JEE Advanced · Physics · 2. Units & Dimensions
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 \(\left[M^0 L^0 T^0 I^0 K^0\right]\)
- B \(\left[M^0 L^0 T^0 I^0 K^{-1}\right]\)
- C \(\left[M^1 L^2 T^{-2} I^{-1} K^{-1}\right]\)
- D \(\left[M^1 L^2 T^{-4} I^{-1} K^{-1}\right]\)
Answer & Solution
Correct Answer
(B) \(\left[M^0 L^0 T^0 I^0 K^{-1}\right]\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{S}=\text { emf per unit temperature difference } \\ & \sigma=\text { Electrical conductivity } \\ & \mathrm{k}=\text { Thermal conductivity } \\ & {[\mathrm{S}]=\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{I}^{-1} \mathrm{~K}^{-1}\right]} \\ & {[\sigma]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^3 \mathrm{I}^2\right]} \\ & {[\mathrm{K}]=\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-3} \mathrm{~K}^{-1}\right]} \\ & {[\mathrm{Z}]=\frac{\mathrm{S}^2 \sigma}{\mathrm{~K}}=\frac{\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-3} \mathrm{~K}^{-2}\right]}{\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-3} \mathrm{~K}^{-1}\right]}}\end{aligned}\)
\([\mathrm{Z}]=\left[\mathrm{K}^{-1}\right]\)
\([\mathrm{Z}]=\left[\mathrm{K}^{-1}\right]\)
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