Question

Two copper wires a n b of equal masses r taken.. The lenghts a is doubled the length of b.. If the resistance of the wire a is 160ohm.... Calculate the resistance b

Answer 1

Here is your answer !! Hope it helps!!

Answer 2

Answer:

The resistance of b is 40 ohm.

Explanation:

Given that,

Two copper wires  a and b of equal masses are taken..

The length a is doubled the length of b.

$l_{a}=2l_{b}$

Resistance of wire a = 160 ohm

We need to calculate the resistance of b

The resistance is defined as,

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

Where, R = resistance

$\rho$ = resistivity of the wire

The resistance of wire a is

$R_{a}=\dfrac{\rho l_{a}}{A_{a}}$

The resistance of wire b is

$R_{b}=\dfrac{\rho l_{b}}{A_{b}}$

We know that,

The volume of both wires are same.

$V_{a}=V_{b}$

$A_{a}l_{a}=A_{b}l_{b}$

$A_{a}2l_{b}=A_{b}\times l_{b}$

$A_{b}=2A_{a}$

Now, The ratio of the resistance of the wires a and b

$\dfrac{R_{a}}{R_{b}}= \dfrac{\dfrac{\rho l_{a}}{A_{a}}}{\dfrac{\rho l_{b}}{A_{b}}}$

Put the value of $A_{b}$ and $l_{a}$ in the ratio

$\dfrac{R_{a}}{R_{b}}=\dfrac{\dfrac{\rho 2l_{b}}{A_{a}}}{\dfrac{\rho l_{b}}{2A_{a}}}$

$\dfrac{R_{a}}{R_{b}}=\dfrac{4}{1}$

$\dfrac{160}{R_{b}}=\dfrac{4}{1}$

$R_{b}=\dfrac{160}{4}$

$R_{b}=40\Omega$

Hence,  The resistance of b is 40 ohm.