Q33 / 48 · JEE Advanced 2025 (Paper 2) · Physics
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$\left[M^0 L^0 T^0 I^0 K^0\right]$
- Correct answerB$\left[M^0 L^0 T^0 I^0 K^{-1}\right]$
- C$\left[M^1 L^2 T^{-2} I^{-1} K^{-1}\right]$
- D$\left[M^1 L^2 T^{-4} I^{-1} K^{-1}\right]$
Solution
$\begin{aligned} & \mathrm{S}=\text { emf per unit temperature difference } \\ & \sigma=\text { Electrical conductivity } \\ & \mathrm{k}=\text { Thermal conductivity } \\ & {[\mathrm{S}]=\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{I}^{-1} \mathrm{~K}^{-1}\right]} \\ & {[\sigma]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^3 \mathrm{I}^2\right]} \\ & {[\mathrm{K}]=\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-3} \mathrm{~K}^{-1}\right]} \\ & {[\mathrm{Z}]=\frac{\mathrm{S}^2 \sigma}{\mathrm{~K}}=\frac{\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-3} \mathrm{~K}^{-2}\right]}{\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-3} \mathrm{~K}^{-1}\right]}}\end{aligned}$
$[\mathrm{Z}]=\left[\mathrm{K}^{-1}\right]$
Hence, the correct answer is option: 'B'.