The copper wires whose masses are 8 gm and 12 gm have lengths in the ratio 4:3. Find the ratio of their resistances.

As known the resistance formula $R=\rho\frac{l}{A}$

Where $\rho\rightarrow$ resistivity of the wire

$R\rightarrow$ resistance of the wire

$l\rightarrow$ length of the wire 

$A\rightarrow$ cross section area of the wire

For the first wire:

Mass $m_1=8\ gm$, length$\rightarrow l_1$, resistance$\rightarrow R_1$ and cross section area $A_1$

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

For the second wire:

Mass $m_2=8\ gm$, length$\rightarrow l_2$, resistance$\rightarrow R_2$ and cross section area $A_2$

$R_2=\rho\frac{l_2}{A_2}$

Therefore, $\frac{R_1}{R_2}=\frac{\rho\frac{l_1}{A_1}}{\rho\frac{l_2}{A_2}}$

$\frac{R_1}{R_2}=\frac{l_1}{l_2}\times\frac{A_2}{A_1}$   ..... $( i)$

We know $Volume=area\times length=A\times l$

We know the wires are made of the same material, therefore there density will be the same.

Therefore, Density$=\frac{mass}{volume}=\frac{m_1}{A_1\times l_1}=\frac{m_2}{A_2\times l_2}$

Or $\frac{A_2}{A_1}=\frac{m_2}{m_1}\times\frac{l_1}{l_2}$

On substituting value of $\frac{A_2}{A_1}$ in $( i)$

$\frac{R_1}{R_2}=\frac{l_1}{l_2}\times\frac{m_2\times l_1}{m_1\times l_2}=\frac{m_2}{m_1}\times(\frac{l_1}{l_2})^2$

Or $\frac{R_1}{R_2}=\frac{12}{8}\times(\frac{4}{3})^2$    [ratio of the length $l_1:l_2=4:3$]

Or $\frac{R_1}{R_2}=\frac{12}{8}\times\frac{16}{9}$  

Or $\frac{R_1}{R_2}=\frac{8}{3}$

Thus, the ratio of the resistances is $8:3$.

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Updated on: 10-Oct-2022

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