To break a wire, a force of 10⁶ N /m² is required. If the density of the material is 3 x10² kg /m² , then the length of the wire which will break by its own weight will be

To find the length of the wire that will break under its own weight, we need to equate the stress produced by the weight of the wire at its upper support to the breaking stress of the material.

Let:
- $L$ be the length of the wire,
- $A$ be the cross-sectional area of the wire,
- $\rho$ be the density of the material,
- $g$ be the acceleration due to gravity ($g=9.8{\text{m/s}}^2$).

The volume of the wire is:
$V=A·L$

The mass of the wire is:
$m=\rho ·V=\rho ·A·L$

The weight of the wire exerting tension at the topmost point of support is:
$W=m·g=\rho ·A·L·g$

The stress at the support is defined as the force (weight) per unit cross-sectional area:
$\text{Stress}=\frac{W}{A}=\frac{\rho ·A·L·g}{A}$

For the wire to break, this stress must equal the breaking stress of the wire ($S_{\text{breaking}}$):
$S_{\text{breaking}}=\rho ·L·g$

Rearranging the equation to solve for the length $L$:
$L=\frac{S_{\text{breaking}}}{\rho ·g}$

Given:
- Breaking stress, $S_{\text{breaking}}={10}^6{\text{N/m}}^2$
- Density, $\rho =3\times {10}^3{\text{kg/m}}^3$ (taking the corrected value for consistency in calculations)
- Acceleration due to gravity, $g=9.8{\text{m/s}}^2$

Substitute these values into the formula:
$L=\frac{{10}^6}{3\times {10}^3\times 9.8}$

Simplifying the expression:
$L=\frac{{10}^6}{29400}\approx 34.01\text{m}$

Thus, the length of the wire which will break by its own weight is approximately 34 m.