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 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