Elastic Limit

A mass of 5 kg is hung from a copper wire of 5 mm diameter and 2 m in length. Calculate the extension produced. What should be the minimum diameter of the wire so that its elastic limit is not exceeded ? Elastic limit for copper =1.5xx10^(9) dyne cm^(-2) and Y for copper =1.1xx10^(12) dyne cm^(-2)

The elastic limit of an elevator cable is 2xx10^(9) N//m^(2) . The maximum upward acceleration that an elevator of amss 2xx10^(3)kg can have been supported by a cable would not exceed half of the elastic limit would be

A wire of uniform cross-sectional area A and young's modulus Y is stretched within the elastic limits. If s is stress in the wire, the elastic energy density stored in the wire in terms of the given parameters is

According to Hooke's law, within the elastic limit stress/strain = constant. This constant depends on the type of strain or the type of force acting. Tensile stress might result in compressional or elongative strain, however, a tangential stress can only cause a shearing strain. After crossing the elastic limit, the material undergoes elongation and beyond a stage beaks. All modulus of elasticity are basically constants for the materials under stress. If stress/strain is x in elastic region and y in the region of yield, then

State whether the following statements are true or false with reasons. a. Elastic forces are always conservative. b. Elastic forces are strictly conservative only when Hooke's law is obeyed. c. When a wire is loaded beyond the elastic limit and then reloaded, the work done disappears completely as heat.