\(\ell \alpha \varepsilon^{a} k ^{ b } T ^{ c } n ^{ d } q ^{ e }\)
(A) \(\ell=\sqrt{\frac{L^{-3} \times A^{2} T^{2}}{M^{-1} A^{2} T^{4} L^{-3} M^{1} L^{2} T^{-2} \theta^{-1} \theta}}\)
\(\ell=\sqrt{\frac{1}{L^{2}}}=\frac{1}{L}\)
(B) \(\ell =\sqrt{\frac{\varepsilon k_{B} T}{n q^{2}}}\)
\(=\sqrt{\frac{\left(M^{-1} A^{2} T^{4} L^{-3}\right) M^{1} L^{2} T^{-2} \theta^{-1} \theta}{L^{-3} A^{1} T^{2}}}\)
\(=\sqrt{L^{2}}=L\)
(C) \(\ell=\sqrt{\frac{A^{2} T^{2}}{M^{-1} A^{2} T^{4} L^{-3} L^{-2} M^{1} L^{2} T^{-2} \theta^{-1} \theta}}\)
(D) \(\ell=\sqrt{\frac{ A ^{2} T ^{2}}{ M ^{-1} A ^{2} T ^{4} L^{-3} L ^{-1} M ^{+1} L ^{2} T ^{-2} \theta^{-1} \theta}}\)
\(=\sqrt{L^{2}}= L\)
So, the correct option is (D): \(\ell=\sqrt{\left(\frac{q^{2}}{\operatorname{sn}^{1 / 3} k_{B} T}\right)}\)
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, $ \sigma $ the electrical conductivity and $ \kappa $ 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:
Two identical concave mirrors each of focal length $ f $ are facing each other as shown. A glass slab of thickness $ t $ and refractive index $ n_0 $ is placed equidistant from both mirrors on the principal axis. A monochromatic point source $ S $ is placed at the center of the slab. For the image to be formed on $ S $ itself, which of the following distances between the two mirrors is/are correct:
Dimensional Analysis is a process which helps verify any formula by the using the principle of homogeneity. Basically dimensions of each term of a dimensional equation on both sides should be the same.
Limitation of Dimensional Analysis: Dimensional analysis does not check for the correctness of value of constants in an equation.
Let us understand this with an example:
Suppose we don’t know the correct formula relation between speed, distance and time,
We don’t know whether
(i) Speed = Distance/Time is correct or
(ii) Speed =Time/Distance.
Now, we can use dimensional analysis to check whether this equation is correct or not.
By reducing both sides of the equation in its fundamental units form, we get
(i) [L][T]-¹ = [L] / [T] (Right)
(ii) [L][T]-¹ = [T] / [L] (Wrong)
From the above example it is evident that the dimensional formula establishes the correctness of an equation.