As shown in the figure, in an experiment to determine Young's modulus of a wire, the extension-load curve is plotted. The curve is a straight line passing through the origin and makes an angle of 45° with the load axis. The length of wire is 62.8 cm and its diameter is 4 mm. The Young's modulus is found to be x×10⁴ Nm⁻². The value of x is ___________.

Solution

Physics sir · 4 steps

Read the slope of the graph as extension per newton, then Y is just L divided by A times that slope.

1. Rearrange Young's modulus. From Y = (F/A)/(Δ L/L) you get Y = L/A·F/(Δ L). The graph gives the ratio Δ L/F directly.
2. Slope of the line. Extension is on the vertical axis and load is on the horizontal axis, and the line makes 45° with the load axis. So (Δ L)/F = tan 45° = 1 m N⁻¹, which means F/(Δ L) = 1 N m⁻¹.
3. Area of cross-section. Diameter 4 mm means radius r = 2 × 10⁻³ m. So A = π r² = 3.14 × 4 × 10⁻⁶ = 1.256 × 10⁻⁵ m².

   Y = L/A·F/(Δ L) = 0.628/(1.256 × 10⁻⁵) × 1 = 5 × 10⁴ N m⁻²

4. So. Y = 5 × 10⁴ N m⁻², so x = 5.

The common slip: 1.25. Using the diameter 4 mm as the radius makes the area four times too large and gives 1.25, not 5.

The idea to keep. On any load-extension graph, first check which quantity is on which axis. The slope gives either Δ L/F or F/Δ L, and Y = L/A·F/(Δ L) does the rest.

Answer from the NTA final answer key. Solution by PrepWiser.