A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10–2 cm2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.
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Concept :- when a load 'w' is suspended from a stretched wire of length l, then depression at the midpoint is got by ,
∆D = wl³/12πr⁴Y
Where,
r = radius of the wire
Y = Young's modulus
l = length of wire
w = load { mg }
Here,
l = 1 m
m = 100g = 0.1 Kg
w = mg = 0.1 ×9.8 = 0.98N
CSA = πr² =0.5 × 10^-2 cm²
0.5 × 10^-2×10^-4 = πr²
r² = 0.5 × 10^-6/π
depression in a wire ,
∆D = wl³/12πr⁴Y
= 0.98×(1)³/12π{0.5×10^-6/π}²×2 × 10¹¹
= 0.98π/12 × 0.25 × 2×10^(-12+11)
= 0.051 m
∆D = wl³/12πr⁴Y
Where,
r = radius of the wire
Y = Young's modulus
l = length of wire
w = load { mg }
Here,
l = 1 m
m = 100g = 0.1 Kg
w = mg = 0.1 ×9.8 = 0.98N
CSA = πr² =0.5 × 10^-2 cm²
0.5 × 10^-2×10^-4 = πr²
r² = 0.5 × 10^-6/π
depression in a wire ,
∆D = wl³/12πr⁴Y
= 0.98×(1)³/12π{0.5×10^-6/π}²×2 × 10¹¹
= 0.98π/12 × 0.25 × 2×10^(-12+11)
= 0.051 m
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