Question

Find the increment in the length of a steel wire of length 5 m and radius 6 mm under its own weight. Density of steel = 8000 kg//m^(3) and young's modulus of steel = 2xx10^(11) N//m^(2). What is the energy stored in the wire ? (Take g = 9.8 m//s^(2))

Answer 1

Answer:

Explanation:

Tensile stress in wire will be

=  Cross section Area  / Tensile force

= 4 * 3.1 π​ / π * 4 * 10^-6

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Answer 2

The energy stored in the wire is 5.43×10−5J

• Let us go through the given parameters in the question

          Length(l) : 5m

           Radius(r) : 6mm

           Density(ρ): 8000 kg//m^(3)

           young's modulus of steel(Y)= 2xx10^(11) N//m^(2)

• By the formula

      Δl=mgl2AY

          =(ρ(πr^(2)l)gl)/2πr^(2)Y

          = (gl^(2)ρ)/2Y

          =(9.8×25×8000)/2×2×10^(11)

          =4.9×10^(-6) m

• Energy stored in wire = U= (1/2)k(Δl)^(2)

                                          = (1/2)×(YA/l)×(Δl)^(2)

                                          =5.43×10^(-5) J

• Energy stored in the wire = U = 5.43×10−5J