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# Two wires of equal lengths , One of aluminium and the other of copper have the same resistance . Which of the two wires is lighter. $[\rho_A =2.63 \times 10^{-8} \Omega m, \rho_{cn}= 1.72 \times 10^{-8} \Omega -m$. Relative density of $Al =2.7$ , Relative density of $Cu= 8.9 ]$

$\begin{array}{1 1} \text{Aluminium wires are 2.16 times heavier than copper wire} \\ \text{Copper wires are 2.16 times heavier than aluminium}\\ \text{Aluminium wires are of same mass as that copper wire } \\ \text{Cannot be determined} \end{array}$

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A)
Solution :
Let $l=l_{Al}$ density $d_{Al}=2.7$
and area $A{Al}=A_1$ and
For copper
Let $l= l_{cn}$
Density $d_{cn} =8.9$
and $A_{cn} =A_2$
Also $l_{Al}= l_{cn}=l$
$R= \rho \large\frac{l}{A}$
Resistance for aluminium wire $R_{Al} =\rho_{Al} \times \large\frac{l_{Al}}{A_{Al}}$
mass of the aluminium wire
$m_{Al} =A{Al} \times l_{Al} \times d_{Al}$
$\quad= A_1 \times l \times 2.7$
Resistance of the copper wire ,
$R_{cn}=\rho_{cn}$$\times \large\frac{l_{cn}}{A_{cn}}$
$\qquad= \large\frac{1.72 \times 10^{-8} \times l}{A_2}$
mass of copper wire $m_{cn} =A_{cn} \times l_{cn} \times d_{cn}$
$\qquad= A_2 \times l \times 8.9$
But it is given that $R_{Al} =R_{cn}$
$\therefore \large\frac{2.63 \times 10^{-8} \times l}{A_2}$
$\large\frac{A_1}{A_2}=\frac{2.63}{1.72}$
$\large\frac{m_{Al}}{m_{cn}}=\frac{ A_1 \times l \times 2.7}{A_2 \times l \times 8.9}$
$\large\frac{m_{Al}}{m_{cn}}=\frac{ 2.63 \times 2.7}{1.72 \times 89}$
$\qquad= 2.16$
Hence copper wires are $2.16$ times heavier than aluminium .
Answer : Copper wires are 2.16 times heavier than aluminium