Answer:
(a) The second wire will be stretched by 2 mm
(b) The third wire will be stretched by 0.25 mm
Explanation:
Tensile stress on every engineering material is given as the ratio of applied force to unit area of the material.
σ = F / A
Tensile strain on every engineering material is given as the ratio of extension of the material to the original length
δ = e / L
The ratio of tensile stress to tensile strain is known as Young's modulus of the material.
[tex]Y = \frac{FL}{Ae}[/tex]
Part A
cross sectional area and applied force are the same as the original but the length is doubled
[tex]\frac{FL_1}{A_1e_1} = \frac{FL_o}{A_oe_o} \\\\\frac{L_1}{e_1} = \frac{L_o}{e_o}\\\\e_1 = \frac{L_1e_o}{L_o} \\\\But, L_1 =2L_o\\\\e_1 = \frac{2L_oe_o}{L_o}\\e_1 = 2e_o[/tex]
The second wire will be stretched by 2 mm
Part B
a third wire with the same length but twice the diameter of the first
[tex]\frac{FL}{A_1e_1} = \frac{FL}{A_oe_o} \\\\\frac{1}{A_1e_1} = \frac{1}{A_oe_o}\\\\\frac{4}{\pi d_1^2e_1} = \frac{4}{\pi d^2_oe_o}\\\\\frac{1}{d_1^2e_1} = \frac{1}{d^2_oe_o}\\\\d_1^2e_1 = d^2_oe_o\\\\e_1 = \frac{d^2_oe_o}{d_1^2} \\\\e_1 =(\frac{d_o}{d_1})^2e_o\\\\But, d_1 = 2d_o\\\\e_1 =(\frac{d_o}{2d_o})^2e_o\\\\e_1 =(\frac{1}{2})^2e_o\\\\e_1 =(\frac{1}{4})e_o[/tex]
e₁ = ¹/₄ x 1 mm = 0.25 mm
The third wire will be stretched by 0.25 mm
