Question #e2ded

1 Answer
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_"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#