Match List-I with List-II.

List-I	List-II
A. Coefficient of viscosity	I. [M L^2T^-2]$[M L^2T^{-2}]$
B. Surface Tension	II. [M L^2T^-1]$[M L^2T^{-1}]$
C. Angular momentum	III. [M L^-1T^-1]$[M L^{-1}T^{-1}]$
D. Rotational kinetic energy	IV. [M L^0T^-2]$[M L^0T^{-2}]$

### Related Formula K_texteq = K_1 + K_2 quad text(Parallel Combination)$$K_{\text{eq}} = K_1 + K_2 \quad \text{(Parallel Combination)}$$ Delta K_texteq = Delta K_1 + Delta K_2$$\Delta K_{\text{eq}} = \Delta K_1 + \Delta K_2$$ textPercentage Error = fracDelta K_texteqK_texteq times 100$$\text{Percentage Error} = \frac{\Delta K_{\text{eq}}}{K_{\text{eq}}} \times 100$$ ### Core Logic For a parallel combination of springs, the equivalent spring constant is simply the sum of individual constants. K_texteq = K_1 + K_2 = 10 + 20 = 30text N/m$$K_{\text{eq}} = K_{1} + K_{2} = 10 + 20 = 30\text{ N/m}$$ When quantities are added, their absolute errors are also added: Delta K_texteq = Delta K_1 + Delta K_2 = 0.2 + 0.3 = 0.5text N/m$$\Delta K_{\text{eq}} = \Delta K_{1} + \Delta K_{2} = 0.2 + 0.3 = 0.5\text{ N/m}$$ ### Step 1: Calculating Percentage Error The percentage error in K$K$ is: text% Error in K = frac0.530 times 100 = frac53\% approx 1.67\%$$\text{% Error in } K = \frac{0.5}{30} \times 100 = \frac{5}{3}\% \approx 1.67\%$$ ### Pattern Recognition Addition operation = add absolute errors. For parallel springs, it's just plain addition. Then to get percentage, divide the absolute error sum by the nominal value sum and multiply by 100. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 11 Physics: Oscillations

### Related Formula textPrinciple of Homogeneity: Only quantities with same dimensions can be added/subtracted.$$\text{Principle of Homogeneity: Only quantities with same dimensions can be added/subtracted.}$$ ### Core Logic From the term (h + Bt)$(h + Bt)$, h$h$ (height/length) and Bt$Bt$ must have the same dimension. [h] = [Bt]$[h] = [Bt]$ [B] = frac[h][t] = frac[L][T] = [L T^-1]$$[B] = \frac{[h]}{[t]} = \frac{[L]}{[T]} = [L T^{-1}]$$ So, B = [M^0 L T^-1]$B = [M^0 L T^{-1}]$. ### Step 1: Finding Dimension of A From the term left(P + fracABt^2right)$\left(P + \frac{A}{Bt^2}\right)$, pressure P$P$ and fracABt^2$\frac{A}{Bt^2}$ must have the same dimension. Dimension of Pressure P = fractextForcetextArea = [M L^-1 T^-2]$P = \frac{\text{Force}}{\text{Area}} = [M L^{-1} T^{-2}]$. [P] = left[fracABt^2right]$$[P] = \left[\frac{A}{Bt^2}\right]$$ [M L^-1 T^-2] = frac[A][L T^-1] times [T^2]$$[M L^{-1} T^{-2}] = \frac{[A]}{[L T^{-1}] \times [T^2]}$$ [M L^-1 T^-2] = frac[A][L T]$$[M L^{-1} T^{-2}] = \frac{[A]}{[L T]}$$ [A] = [M L^-1 T^-2] times [L T] = [M L^0 T^-1]$$[A] = [M L^{-1} T^{-2}] \times [L T] = [M L^0 T^{-1}]$$ So, A = [M L^0 T^-1]$A = [M L^0 T^{-1}]$. ### Pattern Recognition Apply the Principle of Homogeneity across any (X + Y)$(X + Y)$ terms inside brackets. Extract one unknown, plug it into the next additive bracket group. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 11 Physics: Mechanical Properties of Fluids

### 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