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प्रश्न
A steel wire of mass µ per unit length with a circular cross section has a radius of 0.1 cm. The wire is of length 10 m when measured lying horizontal, and hangs from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming the wire to be uniform and lateral strains << longitudinal strains, find the extension in the length of the wire. The density of steel is 7860 kg m–3 (Young’s modules Y = 2 × 1011 Nm–2).
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उत्तर
Consider the diagram when a small element of length dx is considered at x from the load (x = 0).
Let T(x) and T(x + dx) are tensions on the two cross-sections a distance dx apart, then – t(x + dx) + T(x) = dmg = μ dxg (where μ is the mass/length) ......(∵ dm = μdx)
dT = μgdx .....[∵ dT = T(x + dx) – T(x)]
⇒ T(x) = μgx + C .....(On integrating)
At x = 0, T(0) = Mg
⇒ C = mg
∴ T(x) = μgx + Mg
Let the length dx at x increase by dr, then
Young's modulus Y = `"Stress"/"Strain"`
`((T(x))/A)/((dr)/(dx)) = Y`
⇒ `(dr)/(dx) = 1/(YA) T(x)`
⇒ `r = 1/(YA) int_0^L (μgx + Mg)dx`
= `1/(YA) [(μgx^2)/2 + Mgx]_0^L`
= `1/(YA)[(mgL^2)/2 + MgL]` ......(m is the mass of the wire)
`A = pi xx (10^-3)^2 m^2`
`Y = 200 xx 10^9 Nm^-2`
`m = pi xx (10^-3)^2 xx 10 xx 7860` kg
∴ `r = 1/(2 xx 10^11 xx pi xx 10^-6)` ......`[(pi xx 786 xx 10^-3 xx 10 xx 10)/2 + 25 xx 10 xx 10]`
= `[196.5 xx 10^-6 + 3.98 xx 10^-3]`
= 4 × 10–3 m
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