What should be the diameter of a copper wire ( (Y=12 10¹⁰ Nm ⁻² ) ) of length (5 ~m ) to produce the same elongation produced by a (5 ~m ) long aluminium wire ( (Y=7 10¹⁰ Nm ⁻² ) ) of diameter (3 ~mm ) with the same (40 ~kg ) mass ?

Step-by-step solution

For copper wire, (Y=12 10¹⁰ Nm ⁻² ) Length, ( l=5 ~m ), (F=m g=40 10=400 ~N ) We know that, (Y= F l A l ) ( aligned l & = F l A Y = 400 5 r^2 12 10¹⁰ l & = 2000 12 r^2 10¹⁰ (i) aligned ) For aluminium wire, (Y=7 10¹⁰ Nm ⁻² ) Diameter, (d=3 ~mm =3 10⁻³ ~m ) ( r= d 2 = 3 10⁻³ 2 ~m =1.5 10⁻³ ~m ) Similarly, ( Y= F l A l ) ( l= F l A Y = 400 5 (1.5 10⁻³ )^2 7 10¹⁰ ) ( l= 2000 15.75 10^4 ) ...(ii) According to given condition, ( l ) is same in both cases. From Eqs. (i) and (ii), we get ( array rlrl & 2000 12 r^2 10¹⁰ &