Question

a metal bar of 10 mm diameter when subjected to a pull of 23.5 kn give an elongation of 0.3 mm on a gauge length of 200 mm. the young's modulus of elasticity e of the metal will be (kn/mm^2)

Answer 1

Diameter of bar = 10 mm

Area = (π/4)D^2 = 78.54 mm^2

Force, F = 23.5 kN = 23500 N

∆L = 0.3 mm

L = 200 mm

Young's modulus, E = ?

$E= \frac{stress}{strain}= \frac{ \sigma}{ \epsilon} \\ \\ E= \frac{F \times L}{A \times \Delta L}$

$E = \frac{23500 \times 200}{78.54 \times 0.3} \: \frac{N}{ {mm}^{2} } \\ \\ E = 199473.7 \: N/mm^2 \\ \\ E \approx 2 \times {10}^{5}\ N/mm^2$

Answer 2

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Diameter of bar = 10 mm

Area = (π/4)D^2 = 78.54 mm^2

Force, F = 23.5 kN = 23500 N

∆L = 0.3 mm

L = 200 mm

Young's modulus, E = ?

$E= \frac{stress}{strain}= \frac{ \sigma}{ \epsilon} \\ \\ E= \frac{F \times L}{A \times \Delta L}$

$E = \frac{23500 \times 200}{78.54 \times 0.3} \: \frac{N}{ {mm}^{2} } \\ \\ E = 199473.7 \: N/mm^2 \\ \\ E \approx 2 \times {10}^{5}\ N/mm^2$