# Question e2ded

Aug 2, 2017

Given that the resistivity of copper is half that of aluminium and that the density of copper is three times that of aluminium,

Ratio of resistivities

${\rho}_{\text{Cu"/rho_"Al}} = \frac{1}{2}$

and

Rattio of densities

${d}_{\text{Cu"/d_"Al}} = \frac{3}{1}$

Relating resistances with length and area of cross section.

Resistance of Cu

${R}_{\text{Cu"=rho_"Cu"xxl_"Cu"/A_"Cu}}$

$\implies {R}_{\text{Cu"=rho_"Cu"xxl_"Cu"^2/V_"Cu}}$

=>R_"Cu"=rho_"Cu"xxl_"Cu"^2/(M_"Cu"/d_"Cu")#

$\implies {R}_{\text{Cu"=rho_"Cu"xx(l_"Cu"^2xxd_"Cu")/M_"Cu}}$

Similarly resistance of Al

$\implies {R}_{\text{Al"=rho_"Al"xx(l_"Al"^2xxd_"Al")/M_"Al}}$

By the given condition

${R}_{\text{Cu"=R_"Al" and l_"Cu"=l_"Al}}$

So

${R}_{\text{Cu"=R_"Al}}$

${\rho}_{\text{Cu"xx(l_"Cu"^2xxd_"Cu")/M_"Cu"=rho_"Al"xx(l_"Al"^2xxd_"Al")/M_"Al}}$

$\implies {M}_{\text{Cu"/M_"Al"=rho_"Cu"/rho_"Al"xxd_"Cu"/d_"Al}}$

$\implies {M}_{\text{Cu"/M_"Al}} = \frac{1}{2} \times \frac{3}{1} = \frac{3}{2}$