Question

When the length of the wire increase five times then its resistance becomes50ohms calculate the original length

Answer 1

Correct Q:- When the length of the wire increase five times then its resistance becomes 50ohms calculate the original resistance.

Ans:-

R = 2 ohm

Given :-

$R" = 50 \Omega$

Length become 5 times.

To find :-

The original length.

Solution:-

Let the length be l and area be a.

Let the length be l and area be a. Then, After increasing

l" = 5l

Since, the volume of wire remains constant.

V = A × l

V" = A" × l"

But,

V = V"

A × l = A" × l"

A × l = A" × 5l ( l" = 5l )

→$\dfrac{A}{A"}= \dfrac{5l}{l}$

→$\dfrac{A}{A"}= 5$

→$A" = \dfrac{A}{5}$

Now,

The resistance of wire is given by :-

$R = \rho \dfrac{l}{a}$

After increasing its length the resistance become,

$R = \rho \dfrac {l"}{a"}$

→$50 = \rho \dfrac{5l}{\dfrac{a}{5}}$

→$50 = \rho \dfrac{25l}{a}$

→$50 = 25 \rho \dfrac{l}{a}$

→$50 = 25 R$

→$R = \dfrac{50}{25}$

→$R = 2 \Omega$

hence,

Original resistance will be 2 ohm.