Match the LIST-I with LIST-II

LIST-I	LIST-II
A. Boltzmann constant	I. ML^2T^-1$ML^2T^{-1}$
B. Coefficient of viscosity	II. MLT^-3K^-1$MLT^{-3}K^{-1}$
C. Planck's constant	III. ML^2T^-2K^-1$ML^2T^{-2}K^{-1}$
D. Thermal conductivity	IV. ML^-1T^-1$ML^{-1}T^{-1}$

Choose the correct answer from the options given below :

### Related Formula 1text MSD = fractextTotal LengthtextTotal Divisions$$1\text{ MSD} = \frac{\text{Total Length}}{\text{Total Divisions}}$$ textLeast Count (LC) = 1text MSD - 1text VSD = 1text MSD cdot left(1 - frac2425right)$$\text{Least Count (LC)} = 1\text{ MSD} - 1\text{ VSD} = 1\text{ MSD} \cdot \left(1 - \frac{24}{25}\right)$$ ### Core Logic Calculate the value of one main scale division: 1text MSD = frac15text cm300 = 0.05text cm$$1\text{ MSD} = \frac{15\text{ cm}}{300} = 0.05\text{ cm}$$ Given that 25text VSD = 24text MSD$25\text{ VSD} = 24\text{ MSD}$, we find: 1text VSD = frac2425text MSD$$1\text{ VSD} = \frac{24}{25}\text{ MSD}$$ ### Step 1: Compute Least Count textLC = 1text MSD cdot left(frac125right) = frac0.05text cm25 = 0.002text cm$$\text{LC} = 1\text{ MSD} \cdot \left(\frac{1}{25}\right) = \frac{0.05\text{ cm}}{25} = 0.002\text{ cm}$$ ### Pattern Recognition Least count calculation is conventionally textLC = frac1text MSDN$\text{LC} = \frac{1\text{ MSD}}{N}$ where N$N$ represents the total count of divisions on the vernier scale whenever (N-1)text MSD = Ntext VSD$(N-1)\text{ MSD} = N\text{ VSD}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

### Related Formula Electric Dipole Moment: P_e = q cdot d implies [P_e] = textAcdottextTcdottextL$P_e = q \cdot d \implies [P_e] = \text{A}\cdot\text{T}\cdot\text{L}$ Magnetic Dipole Moment: M_m = I cdot A implies [M_m] = textAcdottextL^2$M_m = I \cdot A \implies [M_m] = \text{A}\cdot\text{L}^2$ ### Core Logic Take the dimensional ratio: left[fracP_eM_mright] = fractextLtextTtextAtextL^2textA = textL^-1textT = textM^0textL^-1textT^1textA^0$$\left[\frac{P_e}{M_m}\right] = \frac{\text{L}\text{T}\text{A}}{\text{L}^2\text{A}} = \text{L}^{-1}\text{T} = \text{M}^0\text{L}^{-1}\text{T}^1\text{A}^0$$ Comparing powers with [textM^PtextL^QtextT^RtextA^S]$[\text{M}^P\text{L}^Q\text{T}^R\text{A}^S]$, we find: P = 0$P = 0$ Q = -1$Q = -1$ ### Pattern Recognition Dipole units match up with fundamental currents and charge definitions. Notice that current A$A$ cancels out completely, leaving a simple geometric spatial ratio. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 12 Physics: Moving Charges and Magnetism

### Related Formula Ratio formulation: fracphi_Ephi_M = fracE cdot AB cdot A = fracEB $$ \frac{\phi_E}{\phi_M} = \frac{E \cdot A}{B \cdot A} = \frac{E}{B} $$ From Maxwell's electromagnetic wave equations: E = c cdot B implies fracEB = c$$ E = c \cdot B implies \frac{E}{B} = c$$ where c$c$ is the speed of light. ### Core Logic Since the ratio reduces to the dimension of speed (c$c$): left[fracphi_Ephi_M ight] = [c] = mathrmM^0 mathrmL^1 mathrmT^-1 mathrmA^0$$\left[\frac{\phi_E}{\phi_M} ight] = [c] = \mathrm{M}^0 \mathrm{L}^1 \mathrm{T}^{-1} \mathrm{A}^0$$ ### Step 1: Identify Exponent Values Matching indices with mathrmM^PmathrmL^QmathrmT^RmathrmA^S$\mathrm{M}^{P}\mathrm{L}^{Q}\mathrm{T}^{R}\mathrm{A}^{S}$: * P = 0$P = 0$ * Q = 1$Q = 1$ * R = -1$R = -1$ * S = 0$S = 0$ Hence, (Q, R) = (1, -1)$(Q, R) = (1, -1)$. ### Pattern Recognition Flux areas cancel out immediately. The ratio fracEB$\frac{E}{B}$ always carries the dimension of velocity (LT^-1$LT^{-1}$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 12 Physics: Electromagnetic Waves

### Core Logic Let's derive the dimensional formulas systematically: * **(A) Mass density:** rho = fractextMasstextVolume = fracML^3 = [M^1 L^-3 T^0] implies text(IV)$\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{M}{L^3} = [M^1 L^{-3} T^0] \implies \text{(IV)}$ [cite: 765]. * **(B) Impulse:** I = F cdot Delta t = [M^1 L^1 T^-2] cdot [T] = [M^1 L^1 T^-1] implies text(II)$I = F \cdot \Delta t = [M^1 L^1 T^{-2}] \cdot [T] = [M^1 L^1 T^{-1}] \implies \text{(II)}$ [cite: 767]. * **(C) Power:** P = fractextWorktextTime = frac[M^1 L^2 T^-2][T] = [M^1 L^2 T^-3] implies text(I)$P = \frac{\text{Work}}{\text{Time}} = \frac{[M^1 L^2 T^{-2}]}{[T]} = [M^1 L^2 T^{-3}] \implies \text{(I)}$ [cite: 770]. * **(D) Moment of inertia:** I = M r^2 = [M^1 L^2 T^0] implies text(III)$I = M r^2 = [M^1 L^2 T^0] \implies \text{(III)}$ [cite: 772]. Matching all four pairings establishes the layout: (A)-(IV), (B)-(II), (C)-(I), (D)-(III)[cite: 110, 758]. ### Pattern Recognition Isolating the unique matching option for a straightforward parameter like Mass Density (A-IV) or Moment of Inertia (D-III) easily allows exclusion of multiple invalid answer branches instantly[cite: 765, 772]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

### Related Formula In multiplication and division arithmetic rules, the final product or quotient must be rounded off to retain as many significant figures as are present in the least precise operand. ### Core Logic Let us check the significant digit count of the operands in the expression : * 32.3$32.3$ has 3$3$ significant figures. * 1125$1125$ has 4$4$ significant figures. * 27.4$27.4$ has 3$3$ significant figures. The minimum number of significant figures among the numbers is 3$3$. ### Step 1: Rounding Off Direct calculation yield : y = 1326.186...$y = 1326.186...$ Rounding this value off to contain exactly 3 significant figures means changing it to 1330, since the digit after 2 is 6 (which is greater than 5), updating the hundreds spot upwards. ### Pattern Recognition Never keep unearned precision from automated calculation. The output is bounded strictly by your least precise entry. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements