Question

Question #e2ded

Answer 1

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_"Cu"/rho_"Al"=1/2`

and

Rattio of densities

`d_"Cu"/d_"Al"=3/1`

Relating resistances with length and area of cross section.

Resistance of Cu

`R_"Cu"=rho_"Cu"xxl_"Cu"/A_"Cu"`

`=>R_"Cu"=rho_"Cu"xxl_"Cu"^2/V_"Cu"`

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

`=>R_"Cu"=rho_"Cu"xx(l_"Cu"^2xxd_"Cu")/M_"Cu"`

Similarly resistance of Al

`=>R_"Al"=rho_"Al"xx(l_"Al"^2xxd_"Al")/M_"Al"`

By the given condition

`R_"Cu"=R_"Al" and l_"Cu"=l_"Al"`

So

`R_"Cu"=R_"Al"`

`rho_"Cu"xx(l_"Cu"^2xxd_"Cu")/M_"Cu"=rho_"Al"xx(l_"Al"^2xxd_"Al")/M_"Al"`

`=>M_"Cu"/M_"Al"=rho_"Cu"/rho_"Al"xxd_"Cu"/d_"Al"`

`=>M_"Cu"/M_"Al"=1/2xx3/1=3/2`