{Matches with IV}$$
(B) Surface tension ($S$):
$$S = \frac{F}{L}$$
$$[S] = \frac{[MLT^{-2}]}{[L]} = [MT^{-2}] = [ML^0T^{-2}] \Rightarrow \text{Matches with III}$$
(C) Pressure ($P$):
$$P = \frac{F}{A}$$
$$[P] = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}] \Rightarrow \text{Matches with I}$$
(D) Surface energy ($E$):
Energy has the dimensions of work.
$$[E] = [ML^2T^{-2}] \Rightarrow \text{Matches with II}$$
### Step 1: Final Match
A $\rightarrow$ IV
B $\rightarrow$ III
C $\rightarrow$ I
D $\rightarrow$ II
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Class 11 Physics: Mechanical Properties of Fluids
