Question

What force would be required to stretch a steel wire of 4 mm^2 of cross section so that its length becomes three times its original length?. Given Young's Modulus =2.4×10^12 dyne cm^-2.​

Answer 1

Given:

Area of cross section of wire is 4 mm².

Young's modulus = 2.4 × 10^(12) dyne/cm².

To find:

Force required to stretch the wire such that the length becomes three times its original length.

Conversion:

$\gamma = 2.4 \times {10}^{12} \: dyne \: {cm}^{ - 2}$

$= > \gamma = 2.4 \times {10}^{11} \: N \: {m}^{ - 2}$

Calculation:

Let initial length be l , final length be 3l.

Hence change in length = ∆l = 2l.

$\sf{ \therefore \: \gamma = \dfrac{force \times l}{area \times \Delta l} }$

$\sf{ = > \: 2.4 \times {10}^{11} = \dfrac{force \times l}{area \times 2l} }$

$\sf{ = > \: 2.4 \times {10}^{11} = \dfrac{force \times l}{ {10}^{ - 6} \times 2l} }$

$\sf{ = > \: 2.4 \times {10}^{11} = \dfrac{force }{2 \times {10}^{ - 6} } }$

$\sf{ = > \: force = 2.4 \times {10}^{11} \times 2 \times {10}^{ - 6} }$

$\sf{ = > \: force = 4.8 \times {10}^{5} \: N }$

So, final answer is:

$\red{ \boxed{ \green{\rm{ \: force = 4.8 \times {10}^{5} \: N }}}}$

Answer 2

Given:

A steel wire of 4 mm^2 of cross section. Given Young's Modulus =2.4×10^12 dyne cm^-2.​

To find:

What force would be required to stretch a steel wire of 4 mm^2 of cross section so that its length becomes three times its original length?.

Solution:

From given, we have,

The area of cross section of a steel wire, A = 4 mm² = 4 × 10^{-2} cm²

The increase in length, Δl = 3l - l = 2l

Young's Modulus, Y = 2.4 × 10^12 dyne cm^-2

We use the formula,

Y = (F/A) (l/Δl)

F = YA (Δl/l)

F = (2.4 × 10^12 × 4 × 10^{-2}) × (2l/l)

F = (9.6 × 10^{10}) × 2

F = 19.2 × 10^{10}

F = 1.92 × 10^{11} dyne

F = 1.92 × 10^{-5} N

Therefore, the force that would be required to stretch a steel wire is 1.92 × 10^{11} dyne or 1.92 × 10^{-5} N