JEE MAIN - Physics (2019 - 8th April Morning Slot - No. 14)
A steel wire having a radius of 2.0 mm,
carrying a load of 4 kg, is hanging from a
ceiling. Given that g = 3.1 p ms–2, what will be
the tensile stress that would be developed in the
wire ?
3.1 × 106 Nm–2
6.2 × 106 Nm–2
4.8 × 106 Nm–2
5.2 × 106 Nm–2
Explanation
Tensile stress in wire will be
= $${{Tensile{\rm{ }}force} \over {Cross{\rm{ }}section{\rm{ }}Area}}$$
= $${{mg} \over {\pi {R^2}}} = {{4 \times 3.1\pi } \over {\pi \times 4 \times {{10}^{ - 6}}}}N{m^{ - 2}}$$
= 3.1 × 106 Nm–2
= $${{Tensile{\rm{ }}force} \over {Cross{\rm{ }}section{\rm{ }}Area}}$$
= $${{mg} \over {\pi {R^2}}} = {{4 \times 3.1\pi } \over {\pi \times 4 \times {{10}^{ - 6}}}}N{m^{ - 2}}$$
= 3.1 × 106 Nm–2
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