A wire is stretched by the application of a force of 50 kg wt/sq. cm.What is the percentage increase in the length of the wire?Y=7X1010 N/m2 *
Answers
7×10^(-10)
Y = F/(∆l/l)
(∆l/l) = F/Y
(∆l/l) × 100% = F/Y ×100 %
(∆l/l) × 100% = 50×9.8/(7×10^10) ×100 % = 7×10^(-10)
(∆l/l) × 100% = 7×10^(-10)
Answer:
We can use Hooke's Law to calculate the percentage increase in the length of the wire when a force of 50 kg/sq. Cm is applied.
Explanation:
Hooke's law states that the force required to stretch a wire (F) is directly proportional to the change in length of the wire (ΔL) and the spring constant of the wire (k). Mathematically it is represented as
F = k * ΔL
Y is Young's modulus of the material, which is 7 x 10^10 N/m^2 for this case, and A is the cross-sectional area of the wire.
We know the force applied on the wire is 50 kg/sq. Cm, which is equivalent to 50 x 10^4 N/sq.cm
F = 50 x 10^4 N/sq.cm
We also know that the original length of the wire is L, and the change in length after stretching is ΔL.
By using the formula
F = Y * A * ΔL/L
We can calculate the change in length as
ΔL = F * L / ( Y * A )
Now, we can use this value of ΔL to calculate the percentage increase in length by using the formula:
% increase in length = (ΔL / L) x 100
So the percentage increase in length of the wire is (ΔL / L) x 100 = (F * L / ( Y * A ))*100/ L.
It's worth noting that the cross-sectional area A and original length L are not provided in the question, so we cannot find the exact percentage increase. However, you can use this formula as a guide to calculate the percentage increase in the length of a wire when the force and young's modulus are known.
To learn more about Hooke's Law, from the given link.
https://brainly.in/question/1587479
To learn more about percentage, from the given link.
https://brainly.in/question/200523
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