Question 9: Properties of Matter

Question

Figure represents the extension $(\Delta l)$ of a wire of length 1 meter, suspended from the ceiling of the room at one end with a load $W$ connected to the other end. If the cross-sectional area of the wire is $10^{-5} \mathrm{~m}^2$ then the Young's modulus of the wire is $\_\_\_\_$ $\mathrm{N} / \mathrm{m}^2$.

Accepted Answer

Young's Modulus (Y) is defined as the ratio of tensile stress to tensile strain within the elastic limit.
Stress ( $\sigma$ ) is force (W) per unit cross-sectional area (A).
$$ \sigma=\frac{W}{A} $$
Strain $(\epsilon)$ is extension $(\Delta \mathrm{l})$ per unit original length $(\mathrm{L})$.
$$ \epsilon=\frac{\Delta \mathrm{l}}{\mathrm{~L}} $$
Thus, Young's Modulus is :
$$ Y=\frac{\sigma}{\epsilon}=\frac{W / A}{\Delta l / L} $$
$$ \mathrm{Y}=\frac{\mathrm{WL}}{\mathrm{~A} \Delta \mathrm{l}} $$
The original length is $\mathrm{L}=1 \mathrm{~m}$
The cross-sectional area is $\mathrm{A}=10^{-5} \mathrm{~m}^2$
From the graph, we need a pair of values for W and $\Delta \mathrm{l}$.
Let's take $\mathrm{W}=20 \mathrm{~N}$ and the corresponding extension $\Delta \mathrm{l}=2 	imes 10^{-4} \mathrm{~m}$.
Substituting the values in the expression for Young's modulus;
$$ Y=\frac{20 	imes 1}{10^{-5} 	imes\left(2 	imes 10^{-4}ight)} $$
$$\Rightarrow $$ $Y=\frac{20}{2 	imes 10^{-9}}$
$$\Rightarrow $$ $Y=10 	imes 10^9$
$$ \mathrm{Y}=1.0 	imes 10^{10} \mathrm{~N} / \mathrm{m}^2 $$
Therefore, the Young's modulus of the wire is $1.0 	imes 10^{10} \mathrm{~N} / \mathrm{m}^2$. Thus, the correct option is (C).

Answer

$$ 1.0 	imes 10^{11} $$

Answer

$$ 2.0 	imes 10^{10} $$

Answer

$$ 1.0 	imes 10^{10} $$

Answer

$$ 2.0 	imes 10^{11} $$

Properties of Matter

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