### Related Formula [m] = [M]^1 [L]^0 [T]^0$[m] = [M]^1 [L]^0 [T]^0$ [c] = [L T^-1]$[c] = [L T^{-1}]$ [G] = [M^-1 L^3 T^-2]$[G] = [M^{-1} L^3 T^{-2}]$ [h] = [M L^2 T^-1]$[h] = [M L^2 T^{-1}]$ ### Core Logic By applying the principle of dimensional homogeneity, the dimensions on both sides of the equation must be identical. [M]^1 [L]^0 [T]^0 = [L T^-1]^p [M^-1 L^3 T^-2]^-1/2 [M L^2 T^-1]^1/2$[M]^1 [L]^0 [T]^0 = [L T^{-1}]^p [M^{-1} L^3 T^{-2}]^{-1/2} [M L^2 T^{-1}]^{1/2}$ ### Step 1: Substitute Dimensions [M] = L^p T^-p cdot M^1/2 L^-3/2 T^1 cdot M^1/2 L^1 T^-1/2$[M] = L^p T^{-p} \cdot M^{1/2} L^{-3/2} T^1 \cdot M^{1/2} L^1 T^{-1/2}$ ### Step 2: Collect Powers of L Equating the powers of [L]$[L]$ on both sides: 0 = p - frac32 + 1$0 = p - \frac{3}{2} + 1$ 0 = p - frac12$0 = p - \frac{1}{2}$ p = frac12$p = \frac{1}{2}$ ### Pattern Recognition The expression m propto sqrthc/G$m \propto \sqrt{hc/G}$ is a known fundamental relation representing the Planck mass. The exponent on c$c$ inside the square root gives p = 1/2$p = 1/2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

### Related Formula F = eta A fracdvdy$F = \eta A \frac{dv}{dy}$ textSurface Tension = fracFl$\text{Surface Tension} = \frac{F}{l}$ L = mvr$L = mvr$ K.E = frac12 I omega^2$K.E = \frac{1}{2} I \omega^2$ ### Core Logic Let us determine the dimensional formula for each quantity sequentially: **A. Coefficient of viscosity (eta$\eta$):** Using F = eta A fracdvdy$F = \eta A \frac{dv}{dy}$, we have: [M L T^-2] = eta [L^2] [T^-1]$[M L T^{-2}] = \eta [L^2] [T^{-1}]$ eta = [M L^-1 T^-1] Rightarrow text(III)$\eta = [M L^{-1} T^{-1}] \Rightarrow \text{(III)}$ **B. Surface Tension (S.T.):** textS.T = fracFell = frac[M L T^-2][L] = [M L^0 T^-2] Rightarrow text(IV)$\text{S.T} = \frac{F}{\ell} = \frac{[M L T^{-2}]}{[L]} = [M L^0 T^{-2}] \Rightarrow \text{(IV)}$ **C. Angular momentum (L):** L = mvr = [M] [L T^-1] [L] = [M L^2 T^-1] Rightarrow text(II)$L = mvr = [M] [L T^{-1}] [L] = [M L^2 T^{-1}] \Rightarrow \text{(II)}$ **D. Rotational kinetic energy (K.E.):** textK.E = frac12 I omega^2 = [M L^2 T^-2] Rightarrow text(I)$\text{K.E} = \frac{1}{2} I \omega^2 = [M L^2 T^{-2}] \Rightarrow \text{(I)}$ ### Step 1: Final Matching Matching the derived dimensional formulas: A rightarrow$\rightarrow$ III B rightarrow$\rightarrow$ IV C rightarrow$\rightarrow$ II D rightarrow$\rightarrow$ I ### Pattern Recognition Kinetic energy (whether translational or rotational) always carries the dimension of Work: [M L^2 T^-2]$[M L^2 T^{-2}]$. Surface tension is force per unit length, dropping the L$L$ term. Viscosity commonly includes L^-1$L^{-1}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

### Related Formula T = 2pi sqrtfracellg implies g = frac4pi^2 ellT^2$T = 2\pi \sqrt{\frac{\ell}{g}} \implies g = \frac{4\pi^2 \ell}{T^2}$ ### Core Logic By taking logarithms and differentiating to find relative error (accuracy): fracDelta gg = fracDelta ellell + 2fracDelta TT$\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2\frac{\Delta T}{T}$ ### Step 1: Extrapolating Errors Given values: ell = 20 text cm = 200 text mm$\ell = 20 \text{ cm} = 200 \text{ mm}$ Delta ell = 2 text mm$\Delta \ell = 2 \text{ mm}$ T_texttotal = 40 text s$T_{\text{total}} = 40 \text{ s}$ for 50 oscillations Delta T_texttotal = 1 text s$\Delta T_{\text{total}} = 1 \text{ s}$ Note: The relative error in time period T$T$ is equal to the relative error in total time t$t$: fracDelta TT = fracDelta tt$\frac{\Delta T}{T} = \frac{\Delta t}{t}$. ### Step 2: Substitution fracDelta gg = frac0.2 text cm20 text cm + 2 left(frac1 text s40 text sright)$\frac{\Delta g}{g} = \frac{0.2 \text{ cm}}{20 \text{ cm}} + 2 \left(\frac{1 \text{ s}}{40 \text{ s}}\right)$ fracDelta gg = frac2200 + frac240$\frac{\Delta g}{g} = \frac{2}{200} + \frac{2}{40}$ fracDelta gg = frac1100 + frac5100 = frac6100$\frac{\Delta g}{g} = \frac{1}{100} + \frac{5}{100} = \frac{6}{100}$ ### Step 3: Percentage Conversion Percentage change = fracDelta gg times 100\% = frac6100 times 100\% = 6\%$= \frac{\Delta g}{g} \times 100\% = \frac{6}{100} \times 100\% = 6\%$. Thus, N = 6$N = 6$. ### Pattern Recognition For pendulum gravity error, always use \%g = \%ell + 2(\%T)$\%g = \%\ell + 2(\%T)$. Remember that measuring 50 oscillations reduces absolute error on a single swing, but the relative error Delta t / t$\Delta t / t$ remains unchanged whether you use total time or single period. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 11 Physics: Oscillations

### 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