Question

Match the following two columns:

	Column I		Column II
A.	Electrical resistance	$1.$	$\left[M{L}^3{T}^{-3}{A}^{-2}\right]$
B.	Electrical potential	$2.$	$\left[M{L}^2{T}^{-3}{A}^{-2}\right]$
C.	Specific resistance	$3.$	$\left[M{L}^2{T}^{-3}{A}^{-1}\right]$
D.	Specific conductance	$4.$	None of these

• A. $A-2,B-3,C-1,D-4$
• B. $A-2,B-4,C-3,D-1$
• C. $A-1,B-2,C-3,D-3$
• D. $A-1,B-3,C-2,D-4$

Answer 1

Answer: (A)

$A-2,B-3,C-1,D-4$

The dimensions of electrical resistance
$R=\frac{V}{I}=\frac{\left(\frac{W}{q}\right)}{I}=\frac{\frac{W}{It}}{\left(I\right)}$
$=\frac{W}{I^{2}t}=\left[M{L}^2{T}^{-2}{T}^{-1}{A}^{-2}\right]$
$=\left[M{L}^2{T}^{-3}{A}^{-2}\right]$
Then, $\left(A\right)\to \left(2\right)$
The dimensions of electrical potential
$V=\frac{W}{q}=\frac{W}{It}=\left[M{L}^2{T}^{-2}{A}^{-1}{T}^{-1}\right]$
$=\left[M{L}^2{T}^{-3}{A}^{-1}\right]$
Then, $\left(B\right)\to \left(3\right)$
The dimensions of specific resistance
$\rho =R\frac{l}{A}=\left[M{L}^2{T}^{-3}{A}^{-2}\right]\left[L\right]$
$=\left[M{L}^3{T}^{-3}{A}^{-2}\right]$
Thus, $\left(C\right)\to \left(2\right)$ and the dimensions of specific conductance
$\sigma =\frac{1}{\rho }=\frac{1}{\left[M{L}^3{T}^{-3}{A}^{-2}\right]}$
$=\left[M^{-1}{L}^{-3}{T}^3{A}^2\right]$
$=$ not give in column
Thus, $\left(D\right)\to \left(4\right)$.