JEE Advanced 2025 — Paper 2 — PHY-I solution

JEE Advanced2025Paper 2PHY-I

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, $σ$ the electrical conductivity and $κ$ 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 $Z = \frac{S ^{2} σ}{κ}$ is :-

A
$[M^{0} L^{0} T^{0} I^{0} K^{0}]$
B
$[M^{0} L^{0} T^{0} I^{0} K^{- 1}]$
C
$[M^{1} L^{2} T^{- 2} I^{- 1} K^{- 1}]$
D
$[M^{1} L^{2} T^{- 4} I^{- 1} K^{- 1}]$

Solution

Write the dimensions of each quantity:

$[S] = [emf / K] = [M L^{2} T^{- 3} I^{- 1} K^{- 1}]$

$[σ] = [M^{- 1} L^{- 3} T^{3} I^{2}]$

$[κ] = [M L T^{- 3} K^{- 1}]$

Now compute $[Z] = \frac{[ S ] ^{2} [ σ ]}{[ κ ]}$ :

Numerator: $[S]^{2} [σ] = [M^{2} L^{4} T^{- 6} I^{- 2} K^{- 2}] \cdot [M^{- 1} L^{- 3} T^{3} I^{2}] = [M^{1} L^{1} T^{- 3} K^{- 2}]$

Dividing by $[κ] = [M L T^{- 3} K^{- 1}]$ gives $[Z] = [K^{- 1}]$ , i.e. $[M^{0} L^{0} T^{0} I^{0} K^{- 1}]$ .

$∴$ Option (B) is correct.

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