Question

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.

Answer 1

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$\frac{l}{A}$ where l is the length, A= area of the wire and ρ= specific resistance of the wire.
i.e. l α $\frac{1}{A}$α $\frac{1}{\pi r^2}$ where r is the resistance of the wire.
According to the question, new radius r₁= $\frac{r}{2}$
Then new balancing length l₂ will be proportional to the $\frac{1}{\pi (r/2{)}^2}$, then l₂ will be 4 times than previous balancing length l_1.