Young’s modulus of the material of a wire of length ‘L’ and radius ‘r’ in ‘Y’ \[N/m^{2}\]. If the length in reduced to L/2 and radius to r/2, the Young modulus will be :
The load versus elongation graph for four wires of the same material is shown in the figure. The thickest wire is represented by the line :
The stress versus strain graphs for wires of two materials A and B are as shown in the figure. If \[Y_{A}andY_{B}\] are the Young’s modulus of the materials then :
A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by 1mm, then elastic energy stored in wire is :
The dimensions of two wires A and B are same but their materials are different. Their load extension graphs are shown if \[y_{A} and y_{B}\] are the values of young’s modulus of elasticity of ‘A and ‘B’ respectively then :
The amount of elastic potential energy per unit volume (in SI unit) of a steel wire of length \(100\text{ cm}\) to stretch it by \(1\text{ mm}\) is (if Young’s modulus of the wire \(= 2.0 \times 10^{11}\text{ N m}^{-2}\) )
Energy per unit volume is \(u = \frac{1}{2} \times Y \times (\text{strain})^2\). Strain \(= \frac{Delta l}{l} = \frac{10^{-3}\text{ m}}{1\text{ m}} = 10^{-3}\). Thus, \(u = \frac{1}{2} \times (2.0 \times 10^{11}) \times (10^{-3})^2 = 10^5\text{ J/m}^3\).
Given below are two statements:
Assertion (A): Young’s modulus for a perfectly plastic body is zero.
Reason (R): For a perfectly plastic body, restoring force is zero.
For a perfectly plastic body, there is no tendency to regain its shape, meaning the restoring force (and thus stress) is zero. Since Young's modulus \(Y = \frac{\text{stress}}{\text{strain}}\), it is also zero. Both statements are true and Reason is the correct explanation.
Young’s modulus of rubber is \(10^4\text{ N/m}^2\) and area of cross-section is \(2\text{ cm}^2\). If force of \(2 \times 10^5\text{ dynes}\) is applied along its length, then length of wire becomes how much times of its initial length \(L\):- (Assume stress \(\propto\) strain)
Converting Young's modulus to CGS units gives \(Y = 10^5\text{ dyne/cm}^2\). Stress is \(F/A = 2 \times 10^5 / 2 = 10^5\text{ dyne/cm}^2\). Since \(\text{Strain} = \text{Stress}/Y = 1\), we have \(\Delta L = L\). Thus, the final length becomes \(L + \Delta L = 2L\).
A metal block is experiencing an atmospheric pressure of \(1 \times 10^5\text{ N/m}^2\). When the same block is placed in a vacuum chamber, the fractional change in its volume is (the bulk modulus of metal is \(1.25 \times 10^{11}\text{ N/m}^2\))
The bulk modulus is defined as \(B = \frac{\Delta P}{\Delta V/V}\). Moving to vacuum causes a pressure change of \(\Delta P = 10^5\text{ N/m}^2\). Thus, the fractional volume change is \(\frac{\Delta V}{V} = \frac{\Delta P}{B} = \frac{10^5}{1.25 \times 10^{11}} = 8 \times 10^{-7}\).
When the load on a wire is increased from \(3\text{ kg wt}\) to \(5\text{ kg wt}\) the elongation increases from \(0.61\text{ mm}\) to \(1.02\text{ mm}\). The required work done during the extension of the wire is :
The work done during the extension is \(W = \frac{1}{2} (F_2 x_2 - F_1 x_1)\). Converting values: \(F_1 = 3 \times 9.8\text{ N}\), \(F_2 = 5 \times 9.8\text{ N}\), \(x_1 = 0.61 \times 10^{-3}\text{ m}\), and \(x_2 = 1.02 \times 10^{-3}\text{ m}\) yields \(W \approx 16 \times 10^{-3}\text{ J}\).