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A steel wire of original length 1 m and cross-sectional area 4 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 steel=2 x 10 11 N/m²

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Solution

$I n c a s e o f s t e e l w i r e t h e v e r t i c a l d e p r e s s i o n i s g i v e n b y, d = \sqrt{\frac{F \times 2 l^{2}}{Y A}} W e a r e g i v e n F = 2.16 \times 10 = 21.6 N, l = 1 m, Y = 2 \times 10^{11} N / m^{2} a n d A = 4 \times 10^{- 6} m^{2} s o d = \sqrt{\frac{21.6 \times 2 \times 1 \times 1}{2 \times 10^{11} \times 4 \times 10^{- 6}}} = \sqrt{2.16 \times 10^{- 4}} = 1.47 \times 10^{- 2} m o r d e p r e s s i o n d = 0.0147 m = 14.7 m m$

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In case of steel wire the vertical depression is given by, d=F×2l2Y A We are given F=2.16×10=21.6N, l=1m, Y=2×1011N/m2and A= 4×10-6m2so d= 21.6×2×1×12×1011×4×10-6=2.16×10-4=1.47×10-2mor depression d= 0.0147m=14.7mm