
Resistance of a metal wire of length 1m is $26\Omega $ at ${{20}^{0}}C$ . If the diameter of the wire is $0.3mm$, what will be the resistivity of the metal at that temperature?
Answer
537.6k+ views
Hint: We have to calculate the resistivity of the metal wire for the given values of different parameters. We shall use the formula of resistivity, which is, the resistivity of a wire can be written as the product of its net resistance and its cross-sectional area upon the length of the wire. Also, before putting all the values in the mentioned formula, we will make sure they all have the same units.
Complete answer:
Let the length of the wire be given by $L$.
Then it has been given to us in the problem that the value of $L$ is 1 meter.
Now, let the area of cross-section of the wire be A. Then, it can be calculated using the given formula:
$\Rightarrow A=\pi {{\left( \dfrac{d}{2} \right)}^{2}}$ \[\]
Where, $d$ is the diameter of the metal wire and its value has been given as $0.3mm$ .
Thus, the cross-sectional area can be calculated as:
$\begin{align}
& \Rightarrow A=\pi {{\left( \dfrac{0.3}{2} \right)}^{2}}m{{m}^{2}} \\
& \Rightarrow A=3.14\times 0.0225m{{m}^{2}} \\
& \Rightarrow A=0.07065m{{m}^{2}} \\
& \Rightarrow A\approx 0.07m{{m}^{2}} \\
\end{align}$
Now for conversion of units, since:
$\Rightarrow 1mm={{10}^{-3}}m$
Then,
$\Rightarrow 1m{{m}^{2}}={{10}^{-6}}{{m}^{2}}$
Therefore,
$\Rightarrow 0.07m{{m}^{2}}=0.07\times {{10}^{-6}}{{m}^{2}}$
$\Rightarrow 0.07m{{m}^{2}}=7\times {{10}^{-8}}{{m}^{2}}$
Also, the value of net resistance has been given to us in the problem as:
$\Rightarrow R=26\Omega $
Now, we can calculate the resistivity of the metal wire with the help of the given formula:
$\Rightarrow \rho =\dfrac{RA}{L}$
Putting the values of all the terms in Right Hand Side of the equation, we get:
$\begin{align}
& \Rightarrow \rho =\dfrac{26\times 7\times {{10}^{-8}}}{1}\Omega m \\
& \Rightarrow \rho =182\times {{10}^{-8}}\Omega m \\
& \Rightarrow \rho =1.82\times {{10}^{-6}}\Omega m \\
\end{align}$
Hence, the resistivity of the metal wire comes out to be $1.82\times {{10}^{-6}}\Omega m$.
Note:
These are some basic formulas from one of the most important chapters of Physics, Current Electricity. So, one should very well be aware of these basic formulas. Also, one should always keep check of their calculations at every step of solution or else one small error might end up giving a wrong solution to our question.
Complete answer:
Let the length of the wire be given by $L$.
Then it has been given to us in the problem that the value of $L$ is 1 meter.
Now, let the area of cross-section of the wire be A. Then, it can be calculated using the given formula:
$\Rightarrow A=\pi {{\left( \dfrac{d}{2} \right)}^{2}}$ \[\]
Where, $d$ is the diameter of the metal wire and its value has been given as $0.3mm$ .
Thus, the cross-sectional area can be calculated as:
$\begin{align}
& \Rightarrow A=\pi {{\left( \dfrac{0.3}{2} \right)}^{2}}m{{m}^{2}} \\
& \Rightarrow A=3.14\times 0.0225m{{m}^{2}} \\
& \Rightarrow A=0.07065m{{m}^{2}} \\
& \Rightarrow A\approx 0.07m{{m}^{2}} \\
\end{align}$
Now for conversion of units, since:
$\Rightarrow 1mm={{10}^{-3}}m$
Then,
$\Rightarrow 1m{{m}^{2}}={{10}^{-6}}{{m}^{2}}$
Therefore,
$\Rightarrow 0.07m{{m}^{2}}=0.07\times {{10}^{-6}}{{m}^{2}}$
$\Rightarrow 0.07m{{m}^{2}}=7\times {{10}^{-8}}{{m}^{2}}$
Also, the value of net resistance has been given to us in the problem as:
$\Rightarrow R=26\Omega $
Now, we can calculate the resistivity of the metal wire with the help of the given formula:
$\Rightarrow \rho =\dfrac{RA}{L}$
Putting the values of all the terms in Right Hand Side of the equation, we get:
$\begin{align}
& \Rightarrow \rho =\dfrac{26\times 7\times {{10}^{-8}}}{1}\Omega m \\
& \Rightarrow \rho =182\times {{10}^{-8}}\Omega m \\
& \Rightarrow \rho =1.82\times {{10}^{-6}}\Omega m \\
\end{align}$
Hence, the resistivity of the metal wire comes out to be $1.82\times {{10}^{-6}}\Omega m$.
Note:
These are some basic formulas from one of the most important chapters of Physics, Current Electricity. So, one should very well be aware of these basic formulas. Also, one should always keep check of their calculations at every step of solution or else one small error might end up giving a wrong solution to our question.
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