Question 254: Properties of Matter

Question

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 $$7 	imes 10^{5} \mathrm{Nm}^{-2}$$. The value of mass $$\mathrm{m}$$ is _________ $$\mathrm{kg}$$. (Take, $$\mathrm{g}=9.8 \mathrm{~ms}^{-2}$$ and $$\pi=\frac{22}{7}$$ )

Accepted Answer

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

Properties of Matter

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