Physics, asked by malllokesh6805, 11 months ago

A cylindrical steel wire of 3 m length is to stretch no more than 0.2 cm When a tensile force of 400 N is applied to each end of the wire ? What minimum diameter is required for the wire ?? Y_(steel) = 2.1xx10^(11) N//m^2

Answers

Answered by abdul9838

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$\bf \red{hey \: mate \: here \: is \: answer} \\ \\ \bf \red{according \: to \: question} \\ \\ \\ \bf \red{length(l) = 3m} \\ \\ \bf \red{change \: in \: length \: (\delta \: l) = 0.2cm = } \\ \bf \red{0.2 \times {10}^{ - 2}m } \\ \\ \bf \red{force(f) = 400 \: n} \\ \\ \bf \red{ y =2.1 \times {10}^{11} n \: per \: {s}^{2} } \\ \bf \red{we \: have \: to \: find \: diameter} \\ \\ \bf \red{using \: this \: formula} \\ \\ \\ \bf \red{y = \frac{force \times length}{area \times change \: in \: length} } \\ \\ \bf \red{y = \frac{f \times \: l}{a \times \delta \: l} } \\ \\ \bf \red{2.1 \times {10}^{11} = \frac{400 \times 3}{\pi \: {r}^{2} \times 0.2 {10}^{ - 2} } }$

r=√(1/110×10^4

r=1/1000 approx

r=1. 0×10^-3

As we know that

Radius =Diameter/2

Diameter=2×radius

Diameter=2×10^-3

Answered by Fatimakincsem

Thus the diameter of the wire is d = 1.91 mm

Explanation:

Given data:

length "l" = 3 m
Δl = 0.2 cm = 0.2 x 10^-2 m
F = 400 N
Y(steel) = 2.1 x 10^(11) N//m^2
diameter "d" =?

Solution:

Δl = Fl / A Y

A = Fl / Δl Y

π / 4 . d^2 = 400 x 3 / 0.2 x 10^-2 x 2.1 x 10^11

d = √400 x 3 x 4 / 0.2 x 10^-2 x 2.1 x 10^11 x π

d = 1.91 x 10^-3 m

d = 1.91 mm

Thus the diameter of the wire is d = 1.91 mm

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