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An inductive circuit draws a power 550 W from a 220 V-50 Hz source. The power factor of the circuit is 0.8. The current in the circuit lags behind the voltage. To bring its power factor to unity the capacitor connected in the circuit must have capacitance (b) 4224Tt F 8448 (c) 8448Tt F 4224

Circuit is inductive, i.e., there is no capacitor in the circuit. Given power factor 3 4 2 Power, P= R (220)2 R 2 3 4 550- R= 56.32

Now to bring the power factor to unity, 3 4 4 3>< 2 50 56.32 4224

A neutron ofenergy 1 Me V and mass 1.6 x 10 27 kg passes a proton at such a distance that the angular momentum of neutron relative to proton approximately equals 10 J s. The distance of closest approach neglecting the interaction between particles is (a) 0.44 mm (c) 0.44 A (b) 0.44 nm (d) 0.44 fm

If d is the distance of closest approach given, the angular momentum mvd 1033J s E mv 1 MeV 1.6 x10-13J p=y2m1nE =V2 1.6 10-27 1.6 10-13 = 1.6V2 10-20 kg m s-1 2 Distance of closest approach, 10 x10 13 0.44 fm 1602 10-20-1.6V2