Question

A uniform wire is cut into two pieces such that 1 piece is twice as long as the other. The 2 pieces are connected in parallel in the left gap of meter bridge. When a resistance of $20\Omega$ is connected in the right gap, the neutral point is obtained at a distance of 60 cm from the right end of the wire. Find out the original resistance of wire before it was cut into 2 pieces.

Answer 1

Here $R=20\Omega$

$l=40cm$

$(100-l)=60cm$

$\therefore X=R\frac{l}{(100-l)}=20\frac{4}{0}60$

$13.34\Omega$

It is known $R\alpha l$

$\therefore R=kl$ [$k$ proportional constant]

Let the resistance of shorter wire , $R_1=kx$

$R_1$ and $R_2$ in parallel.

$\therefore \frac{R_1{R}_2}{R_1+R_2}=13.34\Omega ⟹\frac{2}{3}kx=13.34\Omega$

$\therefore kx=20.01\Omega$

$\therefore 2kx=40.02\Omega$

Original resistance of wire=$kx+2kx=3kx=60.03\Omega$