In an experiment to determine the Young’s modulus, steel wires of five different lengths (1, 2, 3, 4 and 5 m) but of same cross section (2 mm²) were taken and curves between extension and load were obtained. The slope (extension/load) of the curves were plotted with the wire length and the following graph is obtained. If the Young’s modulus of given steel wires is x × 10¹¹ Nm^–2, then the value of x is ______.
Fig. Graph

Question

A 2 $\text{kg}$ mass is attached to a spring with spring constant $ k = 200, \text{N/m} $. If the mass is displaced by $ 0.1, \text{m} $, what is the potential energy stored in the spring?

Answer 1

To determine the Young's modulus, $Y$, we use the formula:

$Y = \frac{FL}{A\Delta L}$

Where:

F is the force applied.
L is the original length of the wire.
A is the cross-sectional area.
$\Delta L$ is the extension.

From Hooke's law, the slope of the extension vs. load graph, given as $\frac{\Delta L}{F}$, is proportional to the length of the wire divided by Young's modulus ($L/Y$).

Rearranging the relationship: $Y = \frac{L}{\text{slope} \times A}$.

The graph shows a slope of $0.25 \times 10^{-5}$ m²/N for 1 m of wire.

Substituting the values ($A = 2 \times 10^{-6}$ m²):

$Y = \frac{1}{0.25 \times 10^{-5} \times 2 \times 10^{-6}}$

Calculating:

$Y = \frac{1}{0.5 \times 10^{-11}}$

$Y = 2 \times 10^{11}$ Nm^–2

Thus, $x = 2$.

The value $x$ is confirmed to be within the expected range of 2.

Correct Answer: 2

Solution and Explanation

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