- A$10$
- B$7$
- ✓$2$
- D$73$
$ \mathrm{EMF}=-\mathrm{M} \frac{\mathrm{di}}{\mathrm{dt}}=-0.002\left(\mathrm{i}_0 \omega \cos \omega \mathrm{t}\right) $
$ \mathrm{EMF}_{\max }=\mathrm{i}_0 \omega(0.002)=(5)(50 \pi)(0.002) $
$ \mathrm{EMF}_{\max }=\frac{\pi}{2} \mathrm{~V}$
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[Given: Surface tension of the liquid is $0.075 \mathrm{Nm}^{-1}$, atmospheric pressure is $10^5 \mathrm{~N} \mathrm{~m}^{-2}$, acceleration due to gravity $(g)$ is $10 \mathrm{~m} \mathrm{~s}^{-2}$, density of the liquid is $10^3 \mathrm{~kg} \mathrm{~m}^{-3}$ and contact angle of capillary surface with the liquid is zero]
$(i)$ The energy stored in the capacitor when the battery is disconnected and the separation is doubled ${E_1}$
$(ii)$ The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is ${E_2}.$
Then ${E_1}/{E_2}$ value is
