Question

1 point
0.23) A wire 100 cm long and 2.0 mm diameter has a resistance of 0.7
ohm, the electrical resistivity of the material is :
0 (a) 4.4x10-6 ohm x m
(b) 2.2 x 10-6 ohm x m
O (c) 1.1 x10-6 ohm x m
(d) 0.22 x10-6 ohm x m​

Answer 1

Given:-

• Length of the wire (L) = 100 cm = $\sf{\dfrac{100}{100} = 1\:m}$

• Diameter of the wire (d) = 2 mm

• Resistance of the wire (R) = 0.7 $\Omega$ = $\sf{\dfrac{7}{10} = 7\times 10^{-1}}$

To Find:-

The electrical resistivity of the material.

Solution:-

To find the resistivity we first need calculate the radius of the wire.

Therefore,

$\sf{r = \dfrac{d}{2}}$

= $\sf{r = \dfrac{2}{2}}$

= $\sf{r = 1}$

Radius of wire is 1 mm

Now,

We need to convert mm into m

Therefore,

$\sf{1\:mm = \dfrac{1}{1000}}$

= $\sf{1\:mm = 1\times 10^{-3}}$

Now,

To find the area of the wire,

$\sf{Area = \pi r^2}$

= $\sf{Area = \pi (10^{-3})^2}$

= $\sf{Area = \pi 10^{-6}}$

We know,

$\sf{R = \rho\dfrac{L}{A}}$

$\sf{\implies \rho = \dfrac{RA}{L}}$

Now Substituting the values,

[Taking $\sf{\underline{\pi = 3.14}}$ ]

$\sf{\rho = \dfrac{7\times 10^{-1} \times 3.14 \times 10^{-6}}{1}}$

Taking $\sf{10^{-1}}$ in the denominator using the law of indices:- $\sf{a^{-1} = \dfrac{1}{a}}$

= $\sf{\rho = \dfrac{7\times 3.14\times 10^{-6}}{10}}$

= $\sf{\rho = \dfrac{21.98\times 10^{-6}}{10}}$

= $\sf{\rho = 2.198\times 10^{-6}}$

= $\sf{\rho = 2.2\times 10{-6}}$

Therefore,

Option (b) $\sf{2.2\times 10^{-6}\: \Omega m}$ is the required answer.

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