Question

A wire having a mass of 1 kg possess a resistance of 0.1 0. If the resistivity of
material of wire is 10-7 Om and density of material of wire is 104 kg/mº, then
length of wire is​

Answer 1

The correct question is:

A wire having a mass of 1kg possess a resistance of 0.1 ohm. if the resistivity of material of wire is 10^-7 ohm m and density of material of wire is 10^4 kg/m^3, then the length of the wire is?

Given:

Mass of the wire = 1 kg

Resistance = 0.1 ohm

Resistivity of the wire = 10^-7

Density of the material = 10^4 kg/m³

To find:

Length of the wire.

Solution:

Volume = m/d =1/ 10^4 = 10^-4  m³

Where m = mass of wire

d = density

Volume = A*l = 10^-4  m³

Where A = area

l = length of wire

Resistance = pl/A  ohms

Where p = resistivity

l = length of wire

A = area of wire

0.1 = 10^-7 * l/ A

Since A = 10^-4/l, Putting the value of A we get:

0.1 = 10^-7 * l²/ 10^-4

100 = I²

l = 10 m

Therefore the length of the wire is 10 m.

Answer 2

Given:

A wire having a mass of 1 kg possess a resistance of 0.1 ohm. The resistivity of material of wire is 10^(-7) mho and density of material of wire is 10^(4) kg/m³.

To find:

Length of wire.

Calculation:

First of all, let's calculate the volume of the wire:

$\sf \therefore \: volume = \dfrac{mass}{density} = \dfrac{1}{ {10}^{4} } = {10}^{ - 4} \:{m}^{3}$

Now, we know that for a cylinder ,

$\sf \therefore \: \pi {r}^{2} l = {10}^{ - 4}$

$\sf \implies \: \pi {r}^{2} = \dfrac{ {10}^{ - 4} }{l}$

$\sf \implies \: area = \dfrac{ {10}^{ - 4} }{l} \: \: \: \: \: .......(1)$

Now, resistance is given as :

$\sf \therefore \: R = \rho \times \dfrac{l}{area}$

$\sf \implies \: 0.1 = {10}^{ - 7} \times \dfrac{l}{( \frac{ {10}^{ - 4} }{l}) }$

$\sf \implies \: 0.1 = {10}^{ - 3} \times {l}^{2}$

$\sf \implies \: {l}^{2} = 100$

$\sf \implies \: l = 10 \: m$

So, length of wire is 10 metres.