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Sahay Sir > Question Answers > Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) > The elastic limit and ultimate strength for steel is 2.48×10^8 Pa and 4.89×10^8 Pa respectively. A steel wire of 10 m length and 2 mm cross sectional diameter is subjected to longitudinal tensile stress. Young’s modulus of steel is Y=2×1011Pa (a) Calculate the maximum elongation that can be produced in the wire without permanently deforming it. How much force is needed to produce this extension? (b) Calculate the maximum stretching force that can be applied without breaking the wire.

The elastic limit and ultimate strength for steel is 2.48×10^8 Pa and 4.89×10^8 Pa respectively. A steel wire of 10 m length and 2 mm cross sectional diameter is subjected to longitudinal tensile stress. Young’s modulus of steel is Y=2×1011Pa (a) Calculate the maximum elongation that can be produced in the wire without permanently deforming it. How much force is needed to produce this extension? (b) Calculate the maximum stretching force that can be applied without breaking the wire.

A mild steel wire of length 1.0m and cross sectional area 5.0×10−2cm2 is stretched within its elastic limit horizontally between two pillars. A mass of 100g is suspended form the midpont of the wire. Calculate the depression at the midpoint (Ysteel=200GPa)