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

<p>To find the mass $m$ of the metal block, we need to consider the tensile stress developed in the wire. The formula for tensile stress is:</p> <p>$\text{Tensile Stress} = \frac{\text{Force}}{\text{Area}}$</p> <p>The force acting on the wire is the weight of the metal block, which can be represented as $F = mg$.</p> <p>The cross-sectional area of the wire, given its diameter $d = 14 \, mm$, can be calculated using the formula for the area of a circle:</p> <p>$A = \pi (\frac{d}{2})^2 = \pi (\frac{14}{2})^2 \, mm^2$</p> <p>Now, convert the area to $m^2$:</p> <p>$A = \pi (\frac{14 \times 10^{-3}}{2})^2 \, m^2$</p> <p>We are given that the tensile stress developed in the wire is $7 \times 10^5 \, Nm^{-2}$. Using the tensile stress formula, we can write:</p> <p>$7 \times 10^5 \, Nm^{-2} = \frac{mg}{A}$</p> <p>Now, solve for the mass $m$:</p> <p>$m = \frac{7 \times 10^5 \, Nm^{-2} \cdot A}{g}$</p> <p>Substitute the values of A and g into the equation:</p> <p>$$m = \frac{7 \times 10^5 \, Nm^{-2} \cdot \pi (\frac{14 \times 10^{-3}}{2})^2 \, m^2}{9.8 \, ms^{-2}}$$</p> <p>After calculating, we get:</p> <p>$m \approx 11 \, kg$</p> <p>Therefore, the mass of the metal block is approximately $11 \, kg$.</p>