This problem underlines the concept of continuity of magnetic field lines. They can neither be originated nor be destroyed in space.
The magnetic flux linked with the surface can be considered as the number of magnetic field lines passing through the surface. So, let \(d\phi \) = BA represents magnetic lines in an area A and B.
By the concept of continuity of magnetic field of lines, it cannot end or start in space, therefore the number of lines passing through surface S₁ must be the same as the number of lines passing through the surface S₂. Therefore, in both cases, we get the same answer for flux.

(i) As the magnet falls, the magnetic flux linked with the ring increases. This induces emf in the ring which opposes the motion of the falling magnet, hence a < g.
(ii) When current is suddenly switched ON, magnetic flux linked with the solenoid and thus, with metal ring increases. Current is induced in the ring in an anti-clockwise direction (as seen from the top of the ring). Since the direction of flow of current in the ring is opposite to the current in the solenoid, therefore they will repel each other and the ring jumps up.   

When the switch is thrown from the OFF position (open circuit) to the ON position (closed circuit), then neither B nor A and the angle between B and A does not change. Thus, no change in magnetic flux linked with coil occur, hence no electromotive force is produced and consequently, no current will flow in the circuit.

When the iron core is inserted in the current carrying solenoid, the magnetic field increases due to the magnetisation of iron core and consequently, the flux increases. According to Lenz's law, the emf produced must oppose this increase in flux, which can be done by making decrease in current. So, the current will decrease.

At any instant, flux passing through the ring is given
\(\text { by } \phi=\mathbf{B} \cdot \mathbf{A}=B A \cos \theta=B A \quad[\because \theta=0]\)
\(\text { or } \phi=B_{0}\left(\pi a^{2}\right) \cos \omega t \quad\left[\because B=B_{0} \cos \omega t\right]\)
By Faraday's law of electromagnetic induction, the magnitude of induced emf is given by

\(e=\frac{d \phi}{d t}=B_{0}\left(\pi a^{2}\right) \omega \sin \omega t\)
This causes flow of induced current, which is given by
\(I=B_{0}\left(\pi a^{2}\right) \omega \sin \omega t / R\)
Now, finding the values of current .at different instants.So, we have current at
\(t=\frac{\pi}{2 \omega} \Rightarrow I=\frac{B_{0}\left(\pi a^{2}\right) \omega}{R} \text { along }\)\(\hat{j}\)
Because, \(\sin \omega t=\sin \left(\omega \frac{\pi}{2 \omega}\right)=\sin \frac{\pi}{2}=1\)
At  \( t=\frac{\pi}{\omega} \Rightarrow I=\frac{B_{0}\left(\pi a^{2}\right) \omega}{R} \times \sin \pi=0 \)
Because \(\sin \omega t=\sin \left(\omega \frac{\pi}{\omega}\right)=\sin \pi=0\)
At \(t=\frac{3 \pi}{2 \omega} \Rightarrow I=\frac{B_{0}\left(\pi a^{2}\right) \omega}{R} \text { along }-\hat{j}\)
\(\sin \omega t=\sin \left(\omega \frac{3 \pi}{2 \omega}\right)=\sin \frac{3 \pi}{2}=-1\)