JEE Advanced 2025 Paper 2 Q01 Physics Errors & Measurements Significant Figures Medium

JEE Advanced 2025 Paper 2 · Q01 · Significant Figures

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 $Z = S^2 \sigma/\kappa$ is:

  1. A. $[M^0 L^0 T^0 I^0 K^0]$
  2. B. $[M^0 L^0 T^0 I^0 K^{-1}]$
  3. C. $[M^1 L^2 T^{-2} I^{-1} K^{-1}]$
  4. D. $[M^1 L^2 T^{-4} I^{-1} K^{-1}]$
Reveal answer + step-by-step solution

Correct answer:B

Solution

Dimensions: $[S] = [\text{Volt}/K] = [M L^2 T^{-3} I^{-1} K^{-1}]$. $[\sigma]$ (electrical conductivity) $= [M^{-1} L^{-3} T^3 I^2]$. $[\kappa]$ (thermal conductivity) $= [M L T^{-3} K^{-1}]$. $[S^2] = [M^2 L^4 T^{-6} I^{-2} K^{-2}]$. $[S^2 \cdot \sigma] = [M^2 L^4 T^{-6} I^{-2} K^{-2}] \cdot [M^{-1} L^{-3} T^3 I^2] = [M L T^{-3} K^{-2}]$. Divide by $[\kappa] = [M L T^{-3} K^{-1}]$: $[Z] = [M^0 L^0 T^0 I^0 K^{-1}]$. Answer (B). ($Z$ is essentially $S^2\sigma/\kappa$ = Lorenz number divided by $T$, which has dimension of $1/T$.)

Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →