Question

the area of cross section of a steel wire is 0.5 cm^2 .what is necessary force to increase its length to 1.1 times .if young modulous for steel is 2×10^11​

Answer 1

Y=F/A*Δl/l

F=YAl/Δl

=2*10_11 *0.00005*1/1.1

=9*10_-16N

Answer 2

Given:

Area of cross-section of the steel wire = 0.5cm²

To Find:

force to increase its length

Solution:

Area = 0.5cm²

changing cm into m

= 0.5(1/100)²

A = 5×$10^{-5}$

L1 = L × 1.1

L1 - L = L(1.1-1)

ΔL/L = 0.1

y = 2×$10^{11}$

And we know,

y = F/A× L/ΔL

2 × $10^{11}$ = F/5×$10^{-5}$ × 1/0.1

F = 10N

Therefore, the force to increase in the length = 10N.