Two wires are made of the same material and have the same volume. However wire 1 has cross-sectional area A and wire 2 has cross-sectional area 3A. If the length of wire 1 increases by \[\Delta x\] on applying force F, how much force is needed to stretch wire 2 by the same amount ?
Two wires are made of same material and have same volume. However wire -1 has cross-sectional area A and wire-2 has cross-sectional area 3A. If length of wire -1 increases by \[\Delta x\] on applying force F, how much force is needed to stretch wire-2 by same amount ?
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 :
If \[\rho\] is density of the material of a wire and \[\sigma\] is the breaking stress, the greatest length of the wire that can hang freely without breaking is :
Two wires of the same material & length but diameters in the ratio 1 : 2 are stretched by the same force. The potential energy per unit volume for the two wires when stretched will be in the ratio :
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, \(u = \frac{1}{2} \times (2.0 \times 10^{11}) \times (10^{-3})^2 = 10^5\text{ J/m}^3\).
A uniform rope of density \(rho\) and length \(L\) is hanging from roof. If young’s modulus of material of rope is \(Y\), then elongation produced in rope due to its own weight is:
The elongation of a uniform rope under its own weight is given by \(\Delta L = \frac{MgL}{2AY}\). Substituting mass \(M = \rho A L\), we obtain \(\Delta L = \frac{\rho g L^2}{2Y}\).
A rubber sphere is taken in a lake to a depth \(1800\text{ m}\). If bulk modulus of rubber is \(6 \times 10^8\text{ N/m}^2\), then radius of this rubber sphere will decrease by:
The pressure change is \(dP = \rho g h = 10^3 \times 10 \times 1800 = 1.8 \times 10^7\text{ N/m}^2\). The fractional volume change is \(\frac{dV}{V} = \frac{dP}{B} = \frac{1.8 \times 10^7}{6 \times 10^8} = 3\%\). Since \(\frac{dV}{V} = 3\frac{dr}{r}\), the radius decreases by \(\frac{3\%}{3} = 1\%\).
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.