In a series $$L R$$ circuit $$X_{L}=R$$ and power factor of the circuit is $$P_{1}$$. When capacitor with capacitance $$C$$ such that $$X_{L}=X_{C}$$ is put in series, the power factor becomes $$P_{2}$$. The ratio $$\frac{P_{1}}{P_{2}}$$ is:

A charge particle is moving in a uniform magnetic field $$(2 \hat{i}+3 \hat{j}) \,\mathrm{T}$$. If it has an acceleration of $$(\alpha \hat{i}-4 \hat{j})\, \mathrm{m} / \mathrm{s}^{2}$$, then the value of $$\alpha$$ will be :

$$\mathrm{B}_{X}$$ 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 :