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(B) The magnetic flux through the two loops is in opposite directions because they are wound in opposite senses. Let the area of the smaller loop be $

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$A, D$

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(B) The magnetic flux through the two loops is in opposite directions because they are wound in opposite senses. Let the area of the smaller loop be $A$ and the larger loop be $2A$.The net magnetic flux $\phi$ through the system at time $t$ is given by $\phi = B(2A - A) \cos \omega t = BA \cos \omega t$.The induced emf $\varepsilon$ is given by Faraday's law: $\varepsilon = -\frac{d\phi}{dt} = -\frac{d}{dt}(BA \cos \omega t) = BA \omega \sin \omega t$.$1$. The rate of change of flux is $\frac{d\phi}{dt} = -BA \omega \sin \omega t$. This is maximum when $\sin \omega t = \pm 1$,i.e.,$\omega t = \pi/2, 3\pi/2, \dots$. At these times,the plane of the loops is perpendicular to the plane of the paper. Thus,option $A$ is correct.$2$. The net emf is $\varepsilon = BA \omega \sin \omega t$,which is proportional to $\sin \omega t$,not $\cos \omega t$. Thus,option $B$ is incorrect.$3$. The emf is proportional to the difference of the areas $(2A - A) = A$,not the sum $(2A + A) = 3A$. Thus,option $C$ is incorrect.$4$. The amplitude of the net induced emf is $BA \omega$. The emf induced in the smaller loop alone is $\varepsilon_1 = -\frac{d}{dt}(BA \cos \omega t) = BA \omega \sin \omega t$,which has an amplitude of $BA \omega$. Thus,the amplitudes are equal. Option $D$ is correct.Therefore,the correct options are $A$ and $D$.

List-$I$	List-$II$
$(P)$ At $t = 0.2 \text{ s}$,the magnitude of the induced emf in Volt	$(1)$ $0.07$
$(Q)$ At $t = 0.2 \text{ s}$,the magnitude of the magnetic force in Newton	$(2)$ $0.144$
$(R)$ At $t = 0.2 \text{ s}$,the power dissipated as heat in Watt	$(3)$ $1.20$
$(S)$ The magnitude of terminal velocity of the rod in $\text{m s}^{-1}$	$(4)$ $0.12$
	$(5)$ $2.00$

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