A metal wire having Poisson's ratio 1 / 4 and Young's modulus 8 × 10^{10} ,N / m ^2 is stre

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8.Mechanical Properties of Solids

medium

A metal wire having Poisson's ratio $1 / 4$ and Young's modulus $8 \times 10^{10} \,N / m ^2$ is stretched by a force, which produces a lateral strain of $0.02 \%$ in it. The elastic potential energy stored per unit volume in wire is [in $\left.J / m ^3\right]$

$2.56 \times 10^4$

$1.78 \times 10^6$

$3.72 \times 10^2$

$2.18 \times 10^5$

Solution

(a)

$\frac{\text { Lateral strain }}{\text { Longitudinal strain }}=$ Poisson's ratio

$\frac{0.02 / 100}{\Delta l / l}=\frac{1}{4}$ $\left\{\begin{array}{l}Y=\text { (Young's modulus) } \\ \quad=8 \times 10^{10} \text { (given) } \\ \text { Poission's ratio }=\frac{1}{4} \text { (given) } \\ \text { Lateral strain }=0.02 \% \text { (given) }\end{array}\right.$

$\frac{\Delta l}{l}=\frac{0.08}{100}$

$\Delta U$ (Elastic potential energy per unit volume $=\frac{1}{2} \times Y \times($ Longitudinal strain $\left.)\right)$

Substituting values

$\Delta U=\frac{1}{2} \times 8 \times 10^{10} \times\left(\frac{0.08}{100}\right)^2$

$\Delta U=2.56 \times 10^4 \,J / m ^3$

Standard 11

Physics

A metal wire of length $'L'$ is suspended vertically from a rigid support. When a body of mass $M$ is attached to the lower end of wire, the elongation in wire is $'l'$, consider the following statements

$(I)$ the loss of gravitational potential energy of mass $M$ is $Mgl$

$(II)$ the elastic potential energy stored in the wire is $Mgl$

$(III)$ the elastic potential energy stored in wire is $\frac{1}{2}\, Mg l$

$(IV)$ heat produced is $\frac{1}{2}\, Mg l$

Correct statement are :-

hard

A steel rod of length $\ell$, cross sectional area $A$, young's modulus of elasticity $Y$, and thermal coefficient of linear expansion $'a'$ is heated so that its temperature increases by $t\,^oC$. Work that can be done by rod on heating will be

hard

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 $1\ mm$. Then the elastic energy stored in the wire ……… $J$

medium

A wire fixed at the upper end stretches by length $l$ by applying a force $F$. The work done in stretching is

easy

(AIEEE-2004)

An $8\,m$ long copper wire and $4\,m$ long steel wire, each of cross section $0.5\,cm^2$ are fastened end to end and stretched by $500\,N$ force. The elastic potential energy of the system is (Youngs mod $: Y_{cu}= 1\times 10^{11}\,N/m^2,$ $Y_{steel} = 2\times 10^{11}\,N/m^2$ ) :

hard

A metal wire having Poisson's ratio $1 / 4$ and Young's modulus $8 \times 10^{10} \,N / m ^2$ is stretched by a force, which produces a lateral strain of $0.02 \%$ in it. The elastic potential energy stored per unit volume in wire is [in $\left.J / m ^3\right]$

Solution

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