A potentiometer has 400 cm long wire which is connected to an auxiliary of steady voltage 4 V. A Leclanche cell gives null point at 140 cm and Daniel cell at 100 cm. i)Compare the emfs of the two cells. ii) If the length of wire is increased by 100 cm, find the position of the null point with the first cell.
Answers
Answer:
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Answer:
Explanation:
Let the resistance per unit length of potentiometer wire be r and EMF of battery connected across it's length is E.
the initially the resistance = 4r
let the balancing length initially, be l
1
=1m
E
1
=I
1
r×l
1
where[I
1
(4r)=E]
final resistance = 5r
let the balancing length =l
2
then E
2
=I
2
×r×l
2
where [I
2
(5r)=E]
⇒
E
2
E
1
=
I
2
l
2
I
1
l
1
as [E
1
=E
2
]
⇒I
2
l
2
=I
1
l
1
l
2
=×1
l
2
=5/4=1.25m