Elasticity Numericals Class 11 Physics

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:

Elasticity Numericals Class 11 Physics

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 = l2l1 = 2ll = 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:

Elasticity Numericals Class 11 Physics

Here,

Work done, W = ? 

Length of wire, l = 100 cm = 1m

Cross-sectional area, A = 0.03 cm

                          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:

Elasticity Numericals Class 11 Physics

Here,

Work done, W = ?

Cross-sectional area, A = 10–2 cm

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:

Elasticity Numericals Class 11 Physics

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 l1l1) = ?

(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:

Elasticity Numericals Class 11 Physics

Work done, W = ?

Length of wire, l = 100 cm = 1m

Cross-sectional area, A = 0.03 cm

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: 

Elasticity Numericals Class 11 Physics

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:

Elasticity Numericals Class 11 Physics

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:

Elasticity Numericals Class 11 Physics

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:

Elasticity Numericals Class 11 Physics

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:

Elasticity Numericals Class 11 Physics

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:

Elasticity Numericals Class 11 Physics

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 = $\frac{1}{3}$× $\frac{3\times {{10}^{-4}}}{1200}$×2.4×10

$\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:

Elasticity Numericals Class 11 Physics

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, e= 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 = l2l1 

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

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