### Related Formula Significant Figures Rule for Addition/Subtraction: The final result must be rounded off to keep only as many decimal places as there are in the measurement with the **least** number of decimal places. ### Core Logic Let's look at the decimal places of each measurement: - 435.42mathrm~g$435.42\mathrm{~g}$ has **2 decimal places**. - 226.3mathrm~g$226.3\mathrm{~g}$ has **1 decimal place**. - 0.125mathrm~g$0.125\mathrm{~g}$ has **3 decimal places**. The minimum number of decimal places is **1 decimal place** (from 226.3mathrm~g$226.3\mathrm{~g}$). ### Step 1: Addition and Rounding First, perform the standard mathematical addition: textSum = 435.42 + 226.3 + 0.125 = 661.845mathrm~g$\text{Sum} = 435.42 + 226.3 + 0.125 = 661.845\mathrm{~g}$ Now, round this raw sum off to **1 decimal place**: - The digit after tenths place is 4 (4 < 5$4 < 5$), so we round down. - Net rounded sum = 661.8mathrm~g$661.8\mathrm{~g}$. ### Pattern Recognition Bust the common myth: Addition depends on the least number of decimal places, whereas multiplication/division depends on the least number of significant figures. Always distinguish between these two rules during exams! ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements: Error Analysis and Significant Figures

### Related Formula E = hf implies [h] = frac[E][f] = fractextMtextL^2textT^-2textT^-1 = textMtextL^2textT^-1$E = hf \implies [h] = \frac{[E]}{[f]} = \frac{\text{M}\text{L}^2\text{T}^{-2}}{\text{T}^{-1}} = \text{M}\text{L}^2\text{T}^{-1}$ L = mvr implies [L] = textM cdot (textLtextT^-1) cdot textL = textMtextL^2textT^-1$L = mvr \implies [L] = \text{M} \cdot (\text{L}\text{T}^{-1}) \cdot \text{L} = \text{M}\text{L}^2\text{T}^{-1}$ L = fracnh2pi$L = \frac{nh}{2\pi}$ ### Core Logic Statement I: Comparing the dimensional formula of Planck's constant (h$h$) and angular momentum (L$L$), both are identical [textMtextL^2textT^-1]$[\text{M}\text{L}^2\text{T}^{-1}]$. Hence, Statement I is correct. Statement II: According to Bohr's second postulate, angular momentum is an integral multiple of frach2pi$\frac{h}{2\pi}$, not an integral multiple of h$h$. Hence, Statement II is incorrect. ### Pattern Recognition Watch out for exact definitions in standard postulates. Bohr's model requires angular momentum to be quantized in units of hbar = frach2pi$\hbar = \frac{h}{2\pi}$, making statement II a classic trap. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 12 Physics: Atoms

### 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