Question 260: Units , Dimensions and Error Analysis

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
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
$Z = \frac{S^2 \sigma}{\kappa}$
is :

Accepted Answer

S is emf per unit temperature Ã¢ÂÂ
$$[S]=\frac{[V]}{[K]}=\frac{M\,L^2\,T^{-3}\,I^{-1}}{K}              =M\,L^2\,T^{-3}\,I^{-1}\,K^{-1}.$$
ÃÂ is electrical conductivity Ã¢ÂÂ
$$[ÃÂ]=\frac{[J]}{[E]}          =\frac{I\,L^{-2}}{M\,L\,T^{-3}\,I^{-1}}          =M^{-1}\,L^{-3}\,T^3\,I^2.$$
ÃÂº is thermal conductivity Ã¢ÂÂ
$$[ÃÂº]=\frac{\dot Q/(A)}{Ã¢ÂÂT}          =\frac{M\,L^2\,T^{-3}/L^2}{K\,L^{-1}}          =M\,L\,T^{-3}\,K^{-1}.$$
Now
$$[Z]      =[S]^2\,[ÃÂ]\,[ÃÂº]^{-1}      =(M\,L^2\,T^{-3}\,I^{-1}\,K^{-1})^2       	imes M^{-1}\,L^{-3}\,T^3\,I^2       	imes (M\,L\,T^{-3}\,K^{-1})^{-1}$$
Simplifying exponents gives
$$[Z]=M^0\,L^0\,T^0\,I^0\,K^{-1}.$$
Answer: Option B.

Answer

[M
0
L
0
T
0
I
0
K
0
]

Answer

[M
0
L
0
T
0
I
0
K
−1
]

Answer

[M
1
L
2
T
−2
I
−1
K
−1
]

Answer

[M
1
L
2
T
−4
I
−1
K
−1
]

Units , Dimensions and Error Analysis

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