Question

Match the following:
Table-1 Table-2

$\left(\text{A}\right)L$	$\left(\text{P}\right)\left[M^0{L}^0{T}^{-2}{A}^0\right]$
$\left(\text{B}\right)$ Magnetic flux	$\left(\text{Q}\right)\left[M{L}^2{T}^{-2}{A}^{-1}\right]$
$\left(\text{C}\right)LC$	$\left(\text{R}\right)\left[M{L}^2{T}^{-2}{A}^{-2}\right]$
$\left(\text{D}\right)C{R}^2$	$\left(\text{S}\right)$ None

• A. $\text{A}\to \text{R};\text{B}\to \text{Q};\text{C}\to \text{S};\text{D}\to \text{R}$
• B. $\text{A}\to \text{Q};\text{B}\to \text{Q};\text{C}\to \text{S};\text{D}\to \text{R}$
• C. $\text{A}\to \text{R};\text{B}\to \text{P};\text{C}\to \text{S};\text{D}\to \text{R}$
• D. $\text{A}\to \text{P};\text{B}\to \text{Q};\text{C}\to \text{S};\text{D}\to \text{S}$

Answer 1

The correct option is A $\text{A}\to \text{R};\text{B}\to \text{Q};\text{C}\to \text{S};\text{D}\to \text{R}$

$\text{CASE-I}:$ EMF , $E=L\frac{dI}{dt}$

We know that, $\left[E\right]=\left[M{L}^2{T}^{-3}{A}^{-1}\right]$

$\therefore \left[L\right]=\frac{\left[M{L}^2{T}^{-3}{A}^{-1}\right]\times \left[T\right]}{\left[A\right]}=\left[M{L}^2{T}^{-2}{A}^{-2}\right]$

Thus, $\text{A}\to \text{R}$

$\text{CASE-II}:$ Magnetic flux , $\varphi =LI$

$\therefore \varphi =\left[M{L}^2{T}^{-2}{A}^{-2}\right]\times \left[A\right]=\left[M{L}^2{T}^{-2}{A}^{-1}\right]$

Thus, $\text{B}\to \text{Q}$

$\text{CASE-III}:\left[C\right]=\frac{Q}{E}=\frac{\left[AT\right]}{\left[M{L}^2{T}^{-3}{A}^{-1}\right]}=\left[M^{-1}{L}^{-2}{T}^4{A}^2\right]$

$\therefore \text{[LC]}=\left[\text{L}\right]\left[\text{C}\right]=\left[M{L}^2{T}^{-2}{A}^{-2}\right]\times \left[M^{-1}{L}^{-2}{T}^4{A}^2\right]$

Thus, $\text{C}\to \text{S}$

$\text{CASE-IV}:\left[C{R}^2\right]=\frac{\left[C\right][E{]}^2}{[I{]}^2}=\frac{\left[M^{-1}{L}^{-2}{T}^4{A}^2\right]\times \left[M^2{L}^4{T}^{-6}{A}^{-2}\right]}{\left[A^2\right]}=\left[M{L}^2{T}^{-2}{A}^{-2}\right]$

Thus, $\text{D}\to \text{R}$