Question 1

SCQMEDIUM

A temperature difference can generate e.m.f. in some materials. Let $S$ be the e.m.f. produced per unit temperature difference between the ends of a wire, $\sigma$ the electrical conductivity and $\kappa$ the thermal conductivity of the material of the wire. Taking $M, L, T, I$ and $K$ as dimensions of mass, length, time, current and temperature, respectively, the dimensional formula of the quantity $Z = \frac{S^2 \sigma}{\kappa}$ is:

(A)

$[M^0 L^0 T^0 I^0 K^0]$

(B)

$[M^0 L^0 T^0 I^0 K^{-1}]$

(C)

$[M^1 L^2 T^{-2} I^{-1} K^{-1}]$

(D)

$[M^1 L^2 T^{-4} I^{-1} K^{-1}]$

Detailed Solution

We first find the dimensions of each quantity:

e.m.f. ( $V$ ): $\text{work} / \text{charge} = [ML^2T^{-2}] / [IT] = [ML^2T^{-3}I^{-1}]$ .
$S = \text{e.m.f.} / \text{temperature} = [ML^2T^{-3}I^{-1}K^{-1}]$ .
Electrical conductivity $\sigma = \frac{1}{\rho} = \frac{l}{RA} = \frac{l}{(V/I)A} = \frac{[L][I]}{[ML^2T^{-3}I^{-1}][L^2]} = [M^{-1}L^{-3}T^3I^2]$ .
Thermal conductivity $\kappa$ is found from $\frac{dQ}{dt} = \kappa A \frac{\Delta T}{l} \Rightarrow \kappa = \frac{[ML^2T^{-2}/T] [L]}{[L^2][K]} = [MLT^{-3}K^{-1}]$ .

Now, calculate dimensions of $Z = \frac{S^2 \sigma}{\kappa}$ : $[Z] = \frac{([ML^2T^{-3}I^{-1}K^{-1}])^2 \cdot [M^{-1}L^{-3}T^3I^2]}{[MLT^{-3}K^{-1}]}$ $[Z] = \frac{[M^2L^4T^{-6}I^{-2}K^{-2}] \cdot [M^{-1}L^{-3}T^3I^2]}{[MLT^{-3}K^{-1}]}$ $[Z] = \frac{[M^1L^1T^{-3}K^{-2}]}{[MLT^{-3}K^{-1}]} = [K^{-1}]$ Thus, the dimensional formula is $[M^0 L^0 T^0 I^0 K^{-1}]$ .

Question 1

Detailed Solution

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