Question

A uniform steel wire hangs from ceiling and elongate due to its own weight .the ratio of alongation of the upper of wire to lower half of wire is

Answer 1

The ratio is 3Mg / L4YA : MgL / 4YA = 3:1

Step-by-step explanation:

• Let us first calculate the elongation of the element dx, which let us consider is dy.
• Now assuming M is the mass of the wire, L is the length, Y young modulas
• A is the are of the cross section of the wire.
• Elongation of element dx will be due to mass of the lower part (i.e. due to length x). The weight of that part is

w=M / Lxg  

Hence we have dy = Mg / YAL xdx.

• Now to know the elongation of the upper half we have to integrate it from L / 2 to L. Which is

y(upper) = Mg / YAL ∫L-L/2 xdx = Mg / YAL [L^2−L^2 / 4]

= 3MgL / 4YA

• Now for lower half we have to integrate it from 0 to L/2.

y(lower) = Mg / YAL∫L/2-0 xdx

=Mg / YAL[L^24−0] = MgL / 4YA

Hence the ratio is 3Mg / L4YA : MgL / 4YA = 3:1

Also learn more

A wire of length 2 m and cross sectional area 10⁻⁴ m² is stretched by a load 102 kg. The wire is stretched by 0.1 cm. Calculate longitudinal stress, longitudinal strain, Young's modulus of material of wire. (Ans : 1 x 10⁷ N / m², 5 x 10⁻⁴ , 20 x 10⁹ N / m²) ?

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