Question

A 1 metre long steel wire of cross-sectional area 1 mm2 is extended by 1mm. If Y equals to 2x1011N/m2, then the work done in stretching is (1) 0.1J (2) 0.2J (3) 0.3J (4) 0.4J ​

Answer 1

Answer :-

• Option (1) 0.1J is the required answer

Topic :-

• Mechanical Properties of Solids

Explanations :-

$\bf\; Y = \bf\frac{FL}{A\triangle\;L}$

Where,

Y denote Young Modulus

F denote Force

L denote Length

∆L denote change in length

A denote Area of Cross Section

Conversion of Units

• 1 mm = 1 × 10⁻³m
• 1 mm² = 1 × 10⁻⁶ m²

Also,

The above equation can be written in term of force .

$\bf\; F = \bf\frac{YA\triangle\;L}{L}$ ----(i)

As we know that

$\bf\; W = \bf\frac{1}{2}F\triangle\;L$ -----(ii)

Now,

Put the value of equation (i) in equation (ii) we get

$\bf\; W = \bf\frac{1}{2}\frac{YA(\triangle\;L)^{2}}{2L}$

by putting the value we get

$\sf\implies\; W = \sf\frac{2\times\;10^{11}\times\;10^{-6}(10^{-3})^{2}}{2\times\;1}$

$\sf\implies\; W = \sf\;10^{11}(10^{-12})$

$\sf\implies\; W = 0.1\\\\\sf\implies\; W = 0.1 J$

• Hence, the required answer is 0.1 J

Answer 2

Answer:

A 1 metre long steel wire of cross-sectional area 1 mm2 is extended by 1mm. If Y equals to 2x1011N/m2, then the work done in stretching is (1) 0.1J (2) 0.2J (3) 0.3J (4) 0.4J

Explanation:

0.2J ✅