Elasticity Numericals

List of formulae of Elasticity:

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

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

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

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

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

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

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

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

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

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

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

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

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

37. What force is required to stretch a steel wire of cross sectional area 1 cm² to double its length?

$[Y_{Steel} = 2 \times 10^{11} \text{ N/m}]$

Solution:

Force, F = ?

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

$A = (1 \times 10^{-2})^2 \text{ m}^2 = 1 \times 10^{-4} \text{ m}^2$

Let initial length, $l_1 = l$

Then final length, $l_2 = 2l$

Elongation, $e = l_2 - l_1 = 2l - l = l$

$\therefore$ Elongation, e = l

$Y_{Steel} = 2 \times 10^{11} \ N/m^2$

We know,

Young’s modulus of elasticity,

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

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

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

$\therefore F = 2 \times 10^7 \text{ N}$

38. Calculate the work done in stretching a steel wire 100 cm in length and of cross sectional area 0.030 cm² when a load of 100 N is slowly applied before the elastic limit is reached. [ $Y_{Steel} = 2 \times 10^{11} \text{ N/m}$ ]

Solution

Here,

Work done, W = ?

Length of wire, l = 100 cm = 1 m

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

$A = 0.03 \times (1 \times 10^{-2})^2 \text{ m}^2$

$A = 0.03 \times 10^{-4} \text{ m}^2$

Force, F = 100 N

$Y_{Steel} = 2 \times 10^{11} \text{ Nm}^{-2}$

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

44. Find the work done in stretching a wire of cross sectional area $10^{-2}$ cm² and 2m long through 0.1 mm, If Y for the material of wire is $2 \times 10^{11}$ Nm^-2.

Solution

Here,

Work done, W = ?

$W = 8.33 \times 10^{-3} J$

Cross-sectional area, $A = 10^{-2} \text{ cm}^2$

$A = 10^{-2} \times (1 \times 10^{-2})^2 \text{ m}^2$ $A = 1 \times 10^{-6} \text{ m}^2$

Length of wire, l = 2 m

Elongation, $e = 0.1 \text{ mm} = 0.1 \times 10^{-3} \text{ m}$

$Y_{Steel} = 2 \times 10^{11} \text{ Nm}^{-2}$

We know,

Young’s modulus of elasticity

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

$F = \frac{Y e A}{l}$

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

$F = 10 N$

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

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

$W = \frac{1}{2} \times 10 \times 0.1 \times 10^{-3}$

$W = 5 \times 10^{-4} J$

41. 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.

$[Y_{Brass} = 3.5{\times}10^{10}~Nm^{-2}]$

Solution

Here,

Mass on top, m = 5 kg

Length of rod, l = 50 cm = 0.5 m

Radius, $r = 1 \text{ cm} = 1 \times 10^{-2} \text{ m}$

• (i) Contraction, e (or $\Delta l$ or $l_l l_l$ ) = ?
• (ii) Energy stored, E = ?

$Y_{Brass} = 3.5 \times 10^{10} \text{ Nm}^{-2}$

We know,

Young’s modulus of elasticity

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

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

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

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

$\therefore$ Contraction, e = 2.27×10^-6 m Now,

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

$E = \frac{1}{2} \times mg \times e$

$E = \frac{1}{2} \times 5 \times 10 \times 2.27 \times 10^{-6}$

$\therefore E = 5.68 \times 10^{-5} J$

42. Calculate the work done in stretching a steel wire 100 cm in length and of cross sectional area 0.03 cm² when a load of 100 N is slowly applied without the elastic limit being reached. [ $Y_{Steel} = 2 \times 10^{11}$ N/m]

Solution

Work done, W = ?

Length of wire, l = 100 cm = 1 m

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

$A = 0.03 \times (1 \times 10^{-2})^2 \text{ m}^2$

$A = 0.03 \times 10^{-4} \text{ m}^2$

Force, F = 100 N

$Y_{Steel} = 2 \times 10^{11} \ Nm^{-2}$

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)

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

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

$W = 8.33 \times 10^{-3} J$

Q. No. 36. 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, $\rho = 8000 \text{ kgm}^{-3}$

Mass of wire, $m = 20g = 20 \times 10^{-3} kg$

Length of wire l = 2.5m

Elongation, $e = 1mm = 1 \times 10^{-3}m$

Force, F = 80 N

Young’s modulus of steel, Y = ?

Energy stored in the wire, E = ?

We have,

Young’s modulus of elasticity,

$Y = \frac{Fl}{eA} \dots (i)$

Again,

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

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

$A = 1 \times 10^{-6} \text{ m}^2$

From equation (i),

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

$\therefore Y = 2 \times 10^{11} \text{ Nm}^{-2}$

Now.

Energy stored in the stretched wire,

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

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

Or, $E = 40 \times 10^{-3} J$

39. 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, $\rho = 7800 \text{ kgm}^{-3}$

Mass of wire, $m = 16 g = 16 \times 10^{-3} kg$

Length of wire l = 250 cm = 2.5 m

Elongation, $e = 1.2 \text{ mm} = 1.2 \times 10^{-3} \text{ m}$

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

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

From equation (i),

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

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

$Y = 2.03 \times 10^{11} \text{ Nm}^{-2}$

Now,

Energy stored in the stretched wire,

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

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

Or, $E = 4.8 \times 10^{-2} \text{ J}$

46. A steel wire of density 8000 Kgm^-3 weights 24 g 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 of Steel and the energy stored in the wire. Solution:

Here,

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

Mass of wire, $m = 24 g = 24 \times 10^{-3} kg$

Length of wire l = 250 cm = 2.5 m

Elongation, $e = 1.2 \text{ mm} = 1.2 \times 10^{-3} \text{m}$

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

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

From equation (i),

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

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

$Y = 1.389 \times 10^{11} \text{ Nm}^{-2}$

Now,

Energy stored in the stretched wire,

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

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

Or, $E = 4.8 \times 10^{-2} \text{ J}$

45. A uniform steel wire of density 7800 Kgm^-3 weights 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, $\rho = 7800 \text{ kgm}^{-3}$

Mass of wire, $m = 16 g = 16 \times 10^{-3} kg$

Length of wire l = 250 cm = 2.5 m

Elongation, $e = 1.2 \text{ mm} = 1.2 \times 10^{-3} \text{ m}$

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) [F = Mg]

Again,

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

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

From equation (i),

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

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

$Y = 2.03 \times 10^{11} \text{ Nm}^{-2}$

Now.

Energy stored in the stretched wire,

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

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

Or, $E = 4.8 \times 10^{-2} J$

35. A wire of length 2.5 m and area of cross section $1 \times 10^{-6}$ m² 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 $2 \times 10^{11}$ Nm^-2.

Solution:

Here,

Length of wire l = 2.5m

Area of cross-section, $A = 1 \times 10^{-6} \text{ m}^2$

Hanging mass, m = 15 kg

Elongation, e = ?

Force, F = 80 N

Energy stored in the wire, E = ?

Young’s modulus of steel, $Y = 2 \times 10^{11} \text{ Nm}^{-2}$ .

We have,

Young’s modulus of elasticity,

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

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

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

$e = 1.875 \times 10^{-3} \text{ m}$

Again,

Energy stored in the stretched wire,

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

Or, $E = \frac{1}{2} \times mg \times e$

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

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

Here.

Cross-sectional area, $A = 3 \text{ cm}^2 = 3 \times (1 \times 10^{-2})^2$

Or. $A = 3 \times 10^{-4} \text{ m}^2$

Elastic limit = $2.4 \times 10^8$ 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, $\frac{\text{Force (max}^{\text{m}})}{\text{area}} = \frac{1}{3} \times \text{elastic limit}$

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

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

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

Or, $a_{max} = 20 \text{ ms}^{-2}$

43. How much force is required to punch a hole 1 cm in diameter in a steel sheet 5mm thick whose shearing strength is $2.76 \times 10^8 \ Nm^{-2}$ .

Solution

Here,

Force, F = ?

Diameter of hole, d = 1 cm

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

$r = 0.5 \times 10^{-2} \text{ m}$

Thickness of steel sheet, t = 5 mm

$t = 5 \times 10^{-3} \text{ m}$

Shearing strength = $2.76 \times 10^8 \,\text{Nm}^{-2}$ We have,

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

Or, $F = Shearing strength \times C \times t$

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

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

Or, $F = 43353.98 \text{ N}$

33. 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. [ $Y_{copper} = 1.2 \times 10^{11} \text{ N/m}^2$ and [ $Y_{steel} = 2 \times 10^{11} \text{ N/m}^2$ ]

$\therefore$ Elongation in steel wire, $e_s=5.45\times 10^{-3}\,m.$ and elongation in copper wire, $e_{cu}=1\times 10^{-2}-5.45\times 10^{-3}=4.54\times 10^{-3}\,m$

Q. No. 34. A rubber cord of a catapult has a cross-sectional area 1.0 mm² 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. [ $Y_{rubber} = 5 \times 10^8 \text{ N/m}^2$ ]

Solution:

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

$A = (1 \times 10^{-3})^2 \text{ m}^2 = 1 \times 10^{-6} \text{ m}^2$

Un-stretched (original) length, $l_1 = 10 \text{ cm} = 10 \times 10^{-2} \text{ m}$

Stretched (final) length, $l_2 = 12 \text{ cm} = 12 \times 10^{-2} \text{ m}$

Elongation, $e = l_2 - l_1$

$e = 2cm$

$e = 2 \times 10^{-2} \text{ m}$

Mass of missile, m = 5g

$m = 5 \times 10^{-3} \text{ kg}$

Velocity projection, v = ?

$Y_{rubber} = 5 \times 10^8 \text{ N/m}^2$

We have,

Young’s modulus of elasticity,

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

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

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

$F = 100 \text{ N}$

Then,

Elastic energy stored in a stretched rubber

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

$E = \frac{1}{2} \times 100 \times 2 \times 10^{-2}$

$E = 1 J$

According to work energy theorem,

Elastic potential energy, E = K.E

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

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

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

Or, $v = 20 \text{ m/s}$