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JEE Advanced Physics Units and Dimensions 2025 JEE Advanced 2025 (Paper 2)

JEE Advanced Physics Question (2025) — Solution

Question

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, the electrical conductivity and 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 is :-

Options

  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 ]

Answer

B. [M^0 L^0 T^0 I^0 K^ -1 ]

Step-by-step solution

aligned & S = emf per unit temperature difference \\ & = Electrical conductivity \\ & k = Thermal conductivity \\ & [ S ]= [ ML ^2 ~T ^ -3 I ^ -1 ~K ^ -1 ] \\ & [ ]= [ M ^ -1 ~L ^ -3 ~T ^3 I ^2 ] \\ & [ K ]= [ M ^1 ~L ^1 ~T ^ -3 ~K ^ -1 ] \\ & [ Z ]= S ^2 ~K = [ M ^1 ~L ^1 ~T ^ -3 ~K ^ -2 ] [ M ^1 ~L ^1 ~T ^ -3 ~K ^ -1 ] aligned [ Z ]= [ K ^ -1 ]

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Related: Physics — Units and Dimensions · All PYQ Banks