Question

A uniform wire of steel of length 2.5m and density 8.0g/cm³ weighs 50g. When stretched by a force of 10kgf the length increases by 2mm. Calculate Young's modulus of steel

Answer 1

Answer:

here is your answer...

Answer 2

Given :

• Length = 2.5 m = 250 cm
• Density = 8 g/cm³
• Mass = 50 g
• Force = 10 kg f = 10 x 9.8 N = 98 N = 98 x 10^5 dyne
• Δ l = 2 mm = 0.2 cm

To find :

• Young's modulus

Solution :

we know that ,

⇒ ρ = M / V

here ,

• M = mass
• V = volume
• ρ = density

On rearranging ,

⇒ V = M / ρ

⇒ A x l = M / ρ

⇒ A = M / ρ x l

⇒ A = 50 / 8 x 250

⇒ A = 0.025 cm ²

Young's modulus is given by the formula ,

$\leadsto \gamma = \dfrac{stress }{strain}$

$\leadsto \gamma = \dfrac{F \times \Delta l }{l \times A }$

$\leadsto \gamma = \dfrac{98 \times 10^5 \times 250 }{0.2\times 0.025 }$

$\leadsto \gamma = \dfrac{2450 \times 10^5 }{ 0.005 }$

$\leadsto \boxed{\red{\gamma = 4.9 \times 10^{10} \: N / m^2} }$