Question

A wire of 3cm long and 6.25×10-5m2 cross section is found to stretch by 3mm under a tension of 1200Kg. Calculate the Young’s modulus of the material of the wire.

Answer 1

Answer:

• The Young modulus (Y) of the material is 1.920 × 10⁹ Pa

Given:

1. Area of cross-section (A) = 6.25 × 10⁻⁵ m²
2. Length of wire (l) = 3 cm = 3 × 10⁻² m
3. Change in length (Δ l) = 3 mm = 3 × 10⁻³ m
4. Mass of load (M) = 1200 Kg

Explanation:

$\rule{300}{1.5}$

From the formula we know,

$\bullet\;\underline{\boxed{\sf Y=\Bigg\lgroup \dfrac{F}{A}\Bigg\rgroup.\Bigg\lgroup \dfrac{l}{\Delta l}\Bigg\rgroup}}$

Here,

• F Denotes Force.
• A Denotes Area
• l Denotes length
• Δ l Denotes Extension.

Remember, In this case the force will be caused by the weight of the body.

i.e. F = M g.

Substituting the values,

$\displaystyle \Rightarrow\sf Y=\Bigg\lgroup\dfrac{M\;g}{A}\Bigg\rgroup.\Bigg\lgroup \dfrac{l}{\Delta l}\Bigg\rgroup\\\\\\\Rightarrow\sf Y=\Bigg\lgroup\dfrac{1200\times10}{6.25\times10^{-5}}\Bigg\rgroup\times\Bigg\lgroup \dfrac{3\times 10^{-2}}{3\times 10^{-3}} \Bigg\rgroup\;\;\because\bigg( g=10m/s^{\;2}\bigg)\\\\\\\Rightarrow\sf Y=\Bigg\lgroup \dfrac{12000}{6.25\times10^{-5}}\Bigg\rgroup\times\Bigg\lgroup10^{-2}\times10^{\;3}\Bigg\rgroup$

$\displaystyle\Rightarrow\sf Y=\Bigg\lgroup\dfrac{1920}{10^{-5}}\Bigg\rgroup\times\Bigg\lgroup10^{-2+3}\Bigg\rgroup\\\\\\\Rightarrow\sf Y=1920\times 10^{\;5}\times 10\\\\\\\Rightarrow\sf Y=1920\times 10^{\;6}\\\\\\\Rightarrow\sf Y=1.920\times10^{\;9}\\\\\\\Rightarrow \large{\underline{\boxed{\red{\sf Y=1.92\times 10^{\;9}\;Pa}}}}$

∴ The Young modulus (Y) of the material is 1.920 × 10⁹ Pa.

$\rule{300}{1.5}$

Answer 2

$\huge\underline\mathtt\red{Answer:-}$

The Young modulus (Y) of the material is 1.920 × 10⁹ Pa