A metal block of mass $$m$$ is suspended from a rigid support through a metal wire of diameter 14 mm. The tensile stress developed in the wire under equilibrium state is $$7 \times 10^5$$ N m$$^{-2}$$. The value of mass $$m$$ is ______ kg.
(Take $$g = 9.8$$ m s$$^{-2}$$ and $$\pi = \dfrac{22}{7}$$)

A metal block of mass $$m$$ is suspended from a wire of diameter $$14$$ mm, and the tensile stress in the wire is $$7 \times 10^5$$ N/m$$^2$$.

To find the cross-sectional area of the wire, note that the radius is half the diameter so that radius $$= 7$$ mm $$= 7 \times 10^{-3}$$ m. It follows that $$A = \pi r^2 = \dfrac{22}{7} \times (7 \times 10^{-3})^2 = \dfrac{22}{7} \times 49 \times 10^{-6} = 22 \times 7 \times 10^{-6} = 154 \times 10^{-6} \text{ m}^2$$.

Since tensile stress is defined as force per unit area, we have Stress $$= \dfrac{F}{A} = \dfrac{mg}{A}$$, and substituting the given values yields $$7 \times 10^5 = \dfrac{m \times 9.8}{154 \times 10^{-6}}$$.

Solving for $$m$$ gives $$m = \dfrac{7 \times 10^5 \times 154 \times 10^{-6}}{9.8} = \dfrac{107.8}{9.8} = 11 \text{ kg}$$, and hence $$\boxed{11}$$ kg is the mass of the block.