Question

4. A uniform wire is cut into two pieces such that one piece is twice as long as the other. The two pieces are connected in parallel in the left gap of meter bridge. When a resistance of 20 ohm is connected in the right gap, the null point is obtained distance of 60 cm from the right end of the wire. Find the resistance of the wire before it was cut into two pieces ​

Answer 1

Here R=20Ω

l=40cm

(100−l)=60cm

∴X=R

(100−l)

l

=20

0

4

60

13.34Ω

It is known Rαl

∴R=kl [k proportional constant]

Let the resistance of shorter wire , R

1

=kx

R

1

and R

2

in parallel.

∴

R

1

+R

2

R

1

R

2

=13.34Ω⟹

3

2

kx=13.34Ω

∴kx=20.01Ω

∴2kx=40.02Ω

Original resistance of wire=kx+2kx=3kx=60.03Ω