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):

### Related Formula The Least Count (LC) of a screw gauge tool is defined by: textL.C. = fractextPitchtextTotal Number of Circular Divisions (N) $$\text{L.C.} = \frac{\text{Pitch}}{\text{Total Number of Circular Divisions (N)}} $$ ### Core Logic The initial least count is given as [cite: 167, 788]: textL.C.textinitial = fracPN = 0.01text mm$$\text{L.C.}{\text{initial}} = \frac{P}{N} = 0.01\text{ mm}$$ Now, calculate the modified parameters from the text details : * New Pitch: P' = P(1 + 0.75) = 1.75P$P' = P(1 + 0.75) = 1.75P$ * New Divisions: N' = N(1 - 0.50) = 0.5N$N' = N(1 - 0.50) = 0.5N$ ### Step 1: Calculating the New Least Count Set up the updated least count expression ratio : textL.C.textnew = fracP'N' = frac1.75P0.5N = 3.5 times left(fracPN ight)$$\text{L.C.}{\text{new}} = \frac{P'}{N'} = \frac{1.75P}{0.5N} = 3.5 \times \left(\frac{P}{N} ight)$$ Substitute the initial least count value : textL.C.textnew = 3.5 times 0.01text mm = 0.035text mm$$\text{L.C.}{\text{new}} = 3.5 \times 0.01\text{ mm} = 0.035\text{ mm}$$ Converting into the requested scientific prefix units (10^-3text mm$10^{-3}\text{ mm}$) : textL.C.textnew = 35 times 10^-3text mm$$\text{L.C.}{\text{new}} = 35 \times 10^{-3}\text{ mm}$$ Therefore, the requested value is 35. ### Pattern Recognition Least count scales proportionally with pitch increases, and inversely with reductions in circular divisions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

### Related Formula * **Angular Impulse** = Change in Angular Momentum = tau cdot Delta t = [M L^2 T^-2] cdot [T] = [M L^2 T^-1]$\tau \cdot \Delta t = [M L^2 T^{-2}] \cdot [T] = [M L^2 T^{-1}]$ [cite: 670, 672] * **Latent Heat** (L$L$) = fracQm = frac[M L^2 T^-2][M] = [M^0 L^2 T^-2]$\frac{Q}{m} = \frac{[M L^2 T^{-2}]}{[M]} = [M^0 L^2 T^{-2}]$ * **Electrical resistivity** (\rho) = fracR cdot Al = frac[M L^2 T^-3 A^-2] cdot [L^2][L] = [M L^3 T^-3 A^-2]$\frac{R \cdot A}{l} = \frac{[M L^2 T^{-3} A^{-2}] \cdot [L^2]}{[L]} = [M L^3 T^{-3} A^{-2}]$ * **Electromotive force** (V$V$) = fracWq = frac[M L^2 T^-2][A T] = [M L^2 T^-3 A^-1]$\frac{W}{q} = \frac{[M L^2 T^{-2}]}{[A T]} = [M L^2 T^{-3} A^{-1}]$ ### Core Logic By comparing the formulas derived for each physical quantity with the options given in List-II [cite: 670, 673, 674]: * (A) matches with (III) [cite: 670, 672] * (B) matches with (I) * (C) matches with (IV) * (D) matches with (II) Hence, the correct matching is (A)-(III), (B)-(I), (C)-(IV), (D)-(II). ### Pattern Recognition In match-the-column dimensional analysis questions, identifying even one or two straightforward quantities like Latent Heat (L = Q/m$L = Q/m$) often immediately eliminates three incorrect options, securing a quick correct answer. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

### Core Logic Let\'s check the dimensions of each pair : * textTorque = [textEnergy] = [ML^2T^-2]$\text{Torque} = [\text{Energy}] = [ML^2T^{-2}]$ * textSurface Tension = [MT^-2]$\text{Surface Tension} = [MT^{-2}]$ vs textImpulse = [MLT^-1]$\text{Impulse} = [MLT^{-1}]$ * [textAngular Momentum] = [textPlanck\'s Constant] = [ML^2T^-1]$[\text{Angular Momentum}] = [\text{Planck\'s Constant}] = [ML^2T^{-1}]$ * [textPressure] = [textYoung\'s Modulus] = [ML^-1T^-2]$[\text{Pressure}] = [\text{Young\'s Modulus}] = [ML^{-1}T^{-2}]$ ### 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]$[MT^{-2}]$), while impulse is force times time ([MLT^-1]$[MLT^{-1}]$) ### Chapter Mix Class 11 Physics: Units and Measurements

### Related Formula [v] = [At^2] = left[fracBtC+tright]$$[v] = [At^2] = \left[\frac{Bt}{C+t}\right]$$ ### Core Logic By the principle of dimensional homogeneity: 1. [C] = [t] = [T]$[C] = [t] = [T]$ 2. [At^2] = [v] implies [A][T^2] = [LT^-1] implies [A] = [LT^-3]$[At^2] = [v] \implies [A][T^2] = [LT^{-1}] \implies [A] = [LT^{-3}]$ 3. left[fracBtTright] = [v] implies [B] = [LT^-1]$\left[\frac{Bt}{T}\right] = [v] \implies [B] = [LT^{-1}]$ ### Step 1: Calculate Dimensions of ABC [ABC] = [LT^-3] cdot [LT^-1] cdot [T] = [L^2 T^-3]$$[ABC] = [LT^{-3}] \cdot [LT^{-1}] \cdot [T] = [L^2 T^{-3}]$$ ### Pattern Recognition Denominator terms matched first give C$C$, tracking linear velocity units sets the balance for A$A$ and B$B$ ### Chapter Mix Class 11 Physics: Units and Measurements

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