KEY CONCEPT: The magnetic field at the centre of the coil
$$B\left( t \right) = {\mu _0}n{I_1}.$$
As the current increases, $$B$$ will also increase with time till it reaches a maximum value (when the current becomes steady).
The induced emf in the ring
$$e = - \frac{{d\phi }}{{dt}} = - \frac{d}{{dt}}\left( {\overrightarrow B .\overrightarrow A } \right) = - A\frac{d}{{dt}}\left( {{\mu _0}n{I_1}} \right)$$
∴ The induced current in the ring
$${I_2}\left( t \right) = \frac{{\left| e \right|}}{R} = \frac{{{\mu _0}nA}}{R}\frac{{d{I_1}}}{{dt}}$$
[NOTE : $$\frac{{d{I_1}}}{{dt}}$$ decreases with time and hence $${I_2}$$ also decreases with time.]
Where $${I_1} = {I_{\max }}\left( {1 - {e^{ - \frac{t}{\tau }}}} \right)$$
The relevant graphs are

Induced emf in the loop is given by
$$e = - B.\frac{{dA}}{{dt}}$$   where $$A$$ is the area of the loop.
$$\eqalign{ & e = - B.\frac{d}{{dt}}\left( {\pi {r^2}} \right) = - B\,\pi \,2r\frac{{dr}}{{dt}} \cr & r = 2\,cm = 2 \times {10^{ - 2}}m \cr & dr = 2\,mm = 2 \times {10^{ - 3}}m,dt = 1s \cr & e = - 0.04 \times 3.14 \times 2 \times 2 \times {10^{ - 2}} \times \frac{{2 \times {{10}^{ - 3}}}}{1}V \cr & = 0.32\,\pi \times {10^{ - 5}}V \cr & = 3.2\,\pi \times {10^{ - 6}}V \cr & = 3.2\,\pi \,\mu V \cr} $$

Problem Solving Strategy
Differentiate the given equation of current changing in first coil and find out the maximum change in $$\frac{{di}}{{dt}}.$$
The given equation of current changing in the first coil is $$i = {i_0}\sin \omega t\,......\left( {\text{i}} \right)$$
Differentiating Eq. (i) w.r.t. $$t,$$ we have
$$\eqalign{ & \frac{{di}}{{dt}} = \frac{d}{{dt}}\left( {{i_0}\sin \omega t} \right) \cr & {\text{or}}\,\,\frac{{di}}{{dt}} = {i_0}\frac{d}{{dt}}\left( {\sin \omega t} \right) \cr & {\text{or}}\,\,\frac{{di}}{{dt}} = {i_0}\omega \cos \omega t \cr} $$
For maximum $$\frac{{di}}{{dt}},$$  the value of $$\cos \omega t$$  should be equal to 1.
$${\text{So,}}\,\,{\left( {\frac{{di}}{{dt}}} \right)_{\max }} = {i_0}\omega $$
The maximum value of emf is given by
$$\eqalign{ & \therefore {e_{\max }} = M{\left( {\frac{{di}}{{dt}}} \right)_{\max }} = M{i_0}\omega \cr & {\text{As,}}\,\,M = 0.005H,{i_0} = 10\;A,\omega = 100\,\pi \,rad/s \cr & \therefore {e_{\max }} = 0.005 \times 10 \times 100\pi = 5\pi \cr} $$

When the total flux associated with one coil links with the other i.e. a case of maximum flux linkage, then mutual induction in coil 1 due coil 2 is $${M_{12}} = \frac{{{N_2}{\phi _{{B_2}}}}}{{{i_1}}}$$
and mutual induction in coil 2 due to coil 1 is $${M_{21}} = \frac{{{N_1}{\phi _{{B_1}}}}}{{{i_2}}}$$
Similarly, self-inductance in coil 1 is $${L_1} = \frac{{{N_1}{\phi _{{B_1}}}}}{{{i_1}}}$$
and self-inductance in coil 2 is $${L_2} = \frac{{{N_2}{\phi _{{B_2}}}}}{{{i_2}}}$$
If all the flux of coil 2 links coil 1 and vice-versa, then $${\phi _{{B_2}}} = {\phi _{{B_1}}}$$
Since, $${M_{12}} = {M_{21}} = M,$$    hence we have
$$\eqalign{ & {M_{12}}{M_{21}} = {M^2} \cr & = \frac{{{N_1}{N_2}{\phi _{{B_1}}}{\phi _{{B_2}}}}}{{{i_1}{i_2}}} = {L_1}{L_2} \cr & \therefore {M_{\max }} = \sqrt {{L_1}{L_2}} \cr & {\text{Given,}}\,\,{L_1} = 2mH,{L_2} = 8mH \cr & \therefore {M_{\max }} = \sqrt {2 \times 8} = \sqrt {16} \cr & = 4\,mH \cr} $$

Clearly the flux linkage is maximum in case (a) due to the spatial arrangement of the two loops.

Charge on he capacitor at any time t is given by
$$\eqalign{ & q = CV\left( {1 - {e^{\frac{t}{\tau }}}} \right) \cr & {\text{at}}\,t = 2\tau \cr & q = CV\left( {1 - {e^{ - 2}}} \right) \cr} $$

Work done in moving a charge through potential difference $$V$$ is given by $$W = qV$$

Emf induced in the coil or inductor of self-inductance $$L$$ is given by
$$e = - L\frac{{di}}{{dt}}\,\,{\text{or}}\,\,\left| e \right| = L\frac{{di}}{{dt}}$$
Here, $$L = 40\,mH = 40 \times {10^{ - 3}}H$$
$$di = {i_2} - {i_1} = 11 - 1 = 10\,A$$
Time taken to change current from $${i_1}$$ to $${i_2}$$ is
$$\eqalign{ & dt = 4 \times {10^{ - 3}}s \cr & {\text{So,}}\,\,\left| e \right| = 40 \times {10^{ - 3}} \times \left( {\frac{{10}}{{4 \times {{10}^{ - 3}}}}} \right) \cr & = 100\,V \cr} $$

There is no charge in flux through $$PS$$ and so induced emf will be zero. But
$$\eqalign{ & {V_P} - {V_B} = {V_Q} - {V_R} \cr & {\text{Also,}}\,F = Bil\,\,{\text{or}}\,\,i = \frac{F}{{B\ell }} \cr & \therefore {V_P} - {V_B} = {V_Q} - {V_R} = i{r_{QR}} = \frac{F}{{B\ell }} \times \frac{r}{6} = \frac{{Fr}}{{6B\ell }} \cr} $$

Applying Kirchhoff's law of voltage in closed loop
$$ - {V_R} - {V_C} = 0 \Rightarrow \frac{{{V_R}}}{{{V_C}}} = - 1$$