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Consider two physical quantities A and B related to each other as E = fracB - x^2At where E, x and t have dimensions of energy, length and time respectively. The dimension of AB is

Solution & Explanation

### 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 = fracB - x^2At: [A] = frac[B - x^2][E][t] [A] = frac[L^2][ML^2T^-2][T] = frac[L^2][ML^2T^-1] [A] = [M^-1T^1] ### Step 3: Dimension of AB [AB] = [A] times [B] [AB] = [M^-1T^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

Reference Study Guides

Units And Measurements

More Units and Measurements Previous-Year Questions — Page 5

Q jee_main_2025_29_jan_morning Dimensional Analysis
The pair of physical quantities not having same dimensions is :
  • A. textTorque and energy
  • B. textSurface tension and impulse
  • C. textAngular momentum and Planck\'s constant
  • D. textPressure and Young\'s modulus

Solution

### Core Logic Let\'s check the dimensions of each pair : * textTorque = [textEnergy] = [ML^2T^-2] * textSurface Tension = [MT^-2] vs textImpulse = [MLT^-1] * [textAngular Momentum] = [textPlanck\'s Constant] = [ML^2T^-1] * [textPressure] = [textYoung\'s Modulus] = [ML^-1T^-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
Q jee_main_2025_29_jan_morning Dimensional Homogeneity
The expression given below shows the variation of velocity (v) with time (t), v = At^2 + fracBtC + t . The dimension of ABC is:
  • A. left[mathrmM^0 mathrm~L^2 mathrmT^-3right]
  • B. left[mathrmM^0 mathrm~L^1 mathrmT^-3right]
  • C. [mathrmM^0mathrmL^1mathrmT^-2]
  • D. left[mathrmM^0 mathrm~L^2 mathrmT^-2right]

Solution

### Related Formula [v] = [At^2] = left[fracBtC+tright] ### 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[fracBtTright] = [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 jee_main_2024_01_february_morning Vernier Calliper
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:
  • A. frac12
  • B. frac1011
  • C. frac5011
  • D. frac511

Solution

### Related Formula Least Count (LC) definition: textLC = 1text MSD - 1text VSD ### Core Logic From the problem: 10text MSD = 11text VSD implies 1text VSD = frac1011text MSD Substitute into the Least Count formula: textLC = 1text MSD - frac1011text MSD = frac111text MSD ### Step 1: Calculate with Main Scale Units Given that 1text MSD = 5text units: textLC = frac111 times 5 = frac511text units ### Pattern Recognition Watch out for unconventional scale arrangements where textVSD > textMSD. The fundamental difference formula 1text MSD - 1text VSD safely determines magnitudes without needing sign corrections. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements
Q44 jee_main_2024_01_february_morning Error Analysis
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:
  • A. 25.6%
  • B. 39.9%
  • C. 37.3%
  • D. 35.6%

Solution

### Related Formula Resistivity formula: rho = RfracAl = Rfracpi r^2l Maximum relative error equation: fracDeltarhorho = fracDelta RR + 2fracDelta rr + fracDelta ll ### Core Logic Substitute the measured fractions: fracDeltarhorho = frac10100 + 2left(frac0.050.35right) + frac0.215 fracDeltarhorho = frac110 + 2left(frac17right) + frac175 fracDeltarhorho = 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 jee_main_2024_01_february_morning Dimensional Analysis
The dimensional formula of angular impulse is:
  • A. [mathrmM mathrmL^-2mathrmT^-1]
  • B. [mathrmM mathrmL^2mathrmT^-2]
  • C. [mathrmM mathrmL mathrmT^-1]
  • D. [mathrmM mathrmL^2mathrmT^-1]

Solution

### Related Formula Angular Impulse equation: textAngular Impulse = int tau \, dt = Delta L textAngular 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] = [mathrmM] [v] = [mathrmLmathrmT^-1] [r] = [mathrmL] ### Step 1: Combine Dimensions [textAngular Impulse] = [mathrmM] times [mathrmLmathrmT^-1] times [mathrmL] = [mathrmMmathrmL^2mathrmT^-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

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