Question

Two persons pull a rope towards themselves. Each person exerts a force of 100 N on the rope. Find the Young modulus of the material of the rope if it extends in length by 1 cm. Original length of the rope = 2 m and the area of cross section = 2 cm2.

Answer 1

Young modulus of the material of the rope is $10^{8}(\frac{N}{m^{2}})$.

Explanation:

1. Given data

 Force in rope (F) = 100 (N)

 Original length of rope (L) = 2 (m)

 Change in length of rope $(\Delta L)$ = 1 (cm) = 0.01 (m)

 Cross section area of rope (A) = 2$(cm^{2})$ = 0.0002 ($m^{2}$)

2. We know that

   Young modulus of material (Y) =$\frac{Stress(\sigma )}{Strain(\varepsilon )}$    ...1)

3. Where Stress$(\sigma ) = \frac{force(F)}{Area(A)}$    ...2)

   and Strain$(\varepsilon )=\frac{\Delta L}{L}$     ...3)

4. With help of equation 2 and equation 3, Equation 1) can be written as

   Young modulus of material (Y) = $\frac{F\times L}{A\times \Delta L}$

on put the respective value we get

Young modulus of material (Y) = $\frac{100\times 2}{0.0002\times 0.01}$ = 1000000000 ($\frac{N}{m^{2}}$)

Answer 2

Young's Modulus is $10^8$

Explanation:

We know that Young's Modulus is given by -  

$Y = \frac {Fl}{A\Delta l}$

F = $2 \times 100$ = 200 N

l = 1 m

$\Delta l = 0.01 m$

$A = 2 \times 10^-^4 m^2$

So, $Y = \frac {200 \times 1}{2\times 10^-^4 \times 0.01}$

Hence, $Y = 10^8 Nm^-^2$