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(D) The magnetic field $B$ produced by the larger loop (radius $R_1$) at its centre is given by $B = \frac{\mu_0 i_1}{2 R_1}$.Since $R_1 >> R_2$,we ca

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$\frac{R_2^2}{R_1}$

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(D) The magnetic field $B$ produced by the larger loop (radius $R_1$) at its centre is given by $B = \frac{\mu_0 i_1}{2 R_1}$.Since $R_1 >> R_2$,we can assume this magnetic field is approximately uniform over the area of the smaller loop (radius $R_2$).The magnetic flux $\phi_2$ passing through the smaller loop is $\phi_2 = B \cdot A_2$,where $A_2 = \pi R_2^2$ is the area of the smaller loop.Substituting the values,we get $\phi_2 = \left( \frac{\mu_0 i_1}{2 R_1} \right) (\pi R_2^2)$.By definition,the mutual inductance $M$ is given by $M = \frac{\phi_2}{i_1}$.Therefore,$M = \frac{\mu_0 \pi R_2^2}{2 R_1}$.From this expression,it is clear that $M \propto \frac{R_2^2}{R_1}$.

Molecule type	$\frac{C_P}{C_V}$
$A$. Monoatomic	$I$. $\frac{7}{5}$
$B$. Diatomic rigid molecules	$II$. $\frac{9}{7}$
$C$. Diatomic non-rigid molecules	$III$. $\frac{4}{3}$
$D$. Triatomic rigid molecules	$IV$. $\frac{5}{3}$

Two conducting circular loops of radii $R_1$ and $R_2$ are placed in the same plane with their centres coinciding. If $R_1 >> R_2$,the mutual inductance $M$ between them will be directly proportional to

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