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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.
3 years ago

Navjyot Kalra
654 Points
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.
3 years ago
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