Question

Breaking stress for a material is 2*10^8N/m^2. What maximum length of the wire of this material can be taken so that the wire does not break by own weight?[density of material =5*10^3kg/m^3]

Answer 1

Since we know the formula :
$\frac{dl}{l} = \frac{mg}{2ay} = syrain$
$stress = y \times strain$
$stress = y \times \frac{mg}{2ay} = \frac{mg}{2a}$
$2 \times {10}^{8} = \frac{mg}{2a}$
$mass = a \times l \times density = al(5 \times {10}^{3} )$
$2 \times {10}^{8} = a \times l(5 \times {10}^{3} )g \div 2a$
$here \:two \: a \: cancels$
$l = 4081.63m$

Answer 2

Answer:

Explanation: as we know,

F/A =mg/A ( m = dV where d= density and V for volume and we also know that V= A*l )

So therefore...

mg= dAl*g

So,

F/A = dAlg/A

F/A = dlg

Therefore length l = F/A × 1/dg

Here, given F/A is 2* 10^8 N/m^2

Density d = 5* 10^3 kg/m^3

So therefore l = 2 × 10^8/ 5× 10^3* 10

l = 4 × 10^3 m = 4 km