Answer :
Answer:
63.8 V
Explanation:
We are given that
[tex]A_1=1.6 cm^2=1.6\times 10^{-4} m^2[/tex]
[tex]1 cm^2=10^{-4} m^2[/tex]
[tex]A_2=1.2 cm^2=1.2\times 10^{-4} m^2[/tex]
[tex]A_3=4.4 cm^2=4.4\times 10^{-4} m^2[/tex]
[tex]A_4=7 cm^2=7\times 10^{-4} m^2[/tex]
Potential difference,V=140 V
We know that
[tex]R=\frac{\rho l}{A}[/tex]
According to question
[tex]l_1=l_2=l_3=l_4=l[/tex]
In series
[tex]R=R_1+R_2+R_3+R_4[/tex]
[tex]R=\rho l(\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4})[/tex]
[tex]R=\rho l(\frac{1}{1.6\times 10^{-4}}+\frac{1}{1.2\times 10^{-4}}+\frac{1}{4.4\times 10^{-4}}+\frac{1}{7\times 10^{-4}})[/tex]
[tex]R=\rho l(18284.6)[/tex]
[tex]I=\frac{V}{R}=\frac{140}{\rho l\times 18284.6}[/tex]
Potential across 1.2 square cm=[tex]V_1=IR_1=\frac{140}{\rho l\times 18284.6}\times \rho l(\frac{1}{1.2\times 10^{-4}}=63.8 V[/tex]
Hence, the voltage across the 1.2 square cm wire=63.8 V