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(C) The tensile stress $\sigma$ is defined as the force $F$ per unit area $A$,where $F = mg$ and $A = \pi r^2 = \pi (D/2)^2 = \frac{\pi D^2}{4}$.Given

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$11$

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(C) The tensile stress $\sigma$ is defined as the force $F$ per unit area $A$,where $F = mg$ and $A = \pi r^2 = \pi (D/2)^2 = \frac{\pi D^2}{4}$.Given: $\sigma = 7 	imes 10^5\,N m^{-2}$,$D = 14\,mm = 14 	imes 10^{-3}\,m$,$g = 9.8\,m s^{-2}$,and $\pi = \frac{22}{7}$.Substituting these values into the formula $\sigma = \frac{4mg}{\pi D^2}$:$m = \frac{\sigma \pi D^2}{4g}$$m = \frac{(7 	imes 10^5) 	imes (22/7) 	imes (14 	imes 10^{-3})^2}{4 	imes 9.8}$$m = \frac{(7 	imes 10^5) 	imes (22/7) 	imes (196 	imes 10^{-6})}{39.2}$$m = \frac{22 	imes 10^5 	imes 28 	imes 10^{-6}}{39.2} = \frac{616 	imes 10^{-1}}{39.2} = \frac{61.6}{39.2} \approx 1.57$ (Wait,re-calculating: $m = \frac{7 	imes 10^5 	imes (22/7) 	imes 1.96 	imes 10^{-4}}{39.2} = \frac{22 	imes 1.96 	imes 10}{39.2} = \frac{431.2}{39.2} = 11\,kg$).

$A$ 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 $\pi = \frac{22}{7}$)

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$A$ 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 $\pi = \frac{22}{7}$)