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

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, $σ$ the electrical conductivity and $k$ 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 $Z = \frac{S^{2} σ}{k}$ is: [2025]

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(2)

$Emf per unit temperature difference [S] = [M L^{2} T^{- 3} I^{- 1} K^{- 1}]$

$Electrical conductivity [σ] = [M^{- 1} L^{3} T^{3} I^{2}]$

$Thermal conductivity [K] = [M L T^{- 3} K^{- 1}]$

$[Z]$ $= \frac{S^{2} σ}{K} = \frac{{[M L^{2} T^{- 3} I^{- 1} K^{- 1}]}^{2} [M^{- 1} L^{- 3} T^{3} I^{2}]}{[M L T^{- 3} K^{- 1}]}$

$= \frac{[M^{2} L^{4} T^{- 6} I^{- 2} K^{- 2}] [M^{- 1} L^{- 3} T^{3} I^{2}]}{[M L T^{- 3} K^{- 1}]}$

$= \frac{[M L T^{- 3} K^{- 2}]}{[M L T^{- 3} K^{- 1}]}$

$∴$ $[Z] = [M^{0} L^{0} T^{0} I^{0} K^{- 1}]$

Solution

1 $[M^{0} L^{0} T^{0} I^{0} K^{0}]$

2 $[M^{0} L^{0} T^{0} I^{0} K^{- 1}]$

3 $[M^{1} L^{2} T^{- 2} I^{- 1} K^{- 1}]$

4 $[M^{1} L^{2} T^{- 4} I^{- 1} K^{- 1}]$

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Solution

1 [M0L0T0I0K0]

2 [M0L0T0I0K-1]

3 [M1L2T-2I-1K-1]

4 [M1L2T-4I-1K-1]

Ans. (2) Emf per unit temperature difference [S]=[ML2T-3I-1K-1] Electrical conductivity [σ]=[M-1L3T3I2] Thermal conductivity [K]=[MLT-3K-1] [Z]=S2σK=[ML2T-3I-1K-1]2[M-1L-3T3I2][MLT-3K-1] =[M2L4T-6I-2K-2][M-1L-3T3I2][MLT-3K-1] =[MLT-3K-2][MLT-3K-1] ∴ [Z]=[M0L0T0I0K-1]

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1 $[M^{0} L^{0} T^{0} I^{0} K^{0}]$

2 $[M^{0} L^{0} T^{0} I^{0} K^{- 1}]$

3 $[M^{1} L^{2} T^{- 2} I^{- 1} K^{- 1}]$

4 $[M^{1} L^{2} T^{- 4} I^{- 1} K^{- 1}]$