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JEE Advanced 2025Paper 2
Exam
JEE Advanced
Year
2025
Paper
Paper 2
Subject
Physics
Chapter
Units and Dimensions

JEE Advanced 2025 · Paper 2 · Physics · Units and Dimensions

A temperature difference can generate e.m.f. in some materials. Let...

A temperature difference can generate e.m.f. in some materials. Let SS be the e.m.f. produced per unit temperature difference between the ends of a wire, σ\sigma the electrical conductivity and κ\kappa the thermal conductivity of the material of the wire. Taking M,L,T,IM, L, T, I and KK as dimensions of mass, length, time, current and temperature, respectively, the dimensional formula of the quantity Z=S2σκZ=\frac{S^2 \sigma}{\kappa} is :-

Options

  1. A[M0L0T0I0K0]\left[M^0 L^0 T^0 I^0 K^0\right]
  2. B[M0L0T0I0K−1]\left[M^0 L^0 T^0 I^0 K^{-1}\right]
  3. C[M1L2T−2I−1K−1]\left[M^1 L^2 T^{-2} I^{-1} K^{-1}\right]
  4. D[M1L2T−4I−1K−1]\left[M^1 L^2 T^{-4} I^{-1} K^{-1}\right]

Correct answer

B. [M0L0T0I0K−1]\left[M^0 L^0 T^0 I^0 K^{-1}\right]

Step-by-step solution

S= emf per unit temperature difference σ= Electrical conductivity k= Thermal conductivity [S]=[ML2 T−3I−1 K−1][σ]=[M−1 L−3 T3I2][K]=[M1 L1 T−3 K−1][Z]=S2σ K=[M1 L1 T−3 K−2][M1 L1 T−3 K−1]\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]$
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