Question

BAK 1 point A wire of length 6 cm and resistance 15 ohm is stretched so that the length becomes 12 cm and area of cross section is halved. The new resistance of wire is : O 1.25 ohm han O 60 ohm O 180 ohm O 30 ohm​

Answer 1

Answer:

On stretching a wire, both its area of cross-section and length changes but its volume remains constant.

R

1

=3Ω , l

1

=10cm , l

2

=30cm=3l

1

And, A=

l

V

Now, R=ρ

A

l

⟹R=ρ

V

l

2

⟹R∝l

2

On increasing length to 3 times of initial value.

Hence,

R

1

R

2

=

l

1

2

l

2

2

=9

⟹R

2

=9R

1

=9×3Ω

R

2

=27Ω