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For the determination of refractive index of glass slab, a travelling microscope is used whose main scale contains 300 equal divisions equals to 15 cm. The vernier scale attached to the microscope has 25 divisions equals to 24 divisions of main scale. The least count (LC) of the travelling microscope is (in cm):

Solution & Explanation

### Related Formula 1text MSD = fractextTotal LengthtextTotal Divisions textLeast Count (LC) = 1text MSD - 1text VSD = 1text MSD cdot left(1 - frac2425right) ### Core Logic Calculate the value of one main scale division: 1text MSD = frac15text cm300 = 0.05text cm Given that 25text VSD = 24text MSD, we find: 1text VSD = frac2425text MSD ### Step 1: Compute Least Count textLC = 1text MSD cdot left(frac125right) = frac0.05text cm25 = 0.002text cm ### Pattern Recognition Least count calculation is conventionally textLC = frac1text MSDN where N represents the total count of divisions on the vernier scale whenever (N-1)text MSD = Ntext VSD. ### 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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