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For an experimental expression y=frac32.3times112527.4 , where all the digits are significant. Then to report the value of y we should write :-

Solution & Explanation

### Related Formula In multiplication and division arithmetic rules, the final product or quotient must be rounded off to retain as many significant figures as are present in the least precise operand. ### Core Logic Let us check the significant digit count of the operands in the expression : * 32.3 has 3 significant figures. * 1125 has 4 significant figures. * 27.4 has 3 significant figures. The minimum number of significant figures among the numbers is 3. ### Step 1: Rounding Off Direct calculation yield : y = 1326.186... Rounding this value off to contain exactly 3 significant figures means changing it to 1330, since the digit after 2 is 6 (which is greater than 5), updating the hundreds spot upwards. ### Pattern Recognition Never keep unearned precision from automated calculation. The output is bounded strictly by your least precise entry. ### 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

More Units and Measurements Questions — jee_main_2025_24_jan_morning

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