Answer :

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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