11. A cell of emf 'E' is connected a resistance R. The potential difference between the...

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11. A cell of emf 'E' is connected a resistance R. The potential difference between the terminals of the cell is found to be V. The internal resistance of the cell must be : 2(E-V)v 2(E-V)R (1) (2) R E (E-V)R (3) (4) (E-V)R V 12. Two particles are moving as shown (m) 4m/s 2m/s (m) B If e=1 then find velocity of particle B after collision: (1) 4 m/s (2) 2 m/s (3) 6 m/s (4) 0 m/s 13. A particle is released from 24m height, If e = 0.5 then find out maximum height attained by particle after collision on ground: (1) 24m (2) 12m (3) 6m (4) 8m 14. A bullet weighing 50 gm leaves the gun with a velocity of 30 m/s. If the recoil speed imparted to the gun is 1 m/s, the mass of the gun is: (1) 15 kg (2) 30 kg (3) 1.5 kg (4) 20 kg 15. A bullet of mass m moving with a speed v strikes a wooden block of mass M and gets embedded into the block. The final speed is: m M (1) (2) M+m M+m m (3) M+m (4) 2 11. A cell of emf 'E' is connected a resistance R. The potential difference between the terminals of the cell is found to be V. The internal resistance of the cell must be : 2(E-V)v 2(E-V)R (1) (2) R E (E-V)R (3) (4) (E-V)R V 12. Two particles are moving as shown (m) 4m/s 2m/s (m) B If e=1 then find velocity of particle B after collision: (1) 4 m/s (2) 2 m/s (3) 6 m/s (4) 0 m/s 13. A particle is released from 24m height, If e = 0.5 then find out maximum height attained by particle after collision on ground: (1) 24m (2) 12m (3) 6m (4) 8m 14. A bullet weighing 50 gm leaves the gun with a velocity of 30 m/s. If the recoil speed imparted to the gun is 1 m/s, the mass of the gun is: (1) 15 kg (2) 30 kg (3) 1.5 kg (4) 20 kg 15. A bullet of mass m moving with a speed v strikes a wooden block of mass M and gets embedded into the block. The final speed is: m M (1) (2) M+m M+m m (3) M+m (4) 2

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The internal resistance of the cell is given by E VR So the answer is 3 EVRV According to the law of ... View the full answer