Question

A body of mass 3.14 kg is suspended from one end of a wire of length 10 m. The radius of cross-section of the wire is changing uniformly from 5xx10^(-4) m at the top (i.e. point of suspension) to 9.8xx10^(-4) m at the bottom . Young's modulus of elasticity is 2xx10^(11) N//m^(2). The change in length of the wire is

Answer 1

Answer:

The length of the wire is 20012m

${25}^{2}$

Answer 2

Hence the change in length of the wire is Δl = 10^-3 m

Explanation:

Given data:

• Mass "" = 3.14 Kg
• l = 10 m
• r1 = 5 x 10^-4 m
• r2 = 9.8 x 10^-4 m
• Y = 2 x 10^11 N/m^2
• Δl = ?

Solution:

Now for dx A = πr^2

A = π [ r1 + x (r2 - r1) / l ]^2

d (Δl) = F dx /AY = Fdx / π [ r1 + x (r2 - r1) / l ]^2 Y

Δl = ∫l - 0 d Δl = ∫l - 0 Fdx / π [ r1 + x (r2 - r1) / l ]^2 Y

Δl = ∫l - 0 d Δl = F / πY  ∫l - 0 ( dx / ( r1 +  x (r2 - r1) / l]^2

Let r1 + x (r2 - r1) / l  = t

0 + r2 - r1 / l = dt / dx

dx = ldt / r2 - r1

Δl = Fl / πY x ldt / r2 - r1 . 1/t^2

Δl = Fl / πY r1r2

Δl = mgl / πY r1r2

Δl = 3.14 x 9.8 x 10 / 3.14 x 2 x 10^11 x 5 x 10^-4 x 9.8 x 10^-4

Δl = 10^-3 m

Hence the change in length of the wire is Δl = 10^-3 m