A lump of copper is stretched into a wire 5mn in diameter. Another wire of 1 cm diameter is made from another lump of copper of the same mass. Find the ratio of the resistances of the two wires.

Explanation

5mn= 0.5 cm
Resistance , `R=rho1/A=rho1/A^2=(rhom)/(D(pid^2/4)^2)=(16p m)/(piDd^4)`
here, `rho`= resistivity, l= length , d=diameter,
`A=(pid^2)/4` = cross sectional area, V=lA=volume,
m=mass , `D=m/v`= density
Both the lamps have same D and `rho` and according to the question , mass m is equal.
`thereforeR_1/R_2=(d_2/d_1)^4=(1/0.5)^4=(2)^2=16`or,`R_1:R_2=16:1`