Question

define young's modulus of elasticity.how much force is required to have an have an increase of 0.4%in the length of a metallic wire having radius of 0.2mm. y=9.1x10¹⁰ Nm-2​

Answer 1

Answer

• Force required = 45.86 N

Explanation

Young's Modulus (Y): The ratio of tensile or compressive stress 'σ' to the longitudinal strain 'ε' is known as Young's modulus of elasticity.

Given that,

• Increase in the length of wire = 0.4%
• Radius of metallic wire, r = 0.2 mm = 0.2 × 10⁻³ m
• Young's modulus, Y = 9.1 × 10¹⁰ Nm⁻²

To find,

• Force required, F =?

Let's calculate the area of cross section of wire

→ A = πr² = 22/7 × 0.2 × 0.2 × 10⁻⁶ = 1.26 × 10⁻⁷ m²

As given that change in length of the wire is 0.4% so,

→ ΔL/L = 0.4/100 = 0.004

Using formula for Young's modulus

→ Y = (F/A) / (ΔL/L)

→ F = YA × (ΔL/L)

→ F = (9.1 × 10¹⁰) × (1.26 × 10⁻⁷) × (0.004)

→ F = 45.86 N  (Approx.)

Therefore,

• Force required is 45.86 N

Answer 2

$\large{\bold{\green{\underline{Young's\:Modulus\:of\:elasticity\::}}}}$

➣ The ratio of tensile or compressive stress (σ) to the longitudinal strain (ε).

➣Young's modulus denoted by the symbol Y.

E᥊℘ཞɛʂʂıơŋ ơʄ 'Y' ;-

$\red\bigstar\:\:{\underline{\green{\boxed{\bf{\blue{Y\:=\:\dfrac{Stress\:(\sigma)}{Strain\:(\varepsilon)}\:}}}}}}$

$\orange{:\implies}\:\:\bf{Y\:=\:\dfrac{\frac{F}{A}}{\frac{\triangle{l}}{l}}\:}$

$\bf\pink{Where,}$

• F is the force.

• A is area.

• Δl is the change in length.

• l is the original length.

• $\bf{\dfrac{\triangle{l}}{l}}$ is the change in increasing in length.

━─━─━─━─━─━─━─━─━─━─━─━─━─━─━

$\Large{\bold{\orange{\underline{GiVeN\::}}}}$

• $\bf{\dfrac{\triangle{l}}{l}}$ = 0.4 % = $\bf{\dfrac{0.4}{100}}$ = 0.4 × 10⁻²

• Radius (r) of metallic wire is 0.2 mm, i.e. 0.2 × 10⁻³ m

$\longmapsto\:\:\bf{Area\:=\:\pi\:r^2\:}$

$\longmapsto\:\:\bf{Area\:=\:3.14\times{(0.2\times{10^{-3}})^2}\:}$

$\longmapsto\:\:\bf{Area\:=\:3.14\times{0.0.4\times{10^{-6}}}\:}$

$\longmapsto\:\:\bf\purple{Area\:=\:0.125\times{10^{-6}}\:m^2}$

• Y = 9.1 × 10¹⁰ N/m²

➛ Now putting all the above values in the expression of Young's modulus, we get

$:\implies\:\:\bf{9.1\times{10^{10}}\:=\:\dfrac{\frac{F}{0.125\times{10^{-6}}}}{0.4\times{10^{-2}}}\:}$

$:\implies\:\:\bf{F\:=\:9.1\times{10^{10}}\times\:{0.125\times{10^{-6}}}\times{0.4\times{10^{-2}}}\:}$

$:\implies\:\:\bf{F\:=\:0.455\times{10^{2}}\:}$

$:\implies\:\:\bf\green{F\:=\:45.5\:N }$

$\Large\bf\blue{Therefore,}$

The force required on the metallic wire is 45.5 N.