Question

Find the dimensional formula of the quantity \(\alpha\), given: \[ \alpha = \frac{S^2 \cdot \sigma}{k}, \quad S = \frac{\text{EMF}}{\text{time}},\quad \sigma = \text{electric conductivity},\quad k = \text{thermal conductivity} \]

• A. \([M^0L^2T^{-3}K^{-1}]\)
• B. \([M^0L^0T^{-2}K^1]\)
• C. \([M^0L^0T^{-1}K^{-1}]\)
• D. \([M^0L^2T^{-2}K^{-1}]\)

Hint:

Always simplify dimension terms one-by-one and re-verify if intermediate cancellations cause mistakes. Be alert for unit errors!

Answer 1

The Correct Option is A

Step 1: Dimensional formula for EMF (Electromotive Force) \[ \text{EMF} = \frac{\text{Work}}{\text{Charge}} = \frac{ML^2T^{-2}}{AT} = ML^2T^{-3}A^{-1} \] So, \[ S = \frac{\text{EMF}}{\text{time}} = \frac{ML^2T^{-3}A^{-1}}{T} = ML^2T^{-4}A^{-1} \Rightarrow S^2 = M^2L^4T^{-8}A^{-2} \] Step 2: Dimensional formula for electric conductivity \(\sigma\) \[ \sigma = M^{-1}L^{-3}T^3A^2 \] Step 3: Dimensional formula for thermal conductivity \(k\) \[ k = \frac{\text{Energy}}{\text{Time} \cdot \text{Length} \cdot \text{Temperature difference}} = \frac{ML^2T^{-2}}{T \cdot L \cdot K} = MLT^{-3}K^{-1} \] Step 4: Combine to get \(\alpha\) \[ \alpha = \frac{S^2 \cdot \sigma}{k} = \frac{M^2L^4T^{-8}A^{-2} \cdot M^{-1}L^{-3}T^3A^2}{MLT^{-3}K^{-1}} \] Simplify: - Mass: \(M^{2 - 1 - 1} = M^0\) - Length: \(L^{4 - 3 - 1} = L^0\) - Time: \(T^{-8 + 3 + 3} = T^{-2}\) - Current: \(A^{-2 + 2} = A^0\) - Temperature: \(K^1\) \[ \Rightarrow \boxed{[M^0L^0T^{-2}K^1]} \] Wait — looks like a miscalculation. Let's double-check final step: Actually: Numerator: \[ S^2 \cdot \sigma = M^2L^4T^{-8}A^{-2} \cdot M^{-1}L^{-3}T^3A^2 = ML^1T^{-5} \] Denominator: \[ k = M^1L^1T^{-3}K^{-1} \] Now: \[ \alpha = \frac{ML^1T^{-5}}{ML^1T^{-3}K^{-1}} = T^{-2}K^1 \Rightarrow \boxed{[M^0L^0T^{-2}K^1]} \] Final Answer: \([M^0L^0T^{-2}K^1]\) Correct option: (B)