If battery of 6 V is connected to the terminals of three metre long wire of uniform thickness and resistance of the order of 100 ohm, the difference of potential between two points separated by 50 cm in the wire will be

Question

If battery of 6 V is connected to the terminals of three metre long wire of uniform thickness and resistance of the order of 100 ohm, the difference of potential between two points separated by 50 cm in the wire will be (A) 1V (B) 1.5 V (C) 2 V (D) 3 V

Tardigrade Admin · Accepted Answer

Potential drop across 3 m long resistance wire is 6V. Potential drop across 50 cm (= 0.5 m) long resistance wire =(6/3)× 0.5 =(3.0/3) =1V

Q. If battery of 6 V is connected to the terminals of three metre long wire of uniform thickness and resistance of the order of 100 ohm, the difference of potential between two points separated by 50 cm in the wire will be

1V

1.5 V

2 V

3 V

Potential drop across 3 m long resistance wire is 6V. Potential drop across 50 cm (= 0.5 m) long resistance wire $= \frac{6}{3} \times 0.5$ $= \frac{3.0}{3}$ $= 1 V$

Q. If battery of 6 V is connected to the terminals of three metre long wire of uniform thickness and resistance of the order of 100 ohm, the difference of potential between two points separated by 50 cm in the wire will be

1V

1.5 V

2 V

3 V

Potential drop across 3 m long resistance wire is 6V. Potential drop across 50 cm (= 0.5 m) long resistance wire =36​×0.5 =33.0​ =1V

Potential drop across 3 m long resistance wire is 6V. Potential drop across 50 cm (= 0.5 m) long resistance wire $= \frac{6}{3} \times 0.5$ $= \frac{3.0}{3}$ $= 1 V$