Match the LIST-I with LIST-II

List-I	List-II
(A) Spring constant	(I) ML^2T^-2K^-1$ML^{2}T^{-2}K^{-1}$
(B) Thermal conductivity	(II) ML^0T^-2$ML^{0}T^{-2}$
(C) Boltzmann constant	(III) ML^2T^-3A^-2$ML^{2}T^{-3}A^{-2}$
(D) Inductive reactance	(IV) MLT^-3K^-1$MLT^{-3}K^{-1}$

Choose the correct answer from the options given below:

### 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 Addition Rule for Significant Figures: The result cannot have more digits to the right of the decimal point than any of the original numbers. ### Core Logic Let's align the numbers and perform exact addition: 52.01$52.01$ (2 decimal places) 153.2$153.2$ (1 decimal place - least precise) + 0.123$+ 0.123$ (3 decimal places) -------- 205.333$205.333$ ### Step 1: Final Conclusion According to the rules of significant figures for addition, the sum must be rounded to the least number of decimal places present in the given measurements. Here, 153.2$153.2$ has only 1$1$ decimal place. Rounding 205.333$205.333$ to one decimal place yields 205.3 text m$205.3 \text{ m}$. ### Pattern Recognition Addition/Subtraction is bounded by decimal places. Multiplication/Division is bounded by total significant figures. In this case, 1 decimal place rules. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

### Related Formula v_0 = frac29fracr^2 geta (sigma - rho)$$v_0 = \frac{2}{9}\frac{r^2 g}{\eta} (\sigma - \rho)$$ ### Core Logic Rearranging the formula for viscosity eta$\eta$: eta = frac29 fracr^2 gv_0 (sigma - rho)$$\eta = \frac{2}{9} \frac{r^2 g}{v_0} (\sigma - \rho)$$ Since frac29$\frac{2}{9}$, g$g$, sigma$\sigma$, and rho$\rho$ are constants or have negligible error, the relative error depends only on r$r$ and v_0$v_0$. ### Step 1: Error Propagation Taking natural logarithm on both sides: ln(eta) = ln(textconstant) + 2ln(r) - ln(v_0)$$\ln(\eta) = \ln(\text{constant}) + 2\ln(r) - \ln(v_0)$$ Differentiating for maximum possible error (adding absolute errors): fracDelta etaeta = 2fracDelta rr + fracDelta v_0v_0$$\frac{\Delta \eta}{\eta} = 2\frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}$$ ### Step 2: Final Conclusion The maximum fractional error is frac2Delta rr + fracDelta v_0v_0$\frac{2\Delta r}{r} + \frac{\Delta v_0}{v_0}$. ### Pattern Recognition For maximum fractional error of a derived quantity Z = X^a Y^b$Z = X^a Y^b$, the formula is Delta Z/Z = |a|Delta X/X + |b|Delta Y/Y$\Delta Z/Z = |a|\Delta X/X + |b|\Delta Y/Y$. Always sum the absolute relative errors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 11 Physics: Mechanical Properties of Fluids