Question

A wire of diameter 3 mm and length 4 m extends by
2.5 mm when a force of 10 N is applied. Find the
Young's modulus of material of wire.​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{young 's \: modulus}}= 3.4 \times {10}^{7}N/m^{2}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline\bold{Given : }} \\ : \implies \text{Diameter \: of \: wire (d)= 3mm = 3} \times {10}^{ - 4} m \\ \\ : \implies \text{Length \: of \: wire (l)= 4m} \\ \\ : \implies \text{Change \: in \: length}( \Delta \: l) = 2.5mm = 2.5 \times {10}^{ - 4} m \\ \\ : \implies \text{force = 10 N} \\ \\ \red{ \underline\bold{To \: Find : }} \\ : \implies \text{young 's \: modulus = ?}$

• According to given question :

$: \implies \text{young 's \: modulus} = \frac{ \text{strees}}{ \text{strain}} \\ \\ : \implies\text{young 's \: modulus} = \frac{ \frac{force}{area} }{ \frac{change \: in \: length}{original \: length} } \\ \\ : \implies\text{young 's \: modulus} = \frac{f \times l}{A \times \Delta} \\ \\ : \implies\text{young 's \: modulus} = \frac{10 \times 4}{\pi {r}^{2} \times 2.5 \times {10}^{ - 4} } \\ \\ : \implies\text{young 's \: modulus} = \frac{40}{3.14 \times 1.5 \times {10}^{ - 4} \times 2.5 \times {10}^{ - 4} } \\ \\ : \implies\text{young 's \: modulus} = \frac{40 \times {10}^{2} \times 10 \times 10 \times {10}^{8} }{314 \times 15 \times 25 } \\ \\ : \implies\text{young 's \: modulus} = \frac{4 \times {10}^{12} }{117750} \\ \\ : \implies\text{young 's \: modulus} = 0.0000339702 \times {10}^{12} \\ \\ \green{ : \implies\text{young 's \: modulus} = 3.4 \times {10}^{7}N/m^{2} }$

Answer 2

Answer:

3.4 × 10⁷ N / m² .

Explanation:

Given :

Diameter of wire = 3 mm or 3 / 2 × 10⁻³ m

Length of wire = 4 m

Force = 10 N

Change in length = 2.5 mm or 2.5 × 10⁻³ m

Cross section area = π ( 3 / 2 × 10⁻³ )² m²

We know :

Y = F / A × L / Δ L

Y = 10 / π ( 3 / 2 × 10⁻³ )² × 4 / 2.5 × 10⁻³ N / m²

Y = 3.4 × 10⁷ N / m² .