Home
Class 12
PHYSICS
A metal wire of resistance 3Omega is el...

A metal wire of resistance `3Omega` is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle `60^(@)` at the center, the equivalent resistance between these two points will be :

A

`5/2 Omega`

B

`12/5 Omega`

C

`5/3 Omega`

D

`7/2 Omega`

Text Solution

Verified by Experts

The correct Answer is:
C
`R_(1) = 3 Omega`
`R_(f) = 4R_(1) = 12 Omega`
`R_(1) = (5pi)/(3 xx 2pi) xx R_(f) = 10 Omega`
`R_(2) = pi/(3 xx 2pi) xx R_(f) = 2 Omega`
`R = (R_(1)R_(2))/(R_(1) + R_(2)) = 5/3 Omega`
Promotional Banner

Similar Questions

Explore conceptually related problems

A wire of resistance R is elongated n-fold to make a new uniform wire. The resistance of new wire.

A wire of resistance 4 Omega is stretched to four times of its original length resistance of wire now becomes

A metal wire has a resistance of 35Omega . If its length is increased to double by drawing it, then its new resistance will be

A wire has a resistance of 32Omega . It is melted and drawn into a wire of half of its original length. What is the resistance of the new wire?

A uniform wire of resistance =3Omega and length l is stretched to double its length. Now it is bent to form a circular loop and two point P & Q lies on the loop such that they subtend 60^(@) angle at centre. The equivalent resistance between two point P & Q is:

A wire is bent in the form of a triangle now the equivalent resistance between its one end and the mid point of the side is

A wire of resistance 1 Omega is stretched to double its length. What is the new resistance ?

A metallic wire of resistance 12Omega is bent of form a square. The resistance between two diagonal points would be