Given below are two statements: Statement (I): The dimensions of Planck's constant and angular momentum are same. Statement (II): In Bohr's model electron revolve around the nucleus only in those orbits for which angular momentum is integral multiple of Planck's constant. In the light of the above statements, choose the most appropriate answer from the options given below:

### Related Formula Formulas to find dimensional formulas: - Boltzmann constant: k_B = fractextEnergytextTemperature$$k_B = \frac{\text{Energy}}{\text{Temperature}}$$ - Coefficient of viscosity: eta = fracFA fracdvdx$$\eta = \frac{F}{A \frac{dv}{dx}}$$ - Planck's constant: h = fracEnu$$h = \frac{E}{\nu}$$ - Thermal conductivity: fracdQdt = K A fracdTdx Rightarrow K = fractextHeat flow cdot textthicknesstextArea cdot textTemperature difference$$\frac{dQ}{dt} = K A \frac{dT}{dx} \Rightarrow K = \frac{\text{Heat flow} \cdot \text{thickness}}{\text{Area} \cdot \text{Temperature difference}}$$ ### Core Logic Evaluate each constant individually: ### Step 1: Dimensions of Boltzmann constant (k_B$k_B$) [k_B] = frac[ML^2T^-2][K] = [ML^2T^-2K^-1] quad Rightarrow textMatches III$$[k_B] = \frac{[ML^2T^{-2}]}{[K]} = [ML^2T^{-2}K^{-1}] \quad \Rightarrow \text{Matches III}$$ ### Step 2: Dimensions of Coefficient of viscosity (eta$\eta$) [eta] = frac[MLT^-2][L^2] [T^-1] = [ML^-1T^-1] quad Rightarrow textMatches IV$$[\eta] = \frac{[MLT^{-2}]}{[L^2] [T^{-1}]} = [ML^{-1}T^{-1}] \quad \Rightarrow \text{Matches IV}$$ ### Step 3: Dimensions of Planck's constant (h$h$) [h] = frac[ML^2T^-2][T^-1] = [ML^2T^-1] quad Rightarrow textMatches I$$[h] = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}] \quad \Rightarrow \text{Matches I}$$ ### Step 4: Dimensions of Thermal conductivity (K$K$) [K] = frac[ML^2T^-3] [L][L^2] [K] = [MLT^-3K^-1] quad Rightarrow textMatches II$$[K] = \frac{[ML^2T^{-3}] [L]}{[L^2] [K]} = [MLT^{-3}K^{-1}] \quad \Rightarrow \text{Matches II}$$ This sequence yields A-III, B-IV, C-I, D-II, matching Option (1). ### Pattern Recognition To solve matching sets efficiently, search for the most recognizable dimensions first. Planck's constant h$h$ (ML^2T^-1$ML^2T^{-1}$) and viscosity coefficient eta$\eta$ (ML^-1T^-1$ML^{-1}T^{-1}$) are highly unique and usually resolve the options instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

### Related Formula For a physical quantity defined by algebraic powers C = fracp^a q^br^c s^d$C = \frac{p^a q^b}{r^c s^d}$, the maximum fractional error is calculated by summing absolute scaled fractional errors: fracDelta CC = a fracDelta pp + b fracDelta qq + c fracDelta rr + d fracDelta ss$$\frac{\Delta C}{C} = a \frac{\Delta p}{p} + b \frac{\Delta q}{q} + c \frac{\Delta r}{r} + d \frac{\Delta s}{s}$$ Expressed as percentages: \% text error in C = a(\% text error in p) + b(\% text error in q) + c(\% text error in r) + d(\% text error in s)$$\% \text{ error in } C = a(\% \text{ error in } p) + b(\% \text{ error in } q) + c(\% \text{ error in } r) + d(\% \text{ error in } s)$$ ### Core Logic Given expression: C = p^1 q^2 r^-3 s^-1/2$$C = p^1 q^2 r^{-3} s^{-1/2}$$ Max fractional error equation: fracDelta CC = 1 left(fracDelta ppright) + 2 left(fracDelta qqright) + 3 left(fracDelta rrright) + frac12 left(fracDelta ssright)$$\frac{\Delta C}{C} = 1 \left(\frac{\Delta p}{p}\right) + 2 \left(\frac{\Delta q}{q}\right) + 3 \left(\frac{\Delta r}{r}\right) + \frac{1}{2} \left(\frac{\Delta s}{s}\right)$$ ### Step 1: Calculate the total percentage error Substitute the individual percentage errors: - Error in p = 1\%$p = 1\%$ - Error in q = 2\%$q = 2\%$ - Error in r = 3\%$r = 3\%$ - Error in s = 2\%$s = 2\%$ \% text error in C = 1(1\%) + 2(2\%) + 3(3\%) + frac12(2\%)$$\% \text{ error in } C = 1(1\%) + 2(2\%) + 3(3\%) + \frac{1}{2}(2\%)$$ \% text error in C = 1\% + 4\% + 9\% + 1\% = 15\%$$\% \text{ error in } C = 1\% + 4\% + 9\% + 1\% = 15\%$$ The total percentage error in C$C$ is 15\%$15\%$. ### Pattern Recognition In error propagation, individual errors always combine constructively to produce the maximum possible uncertainty limit. Hence, negative powers (like division by r^3$r^3$ or s^1/2$s^{1/2}$) are integrated using positive coefficients during maximum absolute error summation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

### 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