# Two wires of diam

Given data,

Diameter of both the wires, d = 0.25 cm

Hence, radius of both the wires = r = d/2

r = 0.25/2 cm

r = 0.125 cm.

Length of the steel wire, Ls = 1.5 m

Length of the brass wire, Lb = 1.0 m

From the given figure, the total force exerted on the steel wire:

Fs = (4 kg + 6 kg) × 9.8 ms-2

Here, 9.8 ms-2 is the acceleration due to gravity which acts along with the force attached to the wires.

Fs = 10 × 9.8 = 98 N.

Young’s modulus for steel:

Ys = ………………… (1)

Where, = Change in the length of the steel wire. = Area of cross-section of the steel wire = πr2

As = π×(0.125× 10-2)2 m2

Young’s modulus for steel from standard table is, 2.0× 1011 Pa

Substituting the value is (1)

2.0× 1011 Pa = ΔLs = ΔLs = 1.49 × 10-4 m

Hence, elongation of the steel wire is 1.49 × 10-4 m.

From the given figure, the total force exerted on the brass wire:

Fb = 6 kg × 9.8 ms-2

Here, 9.8 ms-2 is the acceleration due to gravity which acts along with the force attached to the wires.

Fb = 6 kg × 9.8 ms-2 = 58.8 N.

Young’s modulus for steel:

Yb = ………………… (1)

Where, = Change in the length of the brass wire. = Area of cross-section of the brass wire = πr2

Ab = π×(0.125× 10-2)2 m2

Young’s modulus for brass from standard table is, 0.91× 1011 Pa

Substituting the value is (1)

0.91× 1011 Pa =  = ΔLb = 1.3 × 10-4 m

Hence, elongation of the steel wire is 1.3 × 10-4 m.

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