Units and Measurements Previous Year Questions — JEE Main Physics
50 past-year Units and Measurements questions from JEE Main (Physics) — page 7.
Q44 (2024)
If mass is written as $m = k c^p G^{-1/2} h^{1/2}$ then the value of $P$ will be: (Constants have their usual meaning with $k$ a dimensionless constant)
- $1/2$
- $1 / 3$
- $2$
- $-1 / 3$
### Related Formula
$$[m] = [M]^1 [L]^0 [T]^0$$
$$[c] = [L T^{-1}]$$
$$[G] = [M^{-1} L^3 T^{-2}]$$
$$[h] = [M L^2 T^{-1}]$$
### Core Logic
By applying the principle of dimensional homogeneity, the dimensions on both sides of the equation must be identical.
$$[M]^1 [L]^0 [T]^0 = [L T^{-1}]^p [M^{-1} L^3 T^{-2}]^{-1/2} [M L^2 T^{-1}]^{1/2}$$
### Step 1: Substitute Dimensions
$$[M] = L^p T^{-p} \cdot M^{1/2} L^{-3/2} T^1 \cdot M^{1/2} L^1 T^{-1/2}$$
### Step 2: Collect Powers of L
Equating the powers of $[L]$ on both sides:
$$0 = p - \frac{3}{2} + 1$$
$$0 = p - \frac{1}{2}$$
$$p = \frac{1}{2}$$
### Pattern Recognition
The expression $m \propto \sqrt{hc/G}$ is a known fundamental relation representing the Planck mass. The exponent on $c$ inside the square root gives $p = 1/2$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Q31 (2024)
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. $[M L^2T^{-2}]$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">B. Surface Tension</td><td style="border: 1px solid #888; padding: 8px;">II. $[M L^2T^{-1}]$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">C. Angular momentum</td><td style="border: 1px solid #888; padding: 8px;">III. $[M L^{-1}T^{-1}]$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">D. Rotational kinetic energy</td><td style="border: 1px solid #888; padding: 8px;">IV. $[M L^0T^{-2}]$</td></tr></tbody></table></div>
- $\text{A-II, B-I, C-IV, D-III}$
- $\text{A-I, B-II, C-III, D-IV}$
- $\text{A-III, B-IV, C-II, D-I}$
- $\text{A-IV, B-III, C-II, D-I}$
### Related Formula
$$F = \eta A \frac{dv}{dy}$$
$$\text{Surface Tension} = \frac{F}{l}$$
$L = mvr$
$$K.E = \frac{1}{2} I \omega^2$$
### Core Logic
Let us determine the dimensional formula for each quantity sequentially:
**A. Coefficient of viscosity ($\eta$):**
Using $F = \eta A \frac{dv}{dy}$, we have:
$$[M L T^{-2}] = \eta [L^2] [T^{-1}]$$
$$\eta = [M L^{-1} T^{-1}] \Rightarrow \text{(III)}$$
**B. Surface Tension (S.T.):**
$$\text{S.T} = \frac{F}{\ell} = \frac{[M L T^{-2}]}{[L]} = [M L^0 T^{-2}] \Rightarrow \text{(IV)}$$
**C. Angular momentum (L):**
$$L = mvr = [M] [L T^{-1}] [L] = [M L^2 T^{-1}] \Rightarrow \text{(II)}$$
**D. Rotational kinetic energy (K.E.):**
$$\text{K.E} = \frac{1}{2} I \omega^2 = [M L^2 T^{-2}] \Rightarrow \text{(I)}$$
### Step 1: Final Matching
Matching the derived dimensional formulas:
A $\rightarrow$ III
B $\rightarrow$ IV
C $\rightarrow$ II
D $\rightarrow$ I
### Pattern Recognition
Kinetic energy (whether translational or rotational) always carries the dimension of Work: $[M L^2 T^{-2}]$. Surface tension is force per unit length, dropping the $L$ term. Viscosity commonly includes $L^{-1}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Q33 (2024)
The measured value of the length of a simple pendulum is $20 \text{ cm}$ with $2 \text{ mm}$ accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is $N\%$. The value of $N$ is:
- $4$
- $8$
- $6$
- $5$
### Related Formula
$$T = 2\pi \sqrt{\frac{\ell}{g}} \implies g = \frac{4\pi^2 \ell}{T^2}$$
### Core Logic
By taking logarithms and differentiating to find relative error (accuracy):
$$\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2\frac{\Delta T}{T}$$
### Step 1: Extrapolating Errors
Given values:
$\ell = 20 \text{ cm} = 200 \text{ mm}$
$\Delta \ell = 2 \text{ mm}$
$T_{\text{total}} = 40 \text{ s}$ for 50 oscillations
$\Delta T_{\text{total}} = 1 \text{ s}$
Note: The relative error in time period $T$ is equal to the relative error in total time $t$: $\frac{\Delta T}{T} = \frac{\Delta t}{t}$.
### Step 2: Substitution
$$\frac{\Delta g}{g} = \frac{0.2 \text{ cm}}{20 \text{ cm}} + 2 \left(\frac{1 \text{ s}}{40 \text{ s}}\right)$$
$$\frac{\Delta g}{g} = \frac{2}{200} + \frac{2}{40}$$
$$\frac{\Delta g}{g} = \frac{1}{100} + \frac{5}{100} = \frac{6}{100}$$
### Step 3: Percentage Conversion
Percentage change $= \frac{\Delta g}{g} \times 100\% = \frac{6}{100} \times 100\% = 6\%$.
Thus, $N = 6$.
### Pattern Recognition
For pendulum gravity error, always use $\%g = \%\ell + 2(\%T)$. Remember that measuring 50 oscillations reduces absolute error on a single swing, but the relative error $\Delta t / t$ remains unchanged whether you use total time or single period.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Class 11 Physics: Oscillations
Q50 (2024)
Consider two physical quantities A and B related to each other as $E = \frac{B - x^2}{At}$ where E, x and t have dimensions of energy, length and time respectively. The dimension of AB is
- $\text{L}^{-2}\text{M}^{1}\text{T}^{0}$
- $\text{L}^2\text{M}^{-1}\text{T}^1$
- $\text{L}^{-2}\text{M}^{-1}\text{T}^{1}$
- $\text{L}^0\text{M}^{-1}\text{T}^1$
### Related Formula
By the Principle of Homogeneity, terms added or subtracted must have the same dimensions:
$[B] = [x^2]$
### Core Logic
Known dimensional formulas:
Length $x \to [L]$
Energy $E \to [ML^2T^{-2}]$
Time $t \to [T]$
### Step 1: Dimension of B
Since $x^2$ is subtracted from $B$:
$$[B] = [x^2] = [L^2]$$
### Step 2: Dimension of A
From the equation $E = \frac{B - x^2}{At}$:
$$[A] = \frac{[B - x^2]}{[E][t]}$$
$$[A] = \frac{[L^2]}{[ML^2T^{-2}][T]} = \frac{[L^2]}{[ML^2T^{-1}]}$$
$$[A] = [M^{-1}T^1]$$
### Step 3: Dimension of AB
$$[AB] = [A] \times [B]$$
$$[AB] = [M^{-1}T^1] \times [L^2]$$
$$[AB] = [L^2 M^{-1} T^1]$$
### Pattern Recognition
Identify sums/differences first to instantly isolate $B$. Once $[B]$ is fixed, the entire numerator is just $L^2$. Swap out variables to isolate $[A]$. Combining is just standard exponent addition.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Q1989 (2024)
If the percentage errors in measuring the length and the diameter of a wire are $0.1\%$ each. The percentage error in measuring its resistance will be:
- $0.2\%$
- $0.3\%$
- $0.1\%$
- $0.144\%$
### Related Formula
$$R = \frac{\rho L}{A} = \frac{\rho L}{\pi \left(\frac{d}{2}\right)^2} = \frac{4\rho L}{\pi d^2}$$
### Core Logic
To find the maximum percentage error in resistance, apply logarithmic differentiation:
$$\frac{\Delta R}{R} = \frac{\Delta L}{L} + 2\frac{\Delta d}{d}$$
Given percentage errors:
* $\frac{\Delta L}{L} \times 100\% = 0.1\%$
* $\frac{\Delta d}{d} \times 100\% = 0.1\%$
### Step 2: Substitution
Substituting the values:
$$\frac{\Delta R}{R} \times 100\% = 0.1\% + 2(0.1\%) = 0.1\% + 0.2\% = 0.3\%$$
### Pattern Recognition
Resistance scales inversely with the square of the diameter. The error multiplier for diameter is 2. Sum the linear components directly: $\text{Error} = L_{\text{error}} + 2 \times d_{\text{error}}$.
### Chapter Mix
Class 12 Physics: Current Electricity
