Units and Measurements Previous Year Questions — JEE Main Physics

50 past-year Units and Measurements questions from JEE Main (Physics) — page 5.

Q1279 (2025)

The pair of physical quantities not having same dimensions is :
  1. $\text{Torque and energy}$
  2. $\text{Surface tension and impulse}$
  3. $\text{Angular momentum and Planck\'s constant}$
  4. $\text{Pressure and Young\'s modulus}$
### Core Logic Let\'s check the dimensions of each pair : * $\text{Torque} = [\text{Energy}] = [ML^2T^{-2}]$ * $\text{Surface Tension} = [MT^{-2}]$ vs $\text{Impulse} = [MLT^{-1}]$ * $[\text{Angular Momentum}] = [\text{Planck\'s Constant}] = [ML^2T^{-1}]$ * $[\text{Pressure}] = [\text{Young\'s Modulus}] = [ML^{-1}T^{-2}]$ ### Step 1: Identify Non-Matching Pair Surface tension and impulse do not share matching dimension frameworks. ### Pattern Recognition Surface tension is force per unit length ($[MT^{-2}]$), while impulse is force times time ($[MLT^{-1}]$) ### Chapter Mix Class 11 Physics: Units and Measurements

Q1281 (2025)

The expression given below shows the variation of velocity (v) with time (t), $v = At^2 + \frac{Bt}{C + t}$ . The dimension of ABC is:
  1. $\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{T}^{-3}\right]$
  2. $\left[\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{T}^{-3}\right]$
  3. $[\mathrm{M}^0\mathrm{L}^1\mathrm{T}^{-2}]$
  4. $\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{T}^{-2}\right]$
### Related Formula $$[v] = [At^2] = \left[\frac{Bt}{C+t}\right]$$ ### Core Logic By the principle of dimensional homogeneity: 1. $[C] = [t] = [T]$ 2. $[At^2] = [v] \implies [A][T^2] = [LT^{-1}] \implies [A] = [LT^{-3}]$ 3. $\left[\frac{Bt}{T}\right] = [v] \implies [B] = [LT^{-1}]$ ### Step 1: Calculate Dimensions of ABC $$[ABC] = [LT^{-3}] \cdot [LT^{-1}] \cdot [T] = [L^2 T^{-3}]$$ ### Pattern Recognition Denominator terms matched first give $C$, tracking linear velocity units sets the balance for $A$ and $B$ ### Chapter Mix Class 11 Physics: Units and Measurements

Q42 (2024)

10 divisions on the main scale of a Vernier calliper coincide with 11 divisions on the Vernier scale. If each division on the main scale is of 5 units, the least count of the instrument is:
  1. $\frac{1}{2}$
  2. $\frac{10}{11}$
  3. $\frac{50}{11}$
  4. $\frac{5}{11}$
### Related Formula Least Count (LC) definition: $$\text{LC} = 1\text{ MSD} - 1\text{ VSD}$$ ### Core Logic From the problem: $$10\text{ MSD} = 11\text{ VSD} \implies 1\text{ VSD} = \frac{10}{11}\text{ MSD}$$ Substitute into the Least Count formula: $$\text{LC} = 1\text{ MSD} - \frac{10}{11}\text{ MSD} = \frac{1}{11}\text{ MSD}$$ ### Step 1: Calculate with Main Scale Units Given that $1\text{ MSD} = 5\text{ units}$: $$\text{LC} = \frac{1}{11} \times 5 = \frac{5}{11}\text{ units}$$ ### Pattern Recognition Watch out for unconventional scale arrangements where $\text{VSD} > \text{MSD}$. The fundamental difference formula $1\text{ MSD} - 1\text{ VSD}$ safely determines magnitudes without needing sign corrections. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

Q44 (2024)

The radius ($r$), length ($l$) and resistance ($R$) of a metal wire was measured in the laboratory as: $$r = (0.35 \pm 0.05)\mathrm{~cm}$$ $$R = (100 \pm 10)\mathrm{~\Omega}$$ $$l = (15 \pm 0.2)\mathrm{~cm}$$ The percentage error in resistivity of the material of the wire is:
  1. 25.6%
  2. 39.9%
  3. 37.3%
  4. 35.6%
### Related Formula Resistivity formula: $$\rho = R\frac{A}{l} = R\frac{\pi r^2}{l}$$ Maximum relative error equation: $$\frac{\Delta\rho}{\rho} = \frac{\Delta R}{R} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}$$ ### Core Logic Substitute the measured fractions: $$\frac{\Delta\rho}{\rho} = \frac{10}{100} + 2\left(\frac{0.05}{0.35}\right) + \frac{0.2}{15}$$ $$\frac{\Delta\rho}{\rho} = \frac{1}{10} + 2\left(\frac{1}{7}\right) + \frac{1}{75}$$ $$\frac{\Delta\rho}{\rho} = 0.1 + 0.2857 + 0.0133 = 0.399$$ ### Step 1: Convert to Percentage $$\% severance = 0.399 \times 100\% = 39.9\%$$ ### Pattern Recognition The power of 2 in $r^2$ doubles its fractional error impact, making it the most dominant source of error in this setup. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 12 Physics: Current Electricity

Q45 (2024)

The dimensional formula of angular impulse is:
  1. $[\mathrm{M} \mathrm{L}^{-2}\mathrm{T}^{-1}]$
  2. $[\mathrm{M} \mathrm{L}^{2}\mathrm{T}^{-2}]$
  3. $[\mathrm{M} \mathrm{L} \mathrm{T}^{-1}]$
  4. $[\mathrm{M} \mathrm{L}^{2}\mathrm{T}^{-1}]$
### Related Formula Angular Impulse equation: $$\text{Angular Impulse} = \int \tau \, dt = \Delta L$$ $$\text{Angular Momentum } (L) = mvr$$ ### Core Logic By the impulse-momentum theorem for rotation, angular impulse equals the net change in angular momentum. Extract dimensions from $[mvr]$: $$[m] = [\mathrm{M}]$$ $$[v] = [\mathrm{L}\mathrm{T}^{-1}]$$ $$[r] = [\mathrm{L}]$$ ### Step 1: Combine Dimensions $$[\text{Angular Impulse}] = [\mathrm{M}] \times [\mathrm{L}\mathrm{T}^{-1}] \times [\mathrm{L}] = [\mathrm{M}\mathrm{L}^2\mathrm{T}^{-1}]$$ ### Pattern Recognition Angular impulse shares the exact same dimensional signature as Planck's constant ($h$) and angular momentum ($L$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 11 Physics: System of Particles and Rotational Motion
Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)

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