
A wire suspended vertically from one of its ends is stretched by attaching a weight of $200{\text{ }}N$ to the lower end. The weight stretches the wire by $1{\text{ }}mm$. Then the elastic energy stored in the wire is:
A. $0.1{\text{ }}J$
B. $0.2{\text{ }}J$
C. $10{\text{ }}J$
D. $20{\text{ }}J$
Answer
480.3k+ views
Hint: From the information given in the above question, we have to find the elastic energy stored in the wire. It is actually equal to the work done by the wire in restoring force. We will use the relation between the elastic energy, stress and strain and hence find the answer by formulating an equation and substituting the values provided in the given question.
Complete step by step answer:
The elastic energy stored in the wire is equal to the work done by the wire in restoring the force that acts on the wire.Thus, to find the elastic energy $E$ of a stretched wire, we use the formula,
$E = \dfrac{1}{2} \times stress \times strain \times volume$
$\text{Stress} = \dfrac{F}{A}$
where $F = $ force and $A = $ area of cross-section of the wire.
$\text{Strain} = \dfrac{{\Delta l}}{l}$
where $\Delta l$ is the change in the length of the wire and $l$ is the original length of the wire.
$\text{Volume} = A \times l$
Hence, we find the value of elastic energy as,
$E = \dfrac{1}{2} \times \dfrac{F}{A} \times \dfrac{{\Delta l}}{l} \times Al \\
\Rightarrow E= \dfrac{1}{2} \times F \times \Delta l - - - - - \left( 1 \right)$
It is given in the question that the force acting on the wire, $F = 200{\text{ }}N$
The stretching of the wire $\Delta l = 1{\text{ }}mm = {10^{ - 3}}{\text{ }}m$
Substituting the values in equation $\left( 1 \right)$ we get,
$E = \dfrac{1}{2} \times 200 \times {10^{ - 3}}$
$ \therefore E = 0.1{\text{ }}J$
The elastic energy stored in the wire is $0.1{\text{ }}J$.
Therefore, the correct option is A.
Note: It must be noted that stress is defined as the force that acts on the object per unit area and strain is defined as the relative change in the dimensions of the body subjected to the force acting on it. The restoring force is defined as the opposing force which tries to compensate for the force acting on it, and restore its original shape.
Complete step by step answer:
The elastic energy stored in the wire is equal to the work done by the wire in restoring the force that acts on the wire.Thus, to find the elastic energy $E$ of a stretched wire, we use the formula,
$E = \dfrac{1}{2} \times stress \times strain \times volume$
$\text{Stress} = \dfrac{F}{A}$
where $F = $ force and $A = $ area of cross-section of the wire.
$\text{Strain} = \dfrac{{\Delta l}}{l}$
where $\Delta l$ is the change in the length of the wire and $l$ is the original length of the wire.
$\text{Volume} = A \times l$
Hence, we find the value of elastic energy as,
$E = \dfrac{1}{2} \times \dfrac{F}{A} \times \dfrac{{\Delta l}}{l} \times Al \\
\Rightarrow E= \dfrac{1}{2} \times F \times \Delta l - - - - - \left( 1 \right)$
It is given in the question that the force acting on the wire, $F = 200{\text{ }}N$
The stretching of the wire $\Delta l = 1{\text{ }}mm = {10^{ - 3}}{\text{ }}m$
Substituting the values in equation $\left( 1 \right)$ we get,
$E = \dfrac{1}{2} \times 200 \times {10^{ - 3}}$
$ \therefore E = 0.1{\text{ }}J$
The elastic energy stored in the wire is $0.1{\text{ }}J$.
Therefore, the correct option is A.
Note: It must be noted that stress is defined as the force that acts on the object per unit area and strain is defined as the relative change in the dimensions of the body subjected to the force acting on it. The restoring force is defined as the opposing force which tries to compensate for the force acting on it, and restore its original shape.
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