Elastic Behaviour, Stress-Strain Relationship, Hook’s Law, Young’s Modulus, Bulk Modulus, Shear Modulus Of Rigidity, Poisson’s Ratio, Elastic Energy (JEE LEVEL)

Illustration

A steel wire 4.0 m in length is stretched through 2.0 mm. The cross-sectional area of the wire is 2.0 mm². If Young’s modulus of steel is 2.0 × 10¹¹ N/m². Find:

(a) the energy density of wire,

(b) the elastic potential energy stored in the wire.

Solution:

Here, l = 4.0 mm, Dl = 2 × 10^–3 m, A = 2.0 × 10^–6 m². Y = 2.0 × 10¹¹ N/m²

(a) The energy density of stretched wire

U=\frac{1}{2}\times \text{stress }\times \text{ strain }=\frac{\text{1}}{\text{2}}\times Y\times {{(\text{strain})}^{2}}

=\frac{1}{2}\times 2.0\times {{10}^{{11}}}\times {{\left( {\frac{{(2\times {{{10}}^{{-3}}})}}{4}} \right)}^{2}} =0.25\times {{10}^{5}}=2.5\times {{10}^{4}}\text{J/}{{\text{m}}^{\text{3}}}

(b) Elastic potential energy = energy density × volume

=2.5\times {{10}^{4}}\times (2.0\times {{10}^{{-6}}})\times 4.0\text{J}

=20\times {{10}^{{-2}}}=0.21\text{J}

Illustration

A cable is replaced by another one of same length and material but twice the diameter. How will this effect the elongation under a given load? How does this affect the maximum load it can support without exceeding the elastic limit?

Solution: Young’s modulus Y=\frac{{Mgl}}{{\pi {{r}^{2}}.\Delta l}}=\frac{{Mgl}}{{\pi {{{\left( {\frac{D}{2}} \right)}}^{2}}.\,\Delta l}}=\frac{{4Mgl}}{{\pi {{D}^{2}}.\,\Delta l}}

(Where D is the diameter of the wire.)

Elongation, \Delta l=\frac{{4Mgl}}{{\pi {{D}^{2}}.Y}}e. \Delta l\propto \frac{1}{{{{D}^{2}}}}

Clearly, if the diameter is doubled, the elongation will become one-fourth.

Also, maximum stress {{\sigma }_{m}}=\frac{{{{M}_{m}}g}}{{(\pi {{D}^{2}}/4)}} or     M_mµD²

Clearly, if the diameter is doubled, the wire can support four times the original load.

Illustration

Find the greatest length of steel wire that can hang vertically without breaking. Breaking stress of steel = 8.0 × 10⁸ N/m². Density of steel = 8.0 × 10³ kg/m³. Take g = 10 m/s².

Solution:

Let l be the length of the wire that can hang vertically without breaking. Then the stretching force on it is equal to its own weight. If therefore, A is the area of cross-section and r be the density, then

Maximum stress ({{\sigma }_{m}})=\frac{{\text{weight}}}{A} \left( {\text{stress}=\frac{{\text{force}}}{{\text{area}}}} \right)

or {{\sigma }_{m}}=\frac{{(Al\rho )}}{A}

  l=\frac{{{{\sigma }_{m}}}}{{\rho g}}

Substituting the values l=\frac{{8.0\times {{{10}}^{8}}}}{{(8.0\times {{{10}}^{3}})\,(10)}}={{10}^{4}}\text{m}

Illustration

Two steel wires of length 1 m and 2 m have diameters 1 mm and 2 mm respectively. If they are stretched by forces of 40 N and 80 N respectively their elongation will be in the ratio of

(1) 1 : 1                       

(2) 2 : 1                  

(3) 4 : 1                        

(4) 1 : 8

Solution:         

Young’s modulus: Y\ \,=\,\ \frac{{{{F}_{1}}/{{A}_{1}}}}{{\Delta {{L}_{1}}/{{L}_{1}}}}\ \,=\ \,\frac{{{{F}_{2}}/{{A}_{2}}}}{{\Delta {{L}_{2}}/{{L}_{2}}}}

\ \frac{{\Delta {{L}_{1}}}}{{\Delta {{L}_{2}}}}\ \,=\ \,\frac{{{{F}_{1}}{{L}_{1}}}}{{{{F}_{2}}{{L}_{2}}}}\ \,\times \ \,\frac{{{{A}_{2}}}}{{{{A}_{1}}}} =\ \,\frac{{40\ \,\times \,\ 1}}{{80\,\ \times \ \,2}}\ \,\times \ \,\frac{{{{2}^{2}}}}{{{{1}^{2}}}}\ \,=\ \,\text{1}

(1)

 Illustration

The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied to all?

(1) L = 50cm;D=0.05mm                      

(2) L = 100 cm; D = 1 mm

(3) L = 200 cm; D = 2 mm

(4) L = 300 cm; D = 3 mm

Solution:         

Extension: \Delta L\,\ =\,\ \frac{{FL}}{{AY}}\ \,\propto \ \,\frac{L}{A}  (QF and Y are common)

\frac{L}{A} is maximum in case of (1) 

(1)

Illustration

Two wires of different materials support a massive object suspended from the centre of a uniform horizontal bar with its ends connected to two vertical wires of equal length. What must be the ratio of the diameters of the wires so that the bar may remain horizontal when the object is removed? [Young’s modulus of the wires are Y₁ and Y₂ respectively]

(1) \frac{{{{d}_{1}}}}{{{{d}_{2}}}}\ \,=\,\ \frac{{{{Y}_{1}}}}{{{{Y}_{2}}}}

 (2) \frac{{{{d}_{1}}}}{{{{d}_{2}}}}\,\ =\,\ \frac{{{{Y}_{2}}}}{{{{Y}_{1}}}}

 (3) \frac{{{{d}_{1}}}}{{{{d}_{2}}}}\ \,=\ \,\sqrt{{\frac{{{{Y}_{1}}}}{{{{Y}_{2}}}}}}

 (4) \frac{{{{d}_{1}}}}{{{{d}_{2}}}}\ \,=\,\ \sqrt{{\frac{{{{Y}_{2}}}}{{{{Y}_{1}}}}}}

 Solution:         

Extension: \Delta l\ \,=\,\ \frac{{Fl}}{{AY}}

For the bar to remain horizontal, D l ₁ = D l ₂

A₁Y₁ = A₂Y₂               (QF and l are common)

d_{1}^{2}{{Y}_{1}}\ \,=\,\ d_{2}^{2}{{Y}_{2}} (QA µ d)

                        \frac{{{{d}_{1}}}}{{{{d}_{2}}}}\ \,=\,\ \sqrt{{\frac{{{{Y}_{\text{2}}}}}{{{{Y}_{\text{1}}}}}}}

(4)

Illustration

By what fraction is the volume of an aluminium sphere be reduced as it is lowered from the surface to the depth of 3 km in ocean? Bulk modulus of Al = 8 ´ 10¹⁰ N/m². At the surface of ocean P = 10⁵ N/m². At a depth 3 km, P = 300 atmospheres.

(1) 20 %                      

(2) 10 %                 

(3) 1 %                 

(4) 0.04 %

Solution:         

\frac{{\Delta V}}{V} = \frac{{\text{Change}\,\,\ \text{in}\,\,\ \text{pressure}}}{{\text{Bulk}\ \,\,\text{modulus}}}\,\ =\,\ \frac{{\text{3}\,\ \times \,\ \text{1}{{\text{0}}^{\text{7}}}\ -\,\ \text{1}{{\text{0}}^{\text{5}}}}}{{\text{8}\ \,\times \ \,\text{1}{{\text{0}}^{{\text{10}}}}}}  = \frac{{299}}{8}\ \,\times \ \,{{10}^{{-5}}}\ \,\times \ \,100\%\ \,=\,\ 0.0374\%»0.04%

(4)

Illustration

A wire of length L, radius r, when stretched with a force F, changes in length by l. What will be the change in length in a wire of same material having length 2L, radius 2r and stretched by a force 2F?

(1) \frac{l}{2}

(2) l                             

(3) 2 l                     

(4) 4L

Solution:         

Y = \frac{{\frac{F}{A}}}{{\left( {\frac{{\Delta \ell }}{\ell }} \right)}}\,\,\,\ \Rightarrow \ \,\frac{{\left( {\frac{F}{{\pi {{r}^{2}}}}} \right)}}{{\left( {\frac{\ell }{L}} \right)}}\ \,\,=\ \,\,\frac{{\left( {\frac{{2F}}{{4\ \pi {{r}^{2}}}}} \right)}}{{\left( {\frac{{\Delta \ell }}{{2L}}} \right)}}

\frac{1}{l}\ =\ \frac{1}{{\Delta l}}

Δl = l

(2)