Question

Under what conditions, a stress is known as breaking stress? A load of 100 kg is suspended

by a wire of length 1.0m. The wire is stretched by 0.20 cm. Calculate its cross sectional

area if tensile strength is 9.8 × 107

Nm&2. Find strain in the wire. Given, g = 9.8 ms–2

.

(See Lesson 8)​

Answer 1

Given :-

Stress = 9.8 × 10⁷

As we know that,

Stress = Force/Area

Area = 100×9.8/9.8×10⁷

$\bf{A\:=\:{10}^{-5}\:{m}^{2}}$

Now,

Strain = ∆l/l

Strain = 0.2×10-²/1

$\bf{Strain\:=\:2\times{10}^{-3}}$

Answer 2

Ques : Under what conditions, a stress is known as breaking stress ?

Ans : The stress which equivalent to breaking point is called as ❝ Breaking stress ❞ .

━━━━━━━━━━━━━━━━━━━━━━━━⠀

Stated that , A load of 100 kg is suspended by a wire of length 1.0m , it's tensile strength or stress is 9.8 × 10⁷ N/m² , Wire is stretched by 0.20 cm and g ( or Acceleration due to gravity ) is 9.8 m/s².

Need To Calculate : It's cross sectional area & strain in wire ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ Finding ❝ Cross Sectional Area ❞ :

⠀⠀▪︎⠀We can Calculate Cross Sectional Area using Formula for Stress ( σ ) and it's Given by :

$\qquad \qquad \star \:\underline {\boxed {\pmb{\sf \:\:\:\ \sigma \:=\: \dfrac{F}{A\:} \:\:}}}$

⠀⠀⠀⠀⠀Where ,

• σ is Stress = 9.8 × 10⁷ N/m²

• A is Area

• F is Force = Mass × Acceleration due to gravity

⠀⠀⠀⠀⠀⠀⠀▪︎⠀Mass = 100 kg

⠀⠀⠀⠀⠀⠀⠀▪︎⠀Acceleration due to gravity = 9.8 m/s².

$\qquad:\implies \sf \:\:\ \sigma \:=\: \dfrac{F}{A\:} \:\\\\ \qquad:\implies \sf \:\:\ 9.8 \:\times \:10^7 \: \:=\: \dfrac{ 9.8 \times 100 }{ \:Area\: } \:\\\\ \qquad:\implies \sf \:\:\ Area \: \:=\: \dfrac{ 9.8 \times 100 }{ \:9.8 \:\times \:10^7 \:\: } \:\\\\ \qquad:\implies \underline {\boxed {\pmb{\frak{ \:\:\ Area \: \:=\: 10^{-5}\:m^2 \:}}}} \:\:\bigstar\:$

$\qquad \therefore \:\underline {\sf Hence, \:The \:Area \; is \: \pmb{\bf 10^{-5}\:m^2 \:}\:\: }$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ Finding ❝ Strain of Wire ❞ :

⠀⠀▪︎⠀We can Calculate Strain in wire using Formula for Strain and it's Given by :

$\qquad \qquad \star \:\underline {\boxed {\pmb{\sf \:\:\:\ Strain \:=\: \dfrac{\triangle\: l \:}{\:l\:} \:\:}}}$

⠀⠀⠀⠀⠀Where ,

• △ l is change in length = 0.20 cm &

• l is the original Length = 1.0 m

$\qquad \dashrightarrow \sf \:\:\:\ Strain \:=\: \dfrac{\triangle\: l \:}{\:l\:} \:\:\\\\ \qquad \dashrightarrow \sf \:\:\:\ Strain \:=\: \dfrac{0.20 \:cm \:}{\:1\:m\:} \:\:\\\\ \qquad \dashrightarrow \sf \:\:\:\ Strain \:=\: \dfrac{ 0.2 \times 10^{-2}\: m \:}{\:1 \:m\:} \:\:\\\\ \qquad \dashrightarrow \sf \:\:\:\ Strain \:=\: \dfrac{ 0.2 \times 10^{-2}\: \:}{\:1\:} \:\:\\\\ \qquad \dashrightarrow \underline {\boxed {\pmb{\frak{ \:\:\ Strain\: \:=\: 2 \:\times 10^{-3}\: \:}}}} \:\:\bigstar\:$

$\qquad \therefore \:\underline {\sf Hence, \:The \:Strain \; is \: \pmb{\bf 2 \times 10^{-3}\:}\:\: }$