Question

the resistance of the conducting wire x of length length 1 metre is 5ohm calculate the resistance of wire made by the same material whose length is 4 times and area of cross section 5 times of the wire x​

Answer 1

Answer:

If you stretch the wire it's resistivity will not change (all because it depends upon nature of wire and temperature only) but also it's volume remains constant so cross section area also changes. As the volume remains constant natural cross section area × natural length = new cross section area × 4× natural length( as the wire is stretched four times) . So new cross section area = natural cross section area / 4

Now assuming that you are aware with the formula for calculating the resistance if length resistivity and cross section area us given which is R = π L/A where π is resistivity , l is length and A is cross section area . Let the natural resistance be R. So

R = πL/A (EQUATION 1)

R (New resistance) = π4L /(A÷4)

= 16πL/A (EQUATION 2)

Divide both the equations and you get will get new resistance equal to 16 times the given resistance.

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Answer 2

The resistance of other wire is $4\ \Omega$ .

Given :

Resistance of wire , $R=5\ \Omega$.

We know, Resistance of an material is :

$R=\dfrac{\rho x}{A}$ {  Here $\rho$ is specific resistance }

So, $5=\dfrac{\rho x}{A}$

Let , resistance of new wire is $R'$.

Therefore , its resistance is given by :

$R'=\dfrac{\rho(4x)}{(5A)}=\dfrac{4}{5}\times \dfrac{\rho x }{A}=\dfrac{4}{5}\times 5=4\ \Omega\ .$    { Here $\rho$ remains same because material is same }

The resistance of other wire is $4\ \Omega$ .

Hence, this is the required solution.

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