$$\mathrm{B}_{X}$$ and $$\mathrm{B}_{\mathrm{Y}}$$ are the magnetic fields at the centre of two coils $$\mathrm{X}$$ and $$\mathrm{Y}$$ respectively each carrying equal current. If coil $$X$$ has 200 turns and $$20 \mathrm{~cm}$$ radius and coil $$Y$$ has 400 turns and $$20 \mathrm{~cm}$$ radius, the ratio of $$B_{X}$$ and $$B_{Y}$$ is :

The current I in the given circuit will be :

The total charge on the system of capacitors $$C_{1}=1 \mu \mathrm{F}, C_{2}=2 \mu \mathrm{F}, \mathrm{C}_{3}=4 \mu \mathrm{F}$$ and $$\mathrm{C}_{4}=3 \mu \mathrm{F}$$ connected in parallel is :

(Assume a battery of $$20 \mathrm{~V}$$ is connected to the combination)

When a particle executes Simple Hormonic Motion, the nature of graph of velocity as a function of displacement will be :