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Match List-I with List-II.
List-IList-II
(A) Mass density(I) [ML^2T^-3]
(B) Impulse(II) [MLT^-1]
(C) Power(III) [ML^2T^0]
(D) Moment of inertia(IV) [ML^-3T^0]
Choose the correct answer from the options given below: [cite: 107]

Solution & Explanation

### Core Logic Let's derive the dimensional formulas systematically: * **(A) Mass density:** rho = fractextMasstextVolume = fracML^3 = [M^1 L^-3 T^0] implies text(IV) [cite: 765]. * **(B) Impulse:** I = F cdot Delta t = [M^1 L^1 T^-2] cdot [T] = [M^1 L^1 T^-1] implies text(II) [cite: 767]. * **(C) Power:** P = fractextWorktextTime = frac[M^1 L^2 T^-2][T] = [M^1 L^2 T^-3] implies text(I) [cite: 770]. * **(D) Moment of inertia:** I = M r^2 = [M^1 L^2 T^0] implies text(III) [cite: 772]. Matching all four pairings establishes the layout: (A)-(IV), (B)-(II), (C)-(I), (D)-(III)[cite: 110, 758]. ### Pattern Recognition Isolating the unique matching option for a straightforward parameter like Mass Density (A-IV) or Moment of Inertia (D-III) easily allows exclusion of multiple invalid answer branches instantly[cite: 765, 772]. ### 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

Q4 2025 Dimensional Analysis
The pair of physical quantities not having same dimensions is : [cite: 1, 5]
  • 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])[cite: 5, 630]. ### Chapter Mix Class 11 Physics: Units and Measurements
Q6 2025 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: [cite: 1, 5]
  • 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] [cite: 597, 598] 3. left[fracBtTright] = [v] implies [B] = [LT^-1] [cite: 597, 599] ### 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[cite: 597, 598]. ### Chapter Mix Class 11 Physics: Units and Measurements

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