### Related Formula Let momentum be P$P$ ([P] = [M L T^-1]$[P] = [M L T^{-1}]$). We assume: P = q^x mu_0^y I^z$P = q^x \mu_0^y I^z$ ### Core Logic Let's find the dimensional formulas of the individual variables: 1. **Charge (q$q$):** [q] = [A T]$[q] = [A T]$ 2. **Current (I$I$):** [I] = [A]$[I] = [A]$ 3. **Permeability of vacuum (mu_0$\mu_0$):** From Biot-Savart law or force between parallel wires: F = fracmu_0 I^2 L2pi d$F = \frac{\mu_0 I^2 L}{2\pi d}$: [mu_0] = frac[F][I]^2 = frac[M L T^-2][A]^2 = [M L T^-2 A^-2]$[\mu_0] = \frac{[F]}{[I]^2} = \frac{[M L T^{-2}]}{[A]^2} = [M L T^{-2} A^{-2}]$ ### Step 1: Apply Dimensional Homogeneity Substitute these into our assumed dimensional equation: [M L T^-1] = [A T]^x [M L T^-2 A^-2]^y [A]^z$[M L T^{-1}] = [A T]^x [M L T^{-2} A^{-2}]^y [A]^z$ [M L T^-1] = [M]^y [L]^y [T]^x - 2y [A]^x - 2y + z$[M L T^{-1}] = [M]^y [L]^y [T]^{x - 2y} [A]^{x - 2y + z}$ Comparing exponents on both sides: - For [M]$[M]$: y = 1$y = 1$ - For [L]$[L]$: y = 1 quad text(consistent)$y = 1 \quad \text{(consistent)}$ - For [T]$[T]$: x - 2y = -1 implies x - 2(1) = -1 implies x = 1$x - 2y = -1 \implies x - 2(1) = -1 \implies x = 1$ - For [A]$[A]$: x - 2y + z = 0 implies 1 - 2(1) + z = 0 implies z = 1$x - 2y + z = 0 \implies 1 - 2(1) + z = 0 \implies z = 1$ Thus, x = 1, y = 1, z = 1$x = 1, y = 1, z = 1$. Therefore, the required quantity is: q^1 mu_0^1 I^1 = q mu_0 I$q^1 \mu_0^1 I^1 = q \mu_0 I$ ### Pattern Recognition Sees: Permeability, charge, and current linked to momentum. Trap: Deriving the dimensions of mu_0$\mu_0$ using complex magnetic formulas. Remember [mu_0] = [textForce]/[textCurrent]^2$[\mu_0] = [\text{Force}]/[\text{Current}]^2$ is the quickest way to get its dimensions. Shortcut: Since [q] = AT$[q] = AT$ and [I] = A$[I] = A$, [q mu_0 I] = [A T] [M L T^-2 A^-2] [A] = [M L T^-1]$[q \mu_0 I] = [A T] [M L T^{-2} A^{-2}] [A] = [M L T^{-1}]$, which is exactly the dimensions of momentum. Hence, option (2) is correct. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 12 Physics: Moving Charges and Magnetism

### Related Formula Definitions of physical quantities: 1. Viscosity: F = -eta A fracdvdx$F = -\eta A \frac{dv}{dx}$ 2. Intensity: I = fractextPowertextArea$I = \frac{\text{Power}}{\text{Area}}$ 3. Pressure gradient: fracdPdx$\frac{dP}{dx}$ 4. Compressibility: K = frac1B = fractextStraintextStress$K = \frac{1}{B} = \frac{\text{Strain}}{\text{Stress}}$ ### Core Logic Let's calculate each dimensional formula: 1. **Coefficient of viscosity (eta$\eta$):** [eta] = frac[F][A]left[fracdvdxright] = fractextM L T^-2textL^2 left(fractextL T^-1textLright) = textM L^-1textT^-1$[\eta] = \frac{[F]}{[A]\left[\frac{dv}{dx}\right]} = \frac{\text{M L T}^{-2}}{\text{L}^2 \left(\frac{\text{L T}^{-1}}{\text{L}}\right)} = \text{M L}^{-1}\text{T}^{-1}$ Matches **(IV)**. 2. **Intensity of wave (I$I$):** [I] = frac[textPower][textArea] = fractextM L^2textT^-3textL^2 = textM T^-3 = textM L^0textT^-3$[I] = \frac{[\text{Power}]}{[\text{Area}]} = \frac{\text{M L}^2\text{T}^{-3}}{\text{L}^2} = \text{M T}^{-3} = \text{M L}^0\text{T}^{-3}$ Matches **(I)**. 3. **Pressure gradient (fracdPdx$\frac{dP}{dx}$):** left[fracdPdxright] = frac[textPressure][textLength] = fractextM L^-1textT^-2textL = textM L^-2textT^-2$\left[\frac{dP}{dx}\right] = \frac{[\text{Pressure}]}{[\text{Length}]} = \frac{\text{M L}^{-1}\text{T}^{-2}}{\text{L}} = \text{M L}^{-2}\text{T}^{-2}$ Matches **(II)**. 4. **Compressibility (K$K$):** Compressibility is the inverse of Bulk Modulus: [K] = frac1[textPressure] = frac1textM L^-1textT^-2 = textM^-1textLtextT^2$[K] = \frac{1}{[\text{Pressure}]} = \frac{1}{\text{M L}^{-1}\text{T}^{-2}} = \text{M}^{-1}\text{L}\text{T}^2$ Matches **(III)**. Thus: (A)-(IV), (B)-(I), (C)-(II), (D)-(III). ### Step 1: Final Conclusion The correct option is (2). ### Pattern Recognition Target basic matching terms first: Compressibility is the reciprocal of pressure, giving [M^-1LT^2]$[M^{-1}LT^2]$. Wave intensity has units of power per unit area, giving [MT^-3]$[MT^{-3}]$. This immediately isolates option (2). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 11 Physics: Mechanical Properties of Fluids Class 11 Physics: Mechanical Properties of Solids

### Related Formula By the principle of dimensional homogeneity, terms added or subtracted must have the same dimensions: [P] = left[fracaV^2right] implies [a] = [P][V]^2$[P] = \left[\frac{a}{V^2}\right] \implies [a] = [P][V]^2$ [V] = [b]$[V] = [b]$ ### Core Logic Let's find the dimensional formula of the quantities: - Pressure P$P$: [P] = textM L^-1textT^-2$[P] = \text{M L}^{-1}\text{T}^{-2}$ - Volume V$V$: [V] = textL^3$[V] = \text{L}^3$ Substituting these to find [a]$[a]$ and [b]$[b]$: [a] = (textM L^-1textT^-2)(textL^6) = textM L^5textT^-2$[a] = (\text{M L}^{-1}\text{T}^{-2})(\text{L}^6) = \text{M L}^5\text{T}^{-2}$ [b] = textL^3 implies [b^-2] = textL^-6$[b] = \text{L}^3 \implies [b^{-2}] = \text{L}^{-6}$ Now, compute the dimensions of ab^-2$ab^{-2}$: [ab^-2] = (textM L^5textT^-2)(textL^-6) = textM L^-1textT^-2$[ab^{-2}] = (\text{M L}^5\text{T}^{-2})(\text{L}^{-6}) = \text{M L}^{-1}\text{T}^{-2}$ This matches the dimensions of pressure. Let's evaluate the options: 1. Planck's constant: [h] = textM L^2textT^-1$[h] = \text{M L}^2\text{T}^{-1}$ 2. Compressibility: [beta] = textM^-1textLtextT^2$[\beta] = \text{M}^{-1}\text{L}\text{T}^2$ 3. Strain: dimensionless 4. Energy density (energy per unit volume): left[fracEVright] = fractextM L^2textT^-2textL^3 = textM L^-1textT^-2$\left[\frac{E}{V}\right] = \frac{\text{M L}^2\text{T}^{-2}}{\text{L}^3} = \text{M L}^{-1}\text{T}^{-2}$ ### Step 1: Final Conclusion Therefore, the dimension of ab^-2$ab^{-2}$ is equivalent to that of Energy density. ### Pattern Recognition By writing the relation directly as [ab^-2] = frac[a][b]^2$[ab^{-2}] = \frac{[a]}{[b]^2}$, and noting [a] = [P][V]^2$[a] = [P][V]^2$ and [b] = [V]$[b] = [V]$, we get [ab^-2] = frac[P][V]^2[V]^2 = [P]$[ab^{-2}] = \frac{[P][V]^2}{[V]^2} = [P]$ (Pressure). Since pressure and energy density have identical dimensions, the answer is immediately Energy density. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 11 Physics: Kinetic Theory of Gases

### Related Formula Formulas to find dimensional formulas: - Boltzmann constant: k_B = fractextEnergytextTemperature$k_B = \frac{\text{Energy}}{\text{Temperature}}$ - Coefficient of viscosity: eta = fracFA fracdvdx$\eta = \frac{F}{A \frac{dv}{dx}}$ - Planck's constant: h = fracEnu$h = \frac{E}{\nu}$ - Thermal conductivity: fracdQdt = K A fracdTdx Rightarrow K = fractextHeat flow cdot textthicknesstextArea cdot textTemperature difference$\frac{dQ}{dt} = K A \frac{dT}{dx} \Rightarrow K = \frac{\text{Heat flow} \cdot \text{thickness}}{\text{Area} \cdot \text{Temperature difference}}$ ### Core Logic Evaluate each constant individually: ### Step 1: Dimensions of Boltzmann constant (k_B$k_B$) [k_B] = frac[ML^2T^-2][K] = [ML^2T^-2K^-1] quad Rightarrow textMatches III$[k_B] = \frac{[ML^2T^{-2}]}{[K]} = [ML^2T^{-2}K^{-1}] \quad \Rightarrow \text{Matches III}$ ### Step 2: Dimensions of Coefficient of viscosity (eta$\eta$) [eta] = frac[MLT^-2][L^2] [T^-1] = [ML^-1T^-1] quad Rightarrow textMatches IV$[\eta] = \frac{[MLT^{-2}]}{[L^2] [T^{-1}]} = [ML^{-1}T^{-1}] \quad \Rightarrow \text{Matches IV}$ ### Step 3: Dimensions of Planck's constant (h$h$) [h] = frac[ML^2T^-2][T^-1] = [ML^2T^-1] quad Rightarrow textMatches I$[h] = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}] \quad \Rightarrow \text{Matches I}$ ### Step 4: Dimensions of Thermal conductivity (K$K$) [K] = frac[ML^2T^-3] [L][L^2] [K] = [MLT^-3K^-1] quad Rightarrow textMatches II$[K] = \frac{[ML^2T^{-3}] [L]}{[L^2] [K]} = [MLT^{-3}K^{-1}] \quad \Rightarrow \text{Matches II}$ This sequence yields A-III, B-IV, C-I, D-II, matching Option (1). ### Pattern Recognition To solve matching sets efficiently, search for the most recognizable dimensions first. Planck's constant h$h$ (ML^2T^-1$ML^2T^{-1}$) and viscosity coefficient eta$\eta$ (ML^-1T^-1$ML^{-1}T^{-1}$) are highly unique and usually resolve the options instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

### Related Formula For a physical quantity defined by algebraic powers C = fracp^a q^br^c s^d$C = \frac{p^a q^b}{r^c s^d}$, the maximum fractional error is calculated by summing absolute scaled fractional errors: fracDelta CC = a fracDelta pp + b fracDelta qq + c fracDelta rr + d fracDelta ss$\frac{\Delta C}{C} = a \frac{\Delta p}{p} + b \frac{\Delta q}{q} + c \frac{\Delta r}{r} + d \frac{\Delta s}{s}$ Expressed as percentages: \% text error in C = a(\% text error in p) + b(\% text error in q) + c(\% text error in r) + d(\% text error in s)$\% \text{ error in } C = a(\% \text{ error in } p) + b(\% \text{ error in } q) + c(\% \text{ error in } r) + d(\% \text{ error in } s)$ ### Core Logic Given expression: C = p^1 q^2 r^-3 s^-1/2$C = p^1 q^2 r^{-3} s^{-1/2}$ Max fractional error equation: fracDelta CC = 1 left(fracDelta ppright) + 2 left(fracDelta qqright) + 3 left(fracDelta rrright) + frac12 left(fracDelta ssright)$\frac{\Delta C}{C} = 1 \left(\frac{\Delta p}{p}\right) + 2 \left(\frac{\Delta q}{q}\right) + 3 \left(\frac{\Delta r}{r}\right) + \frac{1}{2} \left(\frac{\Delta s}{s}\right)$ ### Step 1: Calculate the total percentage error Substitute the individual percentage errors: - Error in p = 1\%$p = 1\%$ - Error in q = 2\%$q = 2\%$ - Error in r = 3\%$r = 3\%$ - Error in s = 2\%$s = 2\%$ \% text error in C = 1(1\%) + 2(2\%) + 3(3\%) + frac12(2\%)$\% \text{ error in } C = 1(1\%) + 2(2\%) + 3(3\%) + \frac{1}{2}(2\%)$ \% text error in C = 1\% + 4\% + 9\% + 1\% = 15\%$\% \text{ error in } C = 1\% + 4\% + 9\% + 1\% = 15\%$ The total percentage error in C$C$ is 15\%$15\%$. ### Pattern Recognition In error propagation, individual errors always combine constructively to produce the maximum possible uncertainty limit. Hence, negative powers (like division by r^3$r^3$ or s^1/2$s^{1/2}$) are integrated using positive coefficients during maximum absolute error summation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements