
(a) Outline the relation between resistance and electrical resistivity of the material of a conductor in the shape of cylinder of length ‘l’ and therefore the area of cross section ‘A’. Thus derive the S.I unit of electrical resistivity.
(b) Resistance of a metal wire of length $5\,m$ is $100\Omega $. If the area of the cross section of the wire is $3 \times {10^{ - 7}}\,{m^2}$. Now calculate the resistivity of the metal.
Answer
424.2k+ views
Hint: Resistance is defined as the property of the conductor which opposes the flow of electric current. It is also defined as the ratio of the voltage applied to the electric current flowing through it.Resistivity is defined as the resistance offered by the material per unit length for unit cross-section. The SI unit of resistivity is Ohm.meter.
Complete step by step answer:
(a) Relation between resistance and electrical resistance within the form of cylinder of length $l$ and area of cross section $A$,
Then we get the formula,
$R = \dfrac{{\rho l}}{A}$
Where $R$ is resistance.
Now, $\rho = \dfrac{{AR}}{l} \to \left( 1 \right)$
currently substitute the units of $R, A$, and $l$, we get
where as, $R=\Omega $
$A={m^2}$ and $l=m$ then
$\Rightarrow \rho = \dfrac{{\Omega \times {m^2}}}{m} \\
\Rightarrow \rho = \Omega m \\ $
$\therefore \rho = \Omega - m$
(b) only if, resistance, R=$100\Omega $, Length $l=5\,m$ and area of cross section $A= 3 \times {10^{ - 7}}\,{m^2}$.
Now we will calculate the resistivity in metal by equation $\rho = \dfrac{{AR}}{l} \to (1)$
Substitute the particularly values with in the equation (1), then we tend to go
$\rho = \dfrac{{100 \times 3 \times {{10}^{ - 7}}}}{5}$
$\Rightarrow \rho = 0.000006\Omega - m \\
\therefore \rho = 6 \times {10^{ - 6}}\Omega - m \\$
Hence the resistivity of a metal is $6 \times {10^{ - 6}}\,\Omega - m$.
Note: Resistance physical phenomenon and related with synchrotron radiation that studies within the atomic layers of carbon. Property of an electric circuit system is ordinarily handling a lot of data apart from any channel in resistance. Hence the resistivity of a metal is $6 \times {10^{ - 6}}\Omega - m$.
Complete step by step answer:
(a) Relation between resistance and electrical resistance within the form of cylinder of length $l$ and area of cross section $A$,
Then we get the formula,
$R = \dfrac{{\rho l}}{A}$
Where $R$ is resistance.
Now, $\rho = \dfrac{{AR}}{l} \to \left( 1 \right)$
currently substitute the units of $R, A$, and $l$, we get
where as, $R=\Omega $
$A={m^2}$ and $l=m$ then
$\Rightarrow \rho = \dfrac{{\Omega \times {m^2}}}{m} \\
\Rightarrow \rho = \Omega m \\ $
$\therefore \rho = \Omega - m$
(b) only if, resistance, R=$100\Omega $, Length $l=5\,m$ and area of cross section $A= 3 \times {10^{ - 7}}\,{m^2}$.
Now we will calculate the resistivity in metal by equation $\rho = \dfrac{{AR}}{l} \to (1)$
Substitute the particularly values with in the equation (1), then we tend to go
$\rho = \dfrac{{100 \times 3 \times {{10}^{ - 7}}}}{5}$
$\Rightarrow \rho = 0.000006\Omega - m \\
\therefore \rho = 6 \times {10^{ - 6}}\Omega - m \\$
Hence the resistivity of a metal is $6 \times {10^{ - 6}}\,\Omega - m$.
Note: Resistance physical phenomenon and related with synchrotron radiation that studies within the atomic layers of carbon. Property of an electric circuit system is ordinarily handling a lot of data apart from any channel in resistance. Hence the resistivity of a metal is $6 \times {10^{ - 6}}\Omega - m$.
Recently Updated Pages
Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 English: Engaging Questions & Answers for Success

Trending doubts
A number is chosen from 1 to 20 Find the probabili-class-10-maths-CBSE

Find the area of the minor segment of a circle of radius class 10 maths CBSE

Distinguish between the reserved forests and protected class 10 biology CBSE

A boat goes 24 km upstream and 28 km downstream in class 10 maths CBSE

A gulab jamun contains sugar syrup up to about 30 of class 10 maths CBSE

Leap year has days A 365 B 366 C 367 D 368 class 10 maths CBSE
