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 $45^{\circ}$ with the load axis The length of wire is $628 \,cm$ and its diameter is $4 \,mm$ The Young's modulus is found to be $x \times 10^4 Nm ^{-2}$The value of $x$ is
To determine Young's modulus \(Y\) of the wire, we consider the slope of the extension-load curve, which is given as \(45^\circ\). The slope is the tangent of the angle, so \(\tan 45^\circ = 1\). Thus, the extension per unit load is constant and equal to 1. Young's modulus is given by: \[ Y = \frac{F \cdot L}{A \cdot \Delta L} \] where \(F\) is the force (load), \(L\) is the original length, \(A\) is the cross-sectional area, and \(\Delta L\) is the extension. The equation can be rearranged to match the slope condition: \( \frac{\Delta L}{F} = \frac{L}{A \cdot Y} = 1 \to A \cdot Y = L \) Given that the diameter \(d = 4\,\text{mm} = 0.4\,\text{cm}\), the area \(A\) is: \[ A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.4}{2}\right)^2 = 0.04\pi\,\text{cm}^2 \] Substitute the values: \( A \cdot Y = 628 \) \( 0.04\pi \cdot Y = 628 \) \( Y = \frac{628}{0.04\pi} \) Calculate \(Y\): \( Y = \frac{628}{0.1256} \approx 5000\,\text{Nm}^{-2} \) So, \(Y = 5 \times 10^4\,\text{Nm}^{-2}\). Thus, \(x = 5\). The calculated value \(x = 5\) falls within the given range \(5,5\). Therefore, the solution is validated.
