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A steel wire of original length 1 m and cross-sectional area 4.00 mm2 is clamped at the two ends so that it lies horizontally and without tension. If a load of 2.16 kg is suspended from the middle point of the wire, what would be its vertical depression? Y of the steel = 2.0 × 1011 N m-2. Take g = 10 m s-2.

A steel wire of original length 1 m and cross-sectional area 4.00 mm2 is clamped at the two ends so that it lies horizontally and without tension. If a load of 2.16 kg is suspended from the middle point of the wire, what would be its vertical depression?
Y of the steel = 2.0 × 1011 N m-2. Take g = 10 m s-2.

Grade:11

1 Answers

Navjyot Kalra
askIITians Faculty 654 Points
6 years ago
Sol. From figure cos θ = x/√x2 + l2 = x/l [1+ x2/l2]-1/2 = x / l ….(1) Increase in length ∆L = (AC + CB) – AB Here, AC = (l2 + x2)1/2 So, ∆L = 2(l2 + x2)1/2 – 100 …(2) Y = F/A l/∆l From equation (1), (2) and (3) and the freebody diagram, 2l cosθ = mg.

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