### Related Formula For a quantity Q = X^a Y^b Z^c$Q = X^a Y^b Z^c$, the maximum fractional error is: fracDelta QQ = |a| fracDelta XX + |b| fracDelta YY + |c| fracDelta ZZ$\frac{\Delta Q}{Q} = |a| \frac{\Delta X}{X} + |b| \frac{\Delta Y}{Y} + |c| \frac{\Delta Z}{Z}$ where, fracDelta XX, fracDelta YY, fracDelta ZZ$\frac{\Delta X}{X}, \frac{\Delta Y}{Y}, \frac{\Delta Z}{Z}$ are fractional errors of individual variables ### Core Logic Given formula: Q = X^-2 Y^3/2 Z^-2/5$Q = X^{-2} Y^{3/2} Z^{-2/5}$. Identify the absolute exponents: - |a| = |-2| = 2$|a| = |-2| = 2$ - |b| = left|frac32right| = frac32$|b| = \left|\frac{3}{2}\right| = \frac{3}{2}$ - |c| = left|-frac25right| = frac25$|c| = \left|-\frac{2}{5}\right| = \frac{2}{5}$ Now write the error expression: fracDelta QQ = 2 fracDelta XX + frac32 fracDelta YY + frac25 fracDelta ZZ$\frac{\Delta Q}{Q} = 2 \frac{\Delta X}{X} + \frac{3}{2} \frac{\Delta Y}{Y} + \frac{2}{5} \frac{\Delta Z}{Z}$ Substitute the given values: - fracDelta XX = 0.1$\frac{\Delta X}{X} = 0.1$ - fracDelta YY = 0.2$\frac{\Delta Y}{Y} = 0.2$ - fracDelta ZZ = 0.5$\frac{\Delta Z}{Z} = 0.5$ ### Step 1: Compute Maximum Fractional Error Calculate term by term: fracDelta QQ = 2 (0.1) + frac32 (0.2) + frac25 (0.5)$\frac{\Delta Q}{Q} = 2 (0.1) + \frac{3}{2} (0.2) + \frac{2}{5} (0.5)$ fracDelta QQ = 0.2 + 0.3 + 0.2 = 0.7$\frac{\Delta Q}{Q} = 0.2 + 0.3 + 0.2 = 0.7$ ### Pattern Recognition Sees: Exponential algebraic relation for errors. Trap: Exponents are negative, but maximum error is cumulative. Always take the *absolute* value of exponents when summing errors! ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

### Related Formula textYoung's Modulus: Y = fracF/ADelta ell / ell$\text{Young's Modulus: } Y = \frac{F/A}{\Delta \ell / \ell}$ textTorque: tau = F cdot r$\text{Torque: } \tau = F \cdot r$ textViscosity Force: F = eta A fracdvdx$\text{Viscosity Force: } F = \eta A \frac{dv}{dx}$ textGravitational Force: F = fracG m_1 m_2r^2$\text{Gravitational Force: } F = \frac{G m_1 m_2}{r^2}$ ### Core Logic Evaluating dimensions component-by-component: - **(A) Young's Modulus**: [Y] = frac[F][A] = fracmathrmMLT^-2mathrmL^2 = mathrmML^-1T^-2 quad rightarrow text(II)$[Y] = \frac{[F]}{[A]} = \frac{\mathrm{MLT^{-2}}}{\mathrm{L^2}} = \mathrm{ML^{-1}T^{-2}} \quad \rightarrow \text{(II)}$ - **(B) Torque**: [tau] = [F][r] = (mathrmMLT^-2)(mathrmL) = mathrmML^2T^-2 quad rightarrow text(IV)$[\tau] = [F][r] = (\mathrm{MLT^{-2}})(\mathrm{L}) = \mathrm{ML^2T^{-2}} \quad \rightarrow \text{(IV)}$ - **(C) Coefficient of Viscosity**: [eta] = frac[F][A][dv/dx] = fracmathrmMLT^-2(mathrmL^2)(mathrmT^-1) = mathrmML^-1T^-1 quad rightarrow text(I)$[\eta] = \frac{[F]}{[A][dv/dx]} = \frac{\mathrm{MLT^{-2}}}{(\mathrm{L^2})(\mathrm{T^{-1}})} = \mathrm{ML^{-1}T^{-1}} \quad \rightarrow \text{(I)}$ - **(D) Gravitational Constant**: [G] = frac[F][r^2][m_1][m_2] = frac(mathrmMLT^-2)(mathrmL^2)mathrmM^2 = mathrmM^-1L^3T^-2 quad rightarrow text(III)$[G] = \frac{[F][r^2]}{[m_1][m_2]} = \frac{(\mathrm{MLT^{-2}})(\mathrm{L^2})}{\mathrm{M^2}} = \mathrm{M^{-1}L^3T^{-2}} \quad \rightarrow \text{(III)}$ Matching path yields: (A)-(II), (B)-(IV), (C)-(I), (D)-(III). ### Pattern Recognition Torque and energy share the identical dimensional formula mathrmML^2T^-2$\mathrm{ML^2T^{-2}}$. Modulus and pressure share mathrmML^-1T^-2$\mathrm{ML^{-1}T^{-2}}$. Spotting these matching associations cuts solving time significantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

### Related Formula fracDelta QQ = fracDelta aa + 4fracDelta bb + fracDelta cc + fracDelta dd$\frac{\Delta Q}{Q} = \frac{\Delta a}{a} + 4\frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{\Delta d}{d}$ ### Core Logic Write down fractional errors from the raw text configurations: - fracDelta aa = frac360 = 0.05$\frac{\Delta a}{a} = \frac{3}{60} = 0.05$ - fracDelta bb = frac0.120 = 0.005$\frac{\Delta b}{b} = \frac{0.1}{20} = 0.005$ - fracDelta cc = frac0.240 = 0.005$\frac{\Delta c}{c} = \frac{0.2}{40} = 0.005$ - fracDelta dd = frac0.150 = 0.002$\frac{\Delta d}{d} = \frac{0.1}{50} = 0.002$ Compute the total fractional error expression: fracDelta QQ = [0.05 + 4(0.005) + 0.005 + 0.002]$\frac{\Delta Q}{Q} = [0.05 + 4(0.005) + 0.005 + 0.002]$ fracDelta QQ = 0.05 + 0.02 + 0.005 + 0.002 = 0.077$\frac{\Delta Q}{Q} = 0.05 + 0.02 + 0.005 + 0.002 = 0.077$ Percentage error expression configuration: \% text Error = fracDelta QQ times 100 = 7.7 \%$\% \text{ Error} = \frac{\Delta Q}{Q} \times 100 = 7.7 \%$ Given that percentage error equals fracx1000$\frac{x}{1000}$: fracx1000 = 7.7 implies x = 7700$\frac{x}{1000} = 7.7 \implies x = 7700$ ### Pattern Recognition Powers scale up error contributions via direct multiplication multipliers. The term b^4$b^4$ contributes exactly 4 times its basic fraction error component to the compilation step. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

### Core Logic First, find the least count of the measurement tool: textLeast Count = fractextPitchtextNumber of circular scale divisions = frac0.75 mathrm~mm15 = 0.05 mathrm~mm$\text{Least Count} = \frac{\text{Pitch}}{\text{Number of circular scale divisions}} = \frac{0.75 \mathrm{~mm}}{15} = 0.05 \mathrm{~mm}$

The area of the rectangular metallic sheet is calculated as: mathrmA = mathrmL cdot mathrmW$\mathrm{A} = \mathrm{L} \cdot \mathrm{W}$ Expressing the absolute error via fractional configuration parts: fracmathrmdAmathrmA = fracmathrmdLmathrmL + fracmathrmdWmathrmW$\frac{\mathrm{dA}}{\mathrm{A}} = \frac{\mathrm{dL}}{\mathrm{L}} + \frac{\mathrm{dW}}{\mathrm{W}}$ Substituting the instrument limits (mathrmdL = mathrmdW = 0.05 mathrm~mm$\mathrm{dL} = \mathrm{dW} = 0.05 \mathrm{~mm}$): fracmathrmdAmathrmA = frac0.055 + frac0.052.5 = frac1100 + frac2100 = frac3100$\frac{\mathrm{dA}}{\mathrm{A}} = \frac{0.05}{5} + \frac{0.05}{2.5} = \frac{1}{100} + \frac{2}{100} = \frac{3}{100}$ ### Step 1: Final Value Match Comparing this to the target format fracmathrmx100$\frac{\mathrm{x}}{100}$ gives: mathrmx = 3$\mathrm{x} = 3$ ### Pattern Recognition The absolute measurement uncertainty matches the instrument's least count value directly. Sum up individual fractional errors to compute the total area uncertainty parameter. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

### Core Logic Let's find the dimensional formula for the ratio of Modulus of Elasticity to Torque: textTarget Dimensions = frac[textModulus of Elasticity][textTorque]$\text{Target Dimensions} = \frac{[\text{Modulus of Elasticity}]}{[\text{Torque}]}$ textTarget Dimensions = frac[mathrmM L^-1 mathrmT^-2][mathrmM L^2 mathrmT^-2] = [mathrmM^0 mathrmL^-3 mathrmT^0]$\text{Target Dimensions} = \frac{[\mathrm{M L}^{-1} \mathrm{T}^{-2}]}{[\mathrm{M L}^2 \mathrm{T}^{-2}]} = [\mathrm{M}^0 \mathrm{L}^{-3} \mathrm{T}^0]$ ### Step 1: Exponent Matching Comparing this output to the target layout formula [mathrmM^mathrma mathrmL^mathrmb mathrmT^mathrmc]$[\mathrm{M}^{\mathrm{a}} \mathrm{L}^{\mathrm{b}} \mathrm{T}^{\mathrm{c}}]$: mathrmc = 0$\mathrm{c} = 0$ ### Pattern Recognition Both dimensions share identical time dependence factors (mathrmT^-2$\mathrm{T}^{-2}$), meaning they cancel out completely. This leaves the time exponent value as exactly zero. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements