A physical quantity Q$Q$ is related to four observables a, b, c, d$a, b, c, d$ as follows: Q = fracab^4cd$Q = \frac{ab^4}{cd}$ where, a = (60 pm 3)mathrm~Pa$a = (60 \pm 3)\mathrm{~Pa}$ ; b = (20 pm 0.1)mathrm~m$b = (20 \pm 0.1)\mathrm{~m}$ ; c = (40 pm 0.2)mathrm~Nsm^-2$c = (40 \pm 0.2)\mathrm{~Nsm}^{-2}$ and d = (50 pm 0.1)mathrm~m$d = (50 \pm 0.1)\mathrm{~m}$ , then the percentage error in Q$Q$ is fracx1000$\frac{x}{1000}$ , where x = $x = $ ______.

### 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