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प्रश्न
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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उत्तर १
Let AB be a mild steel wire of length 2L = lm and its cross-section area A = 0.50 x 10-2 cm2. A mass m = 100 g = 0.1 kg is suspended at mid-point C of wire as shown in figure. Let x be the depression at mid-point i.e., CD = x
`:. AD = DB = sqrt((AC^2 + CD^2)) = sqrt(L^2+x^2)`
:. Increase in length `triangleL = (AD+DB) - AB = 2sqrt(L^2+x^2) - 2L`
`= 2L[(1+x^2/L^2)^(1/2) - 1] = 2L.x^2/(2L^2) = x^2/L`
:. Longitudinal strain = `(triangleL)/(2L) = x^2/(2L^2)`
If T be the tension in the wire as shown figure then in equilibrium `2T cos theta = mg`
or `T = "mg"/(2costheta)`
`= "mg"/(2 x/(sqrt(x^2+L^2))) = (mgsqrt(x^2+L^2))/(2x) = "mgL"/(2x)`
:. Stress = `T/A = "mgL"/(2 x A)`
As Young's modulus Y = `"stress"/"strain"`
`=((mgL)/(2 x A))/((x^2)/(2L^2)) = (mgL)/(2 x A) xx (2L^2)/x^2 = (mgL^3)/(Ax^3)`
`=> x = [(mgL^3)/(YA)]^(1/3) = L[(mg)/(YA)]^(1/3)`
`= 1/2[(0.1xx9.8)/(2xx10^11xx0.50xx10^(-2)xx10^(-4))]^(1/3) = 1.074 xx 10^(-2) m`
= 1.074 cm ≈ 1.07 cm or 0.01 m
उत्तर २
Length of the steel wire = 1.0 m
Area of cross-section, A = 0.50 × 10–2 cm2 = 0.50 × 10–6 m2
A mass 100 g is suspended from its midpoint.
m = 100 g = 0.1 kg
Hence, the wire dips, as shown in the given figure.
Original length = XZ
Depression = l
The length after mass m is attached to the wire = XO + OZ
Increase in the length of the wire:
Δl = (XO + OZ) – XZ
Where,
XO = OZ = `[(0.5)^2 + l^2]^(1/2)`
`:.triangle l = 2[(0.5)^2 + (l)^2]^(1/2) - 1.0`
`= 2 xx 0.5 [1+(l/0.5)^2]^(1/2) - 1.0`
Expanding and neglecting higher terms, we get
`triangle l = l^2/0.5`
`"Strain" = "Increase in length"/"Original length"`
Let T be the tension in the wire.
∴mg = 2T cosθ
Using the figure, it can be written as:
`cos theta = l/((0.5)^2 + l^2)^(1/2)`
`= l/ ((0.5)(1+(l/0.5)^2)^(1/2))`
Expanding the expression and eliminating the higher terms:
`cos theta = l/((0.5)(1+l^2/(2(0.5)^2)))`
`(1+l^2/(0.5))=~1` for small l
`:. cos theta = l/(0.5)`
`:. T = (mg)/(2(l/(0.5))) = (mgxx 0.5)/(2l) = (mg)/(4l)`
`"Stress" = "Tension"/"Area" = (mg)/(4lxxA)`
`Y = (mg xx 0.5)/(4lxxAxxl^2)`
`l = sqrt((mgxx0.5)/(4YA)) `
Young’s modulus of steel, Y = 2 x 1011 Pa
`:.l = sqrt((0.1xx9.8xx0.5)/(4xx2xx10^11xx0.50xx10^(-6)))`
= 0.0106 m
Hence, the depression at the midpoint is 0.0106 m.
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