In an experiment on meter bridge, if the balancing length AC is 'x'. what would be be its value, when the diameter on the meter bridge wire AB is made half ? Justify your answer.
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Solution
As balancing length of the metre bridge l is directly proportional to the resistance R of the wire.
i.e. l α R
l= ρ where l is the length, A= area of the wire and ρ= specific resistance of the wire.
i.e. l α α where r is the resistance of the wire.
According to the question, new radius r1=
Then new balancing length l2will be proportional to the , then l2will be 4 times than previous balancing length l1.