Question

(i) State Ohm's law.
(ii) A metal wire of resistance $6\Omega$ is stretched so that its length is increased to twice its original length. Calculate its new resistance.

Answer 1

(i) Ohm's Law : The electric current flowing through a conductor wire is directly proportional to voltage differences across the conductor wire if there is no change in physical condition (temperature, length, area of cross section etc.) of wire.

(ii) Volume of metal wire remains same.

i.e., volume of the metal wire before stretching $\left(V_1\right)$ = volume of the metal wire after stretching$\left(V_2\right)$

$\because$ $A_1{l}_1=A_2{l}_2$

where, $A_1$ = Area of the metal wire before stretching

$A_2$ = Area of the metal wire after stretching

$l_1$ = Length of the metal wire before stretching

$l_2$ = Length of the metal wire after stretching

$R_1$ = Resistance of the metal wire before stretching

$R_2$ = Resistance of the metal wire after stretching

$\rho$ = resistivity of the metal wire

$\frac{l_2}{l_1}=\frac{A_1}{A_2}$

$R_1=\rho \frac{l_1}{A_1}$

New resistance, $R_2=\rho \frac{l_2}{A_2}$

$\frac{R_2}{R_1}=\frac{l_2}{A_2}\times \frac{A_1}{l_1}=\frac{l_2}{l_1}\times \frac{l_2}{l_1}$

$R_2={\left(\frac{l_2}{l_1}\right)}^2{R}_1$

$={\left(\frac{2l_1}{l_1}\right)}^2{R}_1$

$R_2=4R_1$

$=4\times 6$

$=24\Omega$

Therefore, the new resistance of the metal wire after stretching is $24\Omega$.