# Answer to Question #33079 in Electromagnetism for SAHDEV PATRO

Question #33079

when a wire of lenth 5 m and radius 0.5 mm is stretched by a load of 49 n the elongation produsced in the wire is 0.1 cm . find energy stored per unit volume of the wire

Expert's answer

the volume of stretched wire is length (5m) multiplied by cross-section area (pi r^2 where r is radius):

V = L*S = L*pi * r^2 5.1*3.14*0.0005^2 = 4.0035 * 10^{-6} m^3

The force per unit volume then is total force (49 N) divided by total volume ( 4.0035 * 10^{-6} m^3):

f = F/V = 49/(4.0035*10^{-6}) = 12.2 * 10^6 N/m^3

V = L*S = L*pi * r^2 5.1*3.14*0.0005^2 = 4.0035 * 10^{-6} m^3

The force per unit volume then is total force (49 N) divided by total volume ( 4.0035 * 10^{-6} m^3):

f = F/V = 49/(4.0035*10^{-6}) = 12.2 * 10^6 N/m^3

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