A copper wire of length #L1 = 6m# and cross sectional area of #A1 = 4mm^2# has the same resistance as another copper wire of length #L2 = 18mm.# What is the ratio between the masses of the two wires, m2/m1?

Question

1663 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-08T09:54:12

The ratio is #=9*10^-6#

The mass and the volume are related by

#m=d*v#

The density is #=d#

The ratio of the masses is the same as the ratio of the volumes

#m_2/m_1=(d*v_2)/(d*v_1)=v_2/v_1#

The volume is

#v=l*a#

#v_1=l_1*a_1=6*4*10^-6=24*10^-6m^3#

If the wires have the same resistance,

#R=rho l/a#

#rhol_1/a_1=rho l_2/a_2#

#a_2=l_2/l_1*a_1=0.018/6*4*10^-6=12*10^-9m^2#

#v_2=l_2*a_2=0.018*12*10^-9=216*10^-12m^3#

Therefore,

#m_2/m_1=(216*10^-12)/(24*10^-6)=910^-6#