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 By the Principle of Homogeneity, terms added or subtracted must have the same dimensions: [B] = [x^2]$[B] = [x^2]$ ### Core Logic Known dimensional formulas: Length x to [L]$x \to [L]$ Energy E to [ML^2T^-2]$E \to [ML^2T^{-2}]$ Time t to [T]$t \to [T]$ ### Step 1: Dimension of B Since x^2$x^2$ is subtracted from B$B$: [B] = [x^2] = [L^2]$$[B] = [x^2] = [L^2]$$ ### Step 2: Dimension of A From the equation E = fracB - x^2At$E = \frac{B - x^2}{At}$: [A] = frac[B - x^2][E][t]$$[A] = \frac{[B - x^2]}{[E][t]}$$ [A] = frac[L^2][ML^2T^-2][T] = frac[L^2][ML^2T^-1]$$[A] = \frac{[L^2]}{[ML^2T^{-2}][T]} = \frac{[L^2]}{[ML^2T^{-1}]}$$ [A] = [M^-1T^1]$$[A] = [M^{-1}T^1]$$ ### Step 3: Dimension of AB [AB] = [A] times [B]$$[AB] = [A] \times [B]$$ [AB] = [M^-1T^1] times [L^2]$$[AB] = [M^{-1}T^1] \times [L^2]$$ [AB] = [L^2 M^-1 T^1]$$[AB] = [L^2 M^{-1} T^1]$$ ### Pattern Recognition Identify sums/differences first to instantly isolate B$B$. Once [B]$[B]$ is fixed, the entire numerator is just L^2$L^2$. Swap out variables to isolate [A]$[A]$. Combining is just standard exponent addition. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

### Related Formula R = fracrho LA = fracrho Lpi left(fracd2right)^2 = frac4rho Lpi d^2$$R = \frac{\rho L}{A} = \frac{\rho L}{\pi \left(\frac{d}{2}\right)^2} = \frac{4\rho L}{\pi d^2}$$ ### Core Logic To find the maximum percentage error in resistance, apply logarithmic differentiation: fracDelta RR = fracDelta LL + 2fracDelta dd$$\frac{\Delta R}{R} = \frac{\Delta L}{L} + 2\frac{\Delta d}{d}$$ Given percentage errors: fracDelta LL times 100\% = 0.1\%$\frac{\Delta L}{L} \times 100\% = 0.1\%$ fracDelta dd times 100\% = 0.1\%$\frac{\Delta d}{d} \times 100\% = 0.1\%$ ### Step 2: Substitution Substituting the values: fracDelta RR times 100\% = 0.1\% + 2(0.1\%)\, = 0.1\% + 0.2\% = 0.3\%$$\frac{\Delta R}{R} \times 100\% = 0.1\% + 2(0.1\%)\, = 0.1\% + 0.2\% = 0.3\%$$ ### Pattern Recognition Resistance scales inversely with the square of the diameter. The error multiplier for diameter is 2. Just sum linear components directly: Error = L_error + 2 * d_error. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units And Measurements Class 12 Physics: Current Electricity

### Related Formula textPrinciple of Homogeneity: [F] = [ax^2] = [bt^1/2]$$\text{Principle of Homogeneity: } [F] = [ax^2] = [bt^{1/2}]$$ ### Core Logic By the principle of dimensional homogeneity, each additive term must have the same dimension as the left hand side. Dimension of force F = [M L T^-2]$F = [M L T^{-2}]$. For the term ax^2$ax^2$: [a] = frac[F][x^2] = frac[M L T^-2][L^2] = [M L^-1 T^-2]$$[a] = \frac{[F]}{[x^2]} = \frac{[M L T^{-2}]}{[L^2]} = [M L^{-1} T^{-2}]$$ For the term bt^1/2$bt^{1/2}$: [b] = frac[F][t^1/2] = frac[M L T^-2][T^1/2] = [M L T^-5/2]$$[b] = \frac{[F]}{[t^{1/2}]} = \frac{[M L T^{-2}]}{[T^{1/2}]} = [M L T^{-5/2}]$$ ### Step 2: Computing Required Ratio We need the dimension of fracb^2a$\frac{b^2}{a}$: left[ fracb^2a right] = frac[M L T^-5/2]^2[M L^-1 T^-2]$$\left[ \frac{b^2}{a} \right] = \frac{[M L T^{-5/2}]^2}{[M L^{-1} T^{-2}]}$$ left[ fracb^2a right] = frac[M^2 L^2 T^-5][M L^-1 T^-2]$$\left[ \frac{b^2}{a} \right] = \frac{[M^2 L^2 T^{-5}]}{[M L^{-1} T^{-2}]}$$ left[ fracb^2a right] = [M^2-1 L^2 - (-1) T^-5 - (-2)]$$\left[ \frac{b^2}{a} \right] = [M^{2-1} L^{2 - (-1)} T^{-5 - (-2)}]$$ left[ fracb^2a right] = [M L^3 T^-3]$$\left[ \frac{b^2}{a} \right] = [M L^3 T^{-3}]$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units And Measurements