A steel cable 20 mm diameter and 10 metre long is subjected to a tensile force of 5000 newton . If the modulus of elasticity of the material of the wire is 2 x 105 n/mm2. Then the elongation of the cable is
Answers
Dear Student,
◆ Answer -
l = 0.7957 mm
● Explanation -
# Given -
d = 20 mm = 0.02 m
l = 10 m
F = 5000 N
Y = 2x10^5 N/mm^2 = 2×10^11 N/m^2
# Solution -
Cross sectional area of wire is -
A = πd^2 / 4
A = 22/7 × 0.02^2 / 4
A = 3.142×10^-4 m^2
Young's modulus is given by -
Y = stress / strain
Y = (F/A) / (l/L)
l = FL/YA
Substitute values,
l = (5000 × 10) / (2×10^11 × 3.142×10^-4)
l = 7.957×10^-4 m
l = 0.7957 mm
Hence, the elongation of the cable is 0.7957 mm.
Thanks dear. Hope this helps you..
Answer:
◆ Answer -
l = 0.7957 mm
● Explanation -
# Given -
d = 20 mm = 0.02 m
l = 10 m
F = 5000 N
Y = 2x10^5 N/mm^2 = 2×10^11 N/m^2
# Solution -
Cross sectional area of wire is -
A = πd^2 / 4
A = 22/7 × 0.02^2 / 4
A = 3.142×10^-4 m^2
Young's modulus is given by -
Y = stress / strain
Y = (F/A) / (l/L)
l = FL/YA
Substitute values,
l = (5000 × 10) / (2×10^11 × 3.142×10^-4)
l = 7.957×10^-4 m
l = 0.7957 mm
Hence, the elongation of the cable is 0.7957 mm.