Answer
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Hint: As a very first step, one could read the question well and hence note down the given values from it. Now recall the expression for electrical resistance in terms of resistivity, area and length which is used to solve both sub questions. Now substitute accordingly and hence find the answer.
Formula used:
Resistance,
$R=\dfrac{\rho l}{A}=\dfrac{\rho l}{\pi {{r}^{2}}}$
Complete step by step solution:
In the question, we are given a metal wire of diameter of 0.25mm and an electrical resistivity of $\rho =0.8\times {{10}^{-8}}\Omega m$. This information is followed by two sub-questions.
(a) Here, we are supposed to find the length of this wire that would give a resistance of $5\Omega $.
Radius of the metal wire would be,
$r=\dfrac{0.25}{2}mm=0.125\times {{10}^{-3}}m$
So, we have its area of cross section to be,
$A=\pi {{r}^{2}}=\pi {{\left( 0.125\times {{10}^{-3}} \right)}^{2}}=0.05\times {{10}^{-6}}{{m}^{2}}$
Now we have the expression for resistance in terms of area and length given by,
$R=\dfrac{\rho l}{A}=\dfrac{\rho l}{\pi {{r}^{2}}}$………………………………………………… (1)
Substituting the given values,
$l=\dfrac{5\times 0.05\times {{10}^{-6}}}{0.8\times {{10}^{-8}}}$
$\therefore l=31.25m$
Therefore, we found the required length to be 31.25m.
(b) Here, we have to find the change in resistance, for the diameter of the wire being doubled.
From the equation (1), we see the inverse relationship of resistance with the area of cross section. So, on doubling the diameter doubles the radius and the new resistance would be,
$R'=\dfrac{\rho l}{\pi r{{'}^{2}}}=\dfrac{\rho l}{\pi {{\left( 2r \right)}^{2}}}=\dfrac{\rho l}{4\pi {{r}^{2}}}$
$\therefore R'=\dfrac{R}{4}$
Therefore, we found the value of new resistance to be $\dfrac{1}{4}$times that of the initial value.
Note: We could define the resistivity of a material as the measure of the resistance of a given size of this material to electrical conduction. Another name given to this property is specific electrical resistance or may be volume resistivity. However, the ‘resistivity’ is the term that is used more widely than the other two.
Formula used:
Resistance,
$R=\dfrac{\rho l}{A}=\dfrac{\rho l}{\pi {{r}^{2}}}$
Complete step by step solution:
In the question, we are given a metal wire of diameter of 0.25mm and an electrical resistivity of $\rho =0.8\times {{10}^{-8}}\Omega m$. This information is followed by two sub-questions.
(a) Here, we are supposed to find the length of this wire that would give a resistance of $5\Omega $.
Radius of the metal wire would be,
$r=\dfrac{0.25}{2}mm=0.125\times {{10}^{-3}}m$
So, we have its area of cross section to be,
$A=\pi {{r}^{2}}=\pi {{\left( 0.125\times {{10}^{-3}} \right)}^{2}}=0.05\times {{10}^{-6}}{{m}^{2}}$
Now we have the expression for resistance in terms of area and length given by,
$R=\dfrac{\rho l}{A}=\dfrac{\rho l}{\pi {{r}^{2}}}$………………………………………………… (1)
Substituting the given values,
$l=\dfrac{5\times 0.05\times {{10}^{-6}}}{0.8\times {{10}^{-8}}}$
$\therefore l=31.25m$
Therefore, we found the required length to be 31.25m.
(b) Here, we have to find the change in resistance, for the diameter of the wire being doubled.
From the equation (1), we see the inverse relationship of resistance with the area of cross section. So, on doubling the diameter doubles the radius and the new resistance would be,
$R'=\dfrac{\rho l}{\pi r{{'}^{2}}}=\dfrac{\rho l}{\pi {{\left( 2r \right)}^{2}}}=\dfrac{\rho l}{4\pi {{r}^{2}}}$
$\therefore R'=\dfrac{R}{4}$
Therefore, we found the value of new resistance to be $\dfrac{1}{4}$times that of the initial value.
Note: We could define the resistivity of a material as the measure of the resistance of a given size of this material to electrical conduction. Another name given to this property is specific electrical resistance or may be volume resistivity. However, the ‘resistivity’ is the term that is used more widely than the other two.
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