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 cm2 to double its length?
[YSteel = 2×1011 N/m ]
Solution:
Force, F = ?
Area of cross-section, A = 1 cm2
A = (1×10-2)2 m2 = 1×10-4 m2
Let initial length, l1 = l
Then final length, l2 = 2l
Elongation, e = l2 – l1 = 2l – l = l
$\therefore $ Elongation, e = l
YSteel = 2×1011 N/m2
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 = 2×107 N
2. 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. [YSteel = 2×1011 N/m ]
Solution:
Here,
Work done, W = ?
Length of wire, l = 100 cm = 1m
Cross-sectional area, A = 0.03 cm2
A = 0.03× (1×10–2)2 m2
A = 0.03×10–4 m2
Force, F = 100 N
YSteel = 2×1011 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 cm2 and 2m long through 0.1 mm, If Y for the material of wire is 2×1011 Nm–2.
Solution:
Here,
Work done, W = ?
Cross-sectional area, A = 10–2 cm2
A = 10–2 × (1×10–2)2 m2
A = 1×10–6 m2
Length of wire, l = 2 m
Elongation, e = 0.1 mm = 0.1×10–3 m
YSteel = 2×1011 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.
[YBrass = 3.5×1010 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 l1– l1) = ?
(ii) Energy stored, E = ?
YBrass = 3.5×1010 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 cm2 when a load of 100 N is slowly applied without the elastic limit being reached. [YSteel = 2×1011 N/m ]
Solution:
Work done, W = ?
Length of wire, l = 100 cm = 1m
Cross-sectional area, A = 0.03 cm2
A = 0.03× (1×10–2)2 m2
A = 0.03×10–4 m2
Force, F = 100 N
YSteel = 2×1011 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 = 20×10-3kg
Length of wire l = 2.5m
Elongation, e = 1mm = 1×10-3m
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 m2
From equation (i),
Y = $\frac{Fl}{eA}$= $\frac{80\times 2.5}{1\times {{10}^{-3}}\times 1\times {{10}^{-6}}}$
$\therefore $ Y = 2×1011 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 = 2.03×1011 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 = 24×10-3kg
Length of wire l = 250 cm = 2.5m
Elongation, e = 1.2 mm = 1.2×10-3m
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 = 1.389×1011 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 = 2.03×1011 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 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 2×1011 Nm–2.
Solution:
Here,
Length of wire l = 2.5m
Area of cross-section, A = 1×10–6 m2
Hanging mass, m = 15 kg
Elongation, e = ?
Energy stored in the wire, E = ?
Young’s modulus of steel, Y = 2×1011 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 cm2 has an elastic limit of 2.40×108 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 cm2
Or, A = 3×(1×10–2)2
Or, A = 3×10–4 m2
Elastic limit = 2.4×108 pa (i.e. Nm–2 )
Maximum upward acceleration, amax = ?
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, amax = $\frac{1}{3}$× $\frac{A}{m}$× elastic limit
Or, amax = $\frac{1}{3}$× $\frac{3\times {{10}^{-4}}}{1200}$×2.4×108
$\therefore $ amax = 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 2.76×108 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 ×108 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 $r × t
Or, F = 2.76 ×108 × 2$\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. [Ycopper = 1.2×1011 N/m2 and Ysteel = 2×1011 N/m2]
Solution:
Here,
Length of copper wire, lcu = 1 m
Length of steel wire, lcu = 2 m Total elongation, ecu + es = 1 cm = 1×10–2 m……(i)
(i) Elongation in copper wire, ecu = ?
(ii) Elongation in steel wire, es = ?
For copper, Young’s modulus of elasticity,
Ycu = $\frac{F\,\,{{l}_{cu}}}{{{e}_{cu}}A}$
ecu = $\frac{F\,\,{{l}_{cu}}}{{{Y}_{cu\,\,}}A}$….(ii)
For steel, Young’s modulus of elasticity,
Ys = $\frac{F{{l}_{s}}}{{{e}_{s}}A}$
es = $\frac{F\,{{l}_{S}}}{{{Y}_{S}}A}$….(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 es = 5 es
Or, es = $\frac{6\times 1\times {{10}^{-2}}}{11}$
∴ Elongation in steel wire, es = 5.45×10–3 m.
and elongation in copper wire, ecu = 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 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. [Yrubber = 5×108 N/m2]
Solution:
Area of cross-section, A = 1 mm2
A = (1×10-3)2 m2 = 1×10-6 m2
Un-stretched (original) length, l1 = 10 cm = 10 × 10-2 m
Stretched (final) length, l2 = 12 cm = 12 × 10-2 m
Elongation, e = l2 – l1
e = 2cm
e = 2×10-2 m
Mass of missile, m = 5g
m = 5×10-3 kg
Velocity projection, v = ?
Yrubber = 5×108 N/m2
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 . v2
Or, 1 = $\frac{1}{2}$ × 5×10-3 v2
Or, v = $\sqrt{\frac{2}{5\times {{10}^{-3}}}}$
$\therefore $ v = 20 m/s