Question

16. The area of cross-section and length of a wire are 0.5 mm- and 4.0 m respectively. How much work
will be done to increase its length by 1.0 mm? Young's modulus of elasticity of the material of the
wire is 2.0 x 10" N/m²
(UPB 2007) Ans. 1.25*10kipower-2jole​

Answer 1

GIVEN :-

• Area of cross-section of wire (A) = 0.5mm²
• Length of wire (L) = 4m
• Increase in length (∆L) = 1mm
• Young's modulus (Y) = 2 × 10¹¹ N/m²

TO FIND :-

• Work done (W) for elongation.

TO KNOW :-

$\\ \bigstar\boxed{ \sf \: Y = \dfrac{F \times L}{A \times \Delta \: L} } \\ \\ \bigstar \boxed{ \sf \:W = \dfrac{1}{2} \times F \times \Delta \: L}$

Here ,

• Y → Young's modulus
• F → Force
• L → Length
• A → Area of cross section
• ∆L → Length of Elongation
• W → Work done

SOLUTION :-

We have ,

• Y = 2 × 10¹¹ N/m²
• A = 0.5mm² = 0.5 × 10-⁶m²
• ∆L = 1mm = 1 × 10-³ m²
• L = 4m
• F = ?

$\\ \sf \: Y = \dfrac{F \times L}{A \times \Delta \: L}$

Putting values we get,

$\\ \implies\sf \: 2 \times {10}^{11} = \dfrac{F \times 4}{0.5 \times {10}^{ - 6} \times 1 \times {10}^{ - 3} } \\ \\ \\ \implies\sf \: 2 \times {10}^{11} = \dfrac{4F}{0.5 \times {10}^{ - 9} } \\ \\ \\ \implies \sf \: 2 \times {10}^{11} \times 0.5 \times {10}^{ - 9} = 4F \\ \\ \\\implies \sf \: {10}^{2} = 4F \\ \\ \\ \implies \sf \: 100 = 4F \\ \\ \\ \implies\sf \: F = \dfrac{100}{4} \\ \\ \\ \implies\boxed{ \sf \: F = 25N}$

__________________

Now , we have ,

• F = 25N
• ∆L = 1 × 10-³ m

$\\ \sf \: W = \dfrac{1}{2} \times F \times \Delta \: L$

Putting values we get,

$\\ \sf \: W = \dfrac{1}{2} \times 25 \times 1 \times {10}^{ - 3} \\ \\ \\ \underline{\boxed{ \sf \: W = 12.5 \times {10}^{ - 3} J}}$

Hence , Work done in elongation is 12.5 × 10-³ J.