## Important formulae for solving numericals of elasticity :

(i) Stress = $\frac{Force}{area}$

(ii) Strain = $\frac{Change\,\,in\,\,configuration}{Original\,\,configuration}$

(iii) Longitudinal strain = $\frac{Change\,\,in\,\,length}{Original\,\,length}$

(iv) Volumetric strain = $\frac{Change\,\,in\,\,volume}{Original\,\,volume}$

(v) Modulus of elasticity (E) = $\frac{Stress}{Strain}$

(vi) Young’s modulus of elasticity, Y = $\frac{Fl}{eA}$

(vii) Elastic potential energy stored, E = $\frac{1}{2}$ .F .e

(viii) Work done to stretch the wire = Energy stored

(ix) Energy density, ${{\rho }_{E}}$ = $\frac{1}{2}$ stress × strain

(x) Bulk Modulus, K = $\frac{Normal\,\,stress}{Volumetric\,\,strain}$

(xi) Modulus of rigidity ($\eta $) = $\frac{Tangential\,\,stress}{Shear\,\,strain}$

(xii) Poisson’s ratio ($\sigma $) = $\frac{Lateral\,\,strain}{Longitudinal\,\,strain}$

(xiii) Shearing strength = $\frac{Force}{area}$

(xiv) Density, $\rho $ = $\frac{mass}{volume}$ = $\frac{m}{A.l}$

*1. What force is required to stretch a steel wire of cross sectional area 1 cm*^{2}* to double its length?*

*[Y*_{Steel}* = 2**×**10*^{11}* N/m*^{ }*]*

Solution:

Force, F = ?

Area of cross-section, A = 1 cm^{2}

A = (1×10^{-2})^{2} m^{2} = 1×10^{-4} m^{2}

Let initial length, *l*_{1} = *l*

Then final length, *l*_{2} = 2*l*

Elongation, e = *l*_{2} – *l*_{1} = 2*l* – *l* = *l*

$\therefore $ Elongation, e = *l*

Y_{Steel} = 2×10^{11} N/m^{2}

We know,

Young’s modulus of elasticity,

Y = $\frac{Fl}{eA}$

Or, F = $\frac{YAe}{l}$

Or, F = $\frac{2\times {{10}^{11}}\times l\times 1\times {{10}^{-4}}}{l}$

$\therefore $ F = 2×10^{7} N

*2. Calculate the work done in stretching a steel wire 100 cm in length and of cross sectional area 0.030 cm*^{2}* when a load of 100 N is slowly applied before the elastic limit is reached. ***[Y**_{Steel}** = 2****×****10**^{11}** N/m**^{ }**]**

Solution:

Here,

Work done, W = ?

Length of wire, *l* = 100 cm = 1m

Cross-sectional area, A = 0.03 cm^{2 }

^{ }A^{ }= 0.03× (1×10^{–2})^{2} m^{2}

A = 0.03×10^{–4} m^{2}

Force, F = 100 N

Y_{Steel} = 2×10^{11} Nm^{–2 }

We know,

Young’s modulus of elasticity

Y = $\frac{Fl}{eA}$

Or, e = $\frac{Fl}{YA}$

Or, e = $\frac{100\times 1}{2\times {{10}^{11}}\times 0.03\times {{10}^{-4}}}$

Elongation, e = $\frac{1}{6000}$ m

Now,

Work done (W) = Energy stored (E)

Or, W = $\frac{1}{2}$ .F .e

Or, W = $\frac{1}{2}$ × 100 × $\frac{1}{6000}$

$\therefore $ W = 8.33×10^{-3} J

*3. Find the work done in stretching a wire of cross sectional area 10*^{–2}* cm*^{2}* and 2m long through 0.1 mm, If Y for the material of wire is 2**×**10*^{11}* Nm*^{–2}*.*

Solution:

Here,

Work done, W = ?

Cross-sectional area, A = 10^{–2} cm^{2 }

A^{ }= 10^{–2} × (1×10^{–2})^{2} m^{2}

A = 1×10^{–6} m^{2}

Length of wire, *l* = 2 m

Elongation, e = 0.1 mm = 0.1×10^{–3} m

Y_{Steel} = 2×10^{11} Nm^{–2 }

We know,

Young’s modulus of elasticity

Y = $\frac{Fl}{eA}$

F = $\frac{YeA}{l}$

F = $\frac{2\times {{10}^{11}}\times 0.1\times {{10}^{-3}}\times 1\times {{10}^{-6}}}{2}$

F = 10 N

Now, work done, (W) = Energy stored (E)

W = $\frac{1}{2}$ .F .e

W = $\frac{1}{2}$ × 10 × 0.1×10^{–3}

$\therefore $ W = 5×10^{-4} J

*4. A vertical brass rod of circular section is loaded by placing of 5 kg weight on top of it. If its length is 50 cm and radius of cross section is 1 cm, find the contraction of the rod and the energy stored in it. *

*[Y*_{Brass}* = 3.5**×**10*^{10}* Nm*^{–2}*]*

Solution:

Here,

Mass on top, m = 5 kg

Length of rod, *l* = 50 cm = 0.5 m

Radius, r = 1 cm = 1×10^{–2 }m* *

(i) Contraction, e (or $\Delta l$* or l** _{1}*–

*l*

*) = ?*

_{1}(ii) Energy stored, E = ?

Y_{Brass} = 3.5×10^{10} Nm^{–2 }

We know,

Young’s modulus of elasticity

Y = $\frac{Fl}{eA}$

e = $\frac{Fl}{YA}$

e = $\frac{mgl}{Y\times \pi {{r}^{2}}}$

e = $\frac{5\times 10\times 0.5}{3.5\times {{10}^{10}}\times \pi {{(1\times {{10}^{-2}})}^{2}}}$

$\therefore $ Contraction, e = 2.27×10^{–6} m

Now,

Energy stored, E = $\frac{1}{2}$ .F .e

E = $\frac{1}{2}$ × mg × e

E = $\frac{1}{2}$ × 5 × 10 × 2.27×10^{–6}

$\therefore $ E = 5.68×10^{-5} J

* 5. Calculate the work done in stretching a steel wire 100 cm in length and of cross sectional area 0.03 cm*^{2}* when a load of 100 N is slowly applied without the elastic limit being reached. [Y*_{Steel}* = 2**×**10*^{11}* N/m*^{ }*]*

Solution:

Work done, W = ?

Length of wire, *l* = 100 cm = 1m

Cross-sectional area, A = 0.03 cm^{2 }

A^{ }= 0.03× (1×10^{–2})^{2} m^{2}

A = 0.03×10^{–4} m^{2}

Force, F = 100 N

Y_{Steel} = 2×10^{11} Nm^{–2 }

We know,

Young’s modulus of elasticity

Y = $\frac{Fl}{eA}$

e = $\frac{Fl}{YA}$

e = $\frac{100\times 1}{2\times {{10}^{11}}\times 0.03\times {{10}^{-4}}}$

Elongation, e = $\frac{1}{6000}$ m

Now, work done (W) = Energy stored (E)

W = $\frac{1}{2}$ .F .e

W = $\frac{1}{2}$ × 100 × $\frac{1}{6000}$

$\therefore $ W = 8.33×10^{-3} J

*6. A uniform steel wire of density 8000 Kgm*^{–3}* weight 20 g and is 2.5 m long. It lengthens by 1 mm when trenched by a force of 80 N. Calculate the value of the Young’s modulus of steel and the energy stored in the wire.*

Solution:

Here,

Density of steel wire, $\rho $ = 8000 kgm^{-3}

Mass of wire, m = 20g = 20×10^{-3}kg

Length of wire* l *= 2.5m

Elongation, e = 1mm = 1×10^{-3}m

Force, F = 80 N

Young’s modulus of steel, Y = ?

Energy stored in the wire, E = ?

We have,

Young’s modulus of elasticity,

Y = $\frac{Fl}{eA}$ …….(i)

Again,

Density, $\rho $ = $\frac{mass}{volume}$ = $\frac{m}{A.l}$

A = $\frac{m}{\rho l}$ = $\frac{20\times {{10}^{-3}}}{8000\times 2.5}$

A = 1×10^{-6} m^{2}

From equation (i),

Y = $\frac{Fl}{eA}$= $\frac{80\times 2.5}{1\times {{10}^{-3}}\times 1\times {{10}^{-6}}}$

$\therefore $ Y = 2×10^{11} Nm^{–2}

Now,

Energy stored in the stretched wire,

E = $\frac{1}{2}$ .F .e

Or, E = $\frac{1}{2}$ × 80 × 1×10^{-3}

$\therefore $ E = 40×10^{-3} J

*7. A uniform Steel wire of density 7800 Kgm*^{–3}* weights 16 gm and is 250 cm long. It lengthens by 1.2 mm when is stretched by a force of 80 N. Calculate the Young’s modulus and energy stored in the wire.*

Solution:

Here,

Density of steel wire, $\rho $ = 7800 kgm^{-3}

Mass of wire, m = 16 g = 16×10^{-3 }kg

Length of wire* l *= 250 cm = 2.5 m

Elongation, e = 1.2 mm = 1.2×10^{-3 }m

Force, F = 80 N

Young’s modulus of steel, Y = ?

Energy stored in the wire, E = ?

We have,

Young’s modulus of elasticity,

Y = $\frac{Fl}{eA}$ …….(i)

Again,

Density, $\rho $ = $\frac{mass}{volume}$ = $\frac{m}{A.l}$

A = $\frac{m}{\rho l}$

From equation (i),

Y = $\frac{Fl}{e}$× $\frac{\rho l}{m}$

Or, Y = $\frac{80\times 2.5}{1.2\times {{10}^{-3}}}$× $\frac{7800\times 2.5}{16\times {{10}^{-3}}}$

$\therefore $ Y = 2.03×10^{11} Nm^{–2}

Now,

Energy stored in the stretched wire,

E = $\frac{1}{2}$ .F .e

Or, E = $\frac{1}{2}$ × 80 × 1.2×10^{-3}

$\therefore $ E = 4.8×10^{-2} J

*8. A steel wire of density 8000 Kgm*^{–3}* weights 24 g and is 250 cm long. It lengthens by 1.2 mm when stretched by a force of 80 N. Calculate the Young’s modulus of Steel and the energy stored in the wire.*

Solution:

Here,

Density of steel wire, $\rho $ = 8000 kgm^{-3}

Mass of wire, m = 24 g = 24×10^{-3}kg

Length of wire* l *= 250 cm = 2.5m

Elongation, e = 1.2 mm = 1.2×10^{-3}m

Force, F = 80 N

Young’s modulus of steel, Y = ?

Energy stored in the wire, E = ?

We have,

Young’s modulus of elasticity,

Y = $\frac{Fl}{eA}$ …….(i)

Again,

Density, $\rho $ = $\frac{mass}{volume}$ = $\frac{m}{A.l}$

A = $\frac{m}{\rho l}$

From equation (i),

Y = $\frac{Fl}{e}$× $\frac{\rho l}{m}$

Or, Y = $\frac{80\times 2.5}{1.2\times {{10}^{-3}}}$× $\frac{8000\times 2.5}{24\times {{10}^{-3}}}$

$\therefore $ Y = 1.389×10^{11} Nm^{–2}

Now,

Energy stored in the stretched wire,

E = $\frac{1}{2}$ .F .e

Or, E = $\frac{1}{2}$ × 80 × 1.2×10^{-3}

$\therefore $ E = 4.8×10^{-2} J

*9. A uniform steel wire of density 7800 Kgm*^{–3}* weighs 16 gram and is 250 cm. It lengthens by 1.2 mm when a load of 8 kg is applied. Calculate the value of Young’s modulus for the steel and the energy stored in the wire.*

Solution:

Here,

Density of steel wire, $\rho $ = 7800 kgm^{-3}

Mass of wire, m = 16 g = 16×10^{-3 }kg

Length of wire* l *= 250 cm = 2.5 m

Elongation, e = 1.2 mm = 1.2×10^{-3 }m

Hanging mass, M = 8 Kg

Young’s modulus of steel, Y = ?

Energy stored in the wire, E = ?

We have,

Young’s modulus of elasticity,

Y = $\frac{Fl}{eA}$

Y = $\frac{Mg\times l}{eA}$ …….(i) [$\because $ F = Mg]

Again,

Density, $\rho $ = $\frac{mass}{volume}$ = $\frac{m}{A.l}$

A = $\frac{m}{\rho l}$

From equation (i),

Y = $\frac{Fl}{e}$× $\frac{\rho l}{m}$

Or, Y = $\frac{8\times 10\times 2.5}{1.2\times {{10}^{-3}}}$× $\frac{7800\times 2.5}{16\times {{10}^{-3}}}$

$\therefore $ Y = 2.03×10^{11} Nm^{–2}

Now,

Energy stored in the stretched wire,

E = $\frac{1}{2}$ .F .e

Or, E = $\frac{1}{2}$ × 80 × 1.2×10^{-3}

$\therefore $ E = 4.8×10^{-2} J

*10. A wire of length 2.5 m and area of cross section 1**×**10*^{–6}* m*^{2}* has a mass of 15 kg hanging on it. What is the extension produced? How much is the energy stored in the standard wire if Young’s modulus of wire is 2**×**10*^{11}* Nm*^{–2}*.*

Solution:

Here,

Length of wire* l *= 2.5m

Area of cross-section, A = 1×10^{–6} m^{2}

Hanging mass, m = 15 kg

Elongation, e = ?

Energy stored in the wire, E = ?

Young’s modulus of steel, Y = 2×10^{11} Nm^{–2}.

We have,

Young’s modulus of elasticity,

Y = $\frac{Fl}{eA}$

e = $\frac{Fl}{YA}$ [here, F = mg]

e = $\frac{15\times 10\times 2.5}{2\times {{10}^{11}}\times 1\times {{10}^{-6}}}$

e = 1.875 ×10^{-3} m

Again,

Energy stored in the stretched wire,

E = $\frac{1}{2}$ .F .e

Or, E = $\frac{1}{2}$ × mg × e

Or, E = $\frac{1}{2}$ × 15×10 ×1.875 ×10^{-3}

$\therefore $ E = 0.141 J

*11. A steel cable with cross sectional area 3 cm*^{2}* has an elastic limit of 2.40**×**10*^{8}* pa. Find the maximum upward acceleration that can be given a 1200 kg elevator supported by the cable if the stress does not exceed one third of the elastic limit.*

Solution:

Here,

Cross-sectional area, A = 3 cm^{2}

Or, A = 3×(1×10^{–2})^{2}

Or, A = 3×10^{–4} m^{2}

Elastic limit = 2.4×10^{8} pa (i.e. Nm^{–2 })

Maximum upward acceleration, a_{max }= ?

Mass of elevator, m = 1200 kg

Maximum stress = $\frac{1}{3}$ of elastic limit

Or, $\frac{Force\,\,(ma{{x}^{m}})}{area}$ = $\frac{1}{3}$× elastic limit

Or, $\frac{m\times {{a}_{max}}}{A}$ = $\frac{1}{3}$× elastic limit

Or, a_{max }= $\frac{1}{3}$× $\frac{A}{m}$× elastic limit

Or, a_{max }= $\frac{1}{3}$× $\frac{3\times {{10}^{-4}}}{1200}$×2.4×10^{8 }

$\therefore $ a_{max }= 20 ms^{-2}

*12. How much force is required to punch a hole 1 cm in diameter in a steel sheet 5mm thick whose shearing strength is 2.76**×**10*^{8}* Nm*^{–2}*.*

Solution:

Here,

Force, F = ?

Diameter of hole, d = 1 cm

$\therefore $ Radius of hole, r = 0.5 cm

r = 0.5 × 10^{–2} m

Thickness of steel sheet, t = 5 mm

t = 5 × 10^{–3} m

Shearing strength = 2.76 ×10^{8 }Nm^{–2}

We have,

Shearing strength = $\frac{Force}{area}$

Or, Shearing strength = $\frac{F}{C\times t}$ here, C is circumference.

Or, F = Shearing strength × C × t

Or, F = Shearing strength × 2$\pi $r × t

Or, F = 2.76 ×10^{8 }× 2$\pi $× 0.5 × 10^{–2} × 5 × 10^{–3}

$\therefore $ F = 43353.98 N

*13. A copper wire and a steel wire of the same cross sectional area and of length 1 m and 2 m respectively are connected end to end. A force is applied, which stretches their combined length by 1 cm. Find how much each wire is elongated. [Y*_{copper}* = 1.2**×**10*^{11}* N/m*^{2 }*and Y*_{steel}* = 2**×**10*^{11}* N/m*^{2}*]*

Solution:

Here,

Length of copper wire, *l*_{cu }= 1 m

Length of steel wire, *l*_{cu }= 2 m Total elongation, e_{cu }+ e_{s }= 1 cm = 1×10^{–2 }m……(i)

(i) Elongation in copper wire, e_{cu }= ?

(ii) Elongation in steel wire, e_{s }= ?

For copper, Young’s modulus of elasticity,

Y_{cu} = $\frac{F\,\,{{l}_{cu}}}{{{e}_{cu}}A}$

e_{cu} = $\frac{F\,\,{{l}_{cu}}}{{{Y}_{cu\,\,}}A}$….(ii)

For steel, Young’s modulus of elasticity,

Y_{s} = $\frac{F{{l}_{s}}}{{{e}_{s}}A}$

e_{s} = $\frac{F\,{{l}_{S}}}{{{Y}_{S}}A}$….(iii)

Dividing equation (ii) by (iii)

Or, $\frac{{{e}_{cu}}}{{{e}_{s}}}$= $\frac{{{l}_{cu}}}{{{Y}_{cu}}}$×$\frac{{{Y}_{S}}}{{{l}_{S}}}$

Or, $\frac{1\times {{10}^{-2}}-{{e}_{S}}}{{{e}_{S}}}$= $\frac{1}{1.2\times {{10}^{11}}}$× $\frac{2\times {{10}^{11}}}{2}$

Or, $\frac{1\times {{10}^{-2}}-{{e}_{S}}}{{{e}_{S}}}$= $\frac{5}{6}$

Or, 6×1×10^{–2 }– 6 e_{s}^{ }= 5 e_{s}

Or, e_{s }= $\frac{6\times 1\times {{10}^{-2}}}{11}$

∴ Elongation in steel wire, e_{s }= 5.45×10^{–3 }m.

and elongation in copper wire, e_{cu } _{= }1×10^{–2 }– 5.45×10^{–3} = 4.54 ×10^{–3 }m

*14. A rubber cord of a catapult has a cross-sectional area 1.0 mm*^{2}* and total un-stretched length 10 cm. It is stretched to 12 cm and then released to project a missile of mass 5.0 g. Calculate the velocity of projection. **[Y*_{rubber}* = 5**×**10*^{8}* N/m*^{2}*]*

Solution:

Area of cross-section, A = 1 mm^{2}

A = (1×10^{-3})^{2} m^{2} = 1×10^{-6} m^{2}

Un-stretched (original) length, *l*_{1} = 10 cm = 10 × 10^{-2 }m

Stretched (final) length,* l*_{2} = 12 cm = 12 × 10^{-2 }m

Elongation, e = *l*_{2} – *l*_{1}

e = 2cm

e = 2×10^{-2} m

Mass of missile, m = 5g

m = 5×10^{-3} kg

Velocity projection, v = ?

Y_{rubber} = 5×10^{8} N/m^{2}

We have,

Young’s modulus of elasticity,

Y = $\frac{Fl}{eA}$

F = $\frac{YAe}{l}$

F = $\frac{5\times {{10}^{8}}\times 2\times {{10}^{-2}}\times 1\times {{10}^{-6}}}{10\times {{10}^{-2}}}$

F = 100 N

Then,

Elastic energy stored in a stretched rubber

E = $\frac{1}{2}$ .F .e

E = $\frac{1}{2}$ × 100 × 2 × 10^{-2}

E = 1 J

According to work energy theorem,

Elastic potential energy, E = K.E

Or, K.E = $\frac{1}{2}$ .m . v^{2}

Or, 1 = $\frac{1}{2}$ × 5×10^{-3} v^{2}

Or, v = $\sqrt{\frac{2}{5\times {{10}^{-3}}}}$

$\therefore $ v = 20 m/s

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