Start free

Physics · Units, Dimensions & Error Analysis

JEE Advanced 2025 — Paper 2 — Question 1

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

  1. Option A:

    [M0L0T0I0K0]\left[M^{0} L^{0} T^{0} I^{0} K^{0}\right]

  2. Option B:

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

    Correct
  3. Option C:

    [M1L2T−2I−1K−1]\left[M^{1} L^{2} T^{-2} I^{-1} K^{-1}\right]

  4. Option D:

    [M1L2T−4I−1K−1]\left[M^{1} L^{2} T^{-4} I^{-1} K^{-1}\right]

Answer: B

Step-by-step solution

S=emf\quad \mathrm{S}=\mathrm{emf} per unit temperature difference σ=\boldsymbol{\sigma}= Electrical conductivity

k=\mathrm{k}= Thermal conductivity

[S]=[ML2 T−3I−1 K−1][\mathrm{S}]=\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{I}^{-1} \mathrm{~K}^{-1}\right]

[σ]=[M−1 L−3 T3I2][\sigma]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{3} \mathrm{I}^{2}\right]

[K]=[M1L1T−3K−1][K]=\left[M^{1} L^{1} T^{-3} K^{-1}\right]

[Z]=S2σK=[M1L1T−3K−2][M1L1T−3K−1][Z]=\frac{S^{2} \sigma}{K}=\frac{\left[M^{1} L^{1} T^{-3} K^{-2}\right]}{\left[M^{1} L^{1} T^{-3} K^{-1}\right]}

[Z]=[K−1][\mathrm{Z}]=\left[\mathrm{K}^{-1}\right]

Answer key and solution verified before publishing.

Practise Units, Dimensions & Error Analysis

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Advanced 2025
Paper
Paper 2
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Units and Dimensions Analysis