Two steel wires having same length are suspended from a ceiling under the same load. If the ratio of their energy stored per unit volume is 1: 4 , then the ratio of their diameters is found to be k : 1 . Find the value of k .

Step-by-step solution

Elastic potential energy stored in a loaded'wire, U= 1 2 ( Stress Strain Volume ) Energy stored per unit volume, U= U Volume = 1 2 Stress Strain = 1 2 ( F A )^2 1 Y U_A U_B = 1 2 Y F^2 A_A^2 1 2 Y F^2 A_B^2 = A_B^2 A_A^2 , U_A U_B = d_B^4 d_A^4 Here, U_A U_B = 1 4 , d_B^4 d_A^4 = 1 4 or d_B d_A = 1 2 d_A d_B = 2 : 1= k : 1 (given) k=2