Elasticity – QuantumStudy

A cubical solid aluminium (bulk modulus = -VdP/dV = 70 GPa) block has an edge length 1 m on the surface of earth .

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Q: A cubical solid aluminium (bulk modulus = -VdP/dV = 70 GPa) block has an edge length 1 m on the surface of earth . It is kept on the floor of a 5 km deep ocean . Taking the average density of water and the acceleration due to gravity to be 10³ kg/m^3 and 10 m/s^2 respectively , the change in edge length of the block in mm is …..

Ans: (0.24)

Solution: $\displaystyle \frac{\Delta V}{V} = -\frac{\Delta p}{B}$ ; Where B = Bulk modulus

V = l³

$\frac{\Delta V}{V} = 3 \frac{\Delta l}{l}$

$\displaystyle 3\frac{\Delta l}{l} = |-\frac{\Delta p}{B}|=\frac{\rho g h}{B} $

$\displaystyle \Delta l = \frac{\rho g h l}{3 B} $

A body of mass m = 10 kg is attached to one end of a wire of length 0.3 m . The maximum angular speed …..

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Q: A body of mass m = 10 kg is attached to one end of a wire of length 0.3 m . The maximum angular speed (in rad/s) with which it can be rotated about its other end in space station is (Breaking stress of wire = 4.8 × 10⁷ N/m^2 and area of cross section of the wire = 10^-2 cm²) is …..

Click to See Solution :

Ans: (4)

Sol: Breaking stress of wire = 4.8 × 10⁷ N/m^2

A = 10^-2 cm² = 10^-6 m²

l = 0.3 m

T = m ω² l

$\displaystyle \frac{T}{A} = \frac{m \omega^2 l}{A}$

$\displaystyle Breaking \; Stress = \frac{10 \times 0.3 \omega^2}{10^{-6}}$

$\displaystyle 4.8 \times 10^7 = \frac{10 \times 0.3 \omega^2}{10^{-6}}$

ω = 4 rad/s

A cube of metal is subjected to a hydrostatic pressure of 4 GPa . The percentage change in the length …

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Q: A cube of metal is subjected to a hydrostatic pressure of 4 GPa . The percentage change in the length of the side of the cube is close to (Given bulk modulus of metal , B = 8 × 10¹⁰)

(a) 1.67

(b) 5

(d) 0.6

Click to See Solution :

Ans: (a)

Sol: P = 4 GPa = 4 × 10⁹ Pa , B = 8 × 10⁹ Pa

As Bulk Modulus $\displaystyle B = \frac{P V}{\Delta V} $

$\displaystyle \frac{\Delta V}{V} = \frac{P}{B} = \frac{4 \times 10^9}{8 \times 10^{10}} $

$\displaystyle \frac{\Delta V}{V} = \frac{1}{20} $

$\displaystyle V = a^3 $

$\displaystyle \frac{\Delta V}{V} = \frac{3a^2}{a^3} \Delta a $

$\displaystyle \frac{\Delta a}{a} \times 100 = \frac{1}{3} \times \frac{\Delta V}{V} \times 100 $

$\displaystyle \frac{\Delta a}{a} \times 100 = \frac{1}{3} \times \frac{1}{20} \times 100 $

= 1.67 %

Two steel wires having same length are suspended from a ceiling under the same load . If the ratio of their energy stored ….

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Q: Two steel wires having same length are suspended from a ceiling under the same load . If the ratio of their energy stored per unit volume is 1:4 , the ratio of their diameters is

(a) √2:1

(b) 1:√2

(d) 1:2

Click to See Solution :

Ans: (a)

Sol: Energy stored per unit volume, $\displaystyle u = \frac{U}{V} = \frac{1}{2}Stress \times Strain $

$\displaystyle u = \frac{1}{2} Stress \times \frac{Stress}{Y} $

$\displaystyle u = \frac{1}{2} \frac{F}{A} \times \frac{F}{A Y} $

$\displaystyle u = \frac{1}{2} \frac{F^2}{A^2 Y} $

$\displaystyle u \propto \frac{1}{D^4 } $ (Since A ∝ D²)

$\displaystyle \frac{u_1}{u_2} = \frac{1}{4} $

$\displaystyle \frac{D_2^4}{D_1^4} = \frac{1}{4} $

$\displaystyle \frac{D_2}{D_1} = \frac{1}{\sqrt{2}} $

$\displaystyle \frac{D_1}{D_2} = \frac{\sqrt{2}}{1} $

Steel is more elastic than Rubber . Why ?

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Q: Steel is more elastic than Rubber . Why ?

Sol: Consider a Steel and a rubber wire of same length (L) , Same cross-sectional area (A) .

As we know , Young’s Modulus of Elasticity , $\displaystyle Y = \frac{F L}{A \Delta L}$

If same force (F) be applied on both the wires ,

For Steel , $\displaystyle Y_s = \frac{F L}{A \Delta L_s}$ ….(i)

For Rubber , $\displaystyle Y_r = \frac{F L}{A \Delta L_r}$ ….(ii)

On dividing ,

$\displaystyle \frac{Y_s}{Y_r} = \frac{\Delta L_r}{\Delta L_s}$

As for same force applied , ΔL_r > ΔL_s

$\displaystyle \frac{Y_s}{Y_r} = \frac{\Delta L_r}{\Delta L_s} > 1 $

Thus , Y_s > Y_r

Hence , Steel is more elastic than Rubber .

A cylindrical wire of radius 1 mm , length 1 m , Young’s modulus Y = 2 × 10^11 N/m2 , Poisson’s ratio μ = π/10 is stretched by a force of 100 N . Its radius will become

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Q: A cylindrical wire of radius 1 mm , length 1 m , Young’s modulus Y = 2 × 10¹¹ N/m² , Poisson’s ratio μ = π/10 is stretched by a force of 100 N . Its radius will become

(a) 0.99998 mm

(b) 0.99999 mm

(d) 0.99995 mm

Click to See Answer :

Ans: (d)

Sol: $\displaystyle Stress = \frac{F}{A} = \frac{100}{\pi r^2}$

$\displaystyle Stress = \frac{100}{\pi (10^{-3})^2}= \frac{10^8}{\pi}$

$\displaystyle Strain , \frac{\Delta l}{l}= \frac{Stress}{Y} $

$\displaystyle \frac{\Delta l}{l}= \frac{10^8 / \pi}{2 \times 10^{11}} = \frac{5}{\pi} \times 10^{-4}$

$\displaystyle \mu = – \frac{\Delta r /r}{\Delta l/l} $

$\displaystyle \frac{\pi}{10} = – \frac{\Delta r /r}{\frac{5}{\pi} \times 10^{-4}} $

$\displaystyle \Delta r = -0.00005 mm $

$\displaystyle r_f = 1 – 0.00005 mm $

= 0.99995 mm

A Solid Sphere of radius R made of material of bulk modulus K is surrounded by a liquid in a cylindrical container ….

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Q: A Solid Sphere of radius R made of material of bulk modulus K is surrounded by a liquid in a cylindrical container . A massless piston of area A floats on the surface of liquid . When a mass is placed on the piston to compress the liquid , fractional change in the radius (δR/R) of sphere is

(a) $\displaystyle \frac{m g}{A K} $

(b) $\displaystyle \frac{m g}{3 A K} $

(d) $\displaystyle \frac{m g}{3 A R} $

Click to See Answer :

Ans: (b)

Sol: $\displaystyle | \frac{\Delta V}{V} | = \frac{\Delta P}{K}$

$\displaystyle | \frac{\Delta V}{V} | = \frac{m g}{A K}$

$\displaystyle | \frac{\Delta V}{V} | = 3 \frac{\Delta R}{R}$

$\displaystyle \frac{m g}{A K} = 3 \frac{\Delta R}{R}$

$\displaystyle \frac{\Delta R}{R} = \frac{m g}{3 A K} $

Overall changes in volume and radii of a uniform cylindrical steel wire are 0.2 % and 0.002 % respectively when subjected to some suitable force .

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Q: Overall changes in volume and radii of a uniform cylindrical steel wire are 0.2 % and 0.002 % respectively when subjected to some suitable force . Longitudinal tensile stress acting on the wire is (Y = 2 × 10¹¹ N/m²)

(a) 3.2 × 10⁹ N/m²

(b) 3.2 × 10⁷ N/m²

(d) 4.08 × 10⁸ N/m²

Click to See Answer :

Ans: (d)

Sol: $\displaystyle V = \pi r^2 l $

$\displaystyle l = \frac{V}{\pi r^2 } $

$\displaystyle \frac{\Delta l}{l} = \frac{\Delta V}{V} + 2 \times \frac{\Delta r}{r} $

$\displaystyle \frac{\Delta l}{l} = \frac{0.2}{100} + 2 \times \frac{0.002}{100}= \frac{0.204}{100} $

Longitudinal tensile stress $\displaystyle = Y \times \frac{\Delta l}{l} $

$\displaystyle = 2 \times 10^{11} \times \frac{0.204}{100} $

= 4.08 × 10⁸ N/m²

For two different material it is given that Y1 > Y2 and B1 < B2. Here, Y is Young’s modulus of elasticity and B is the bulk modulus of elasticity. Then, we can conclude that

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Q: For two different material it is given that Y₁ > Y₂ and B₁ < B₂ . Here, Y is Young’s modulus of elasticity and B is the bulk modulus of elasticity. Then, we can conclude that

(a) 1 is more ductile

(b) 2 is more ductile

(d) 2 is more malleable

Click to See Answer :

Ans: (b) , (c)

An aluminium rod having Young’s modulus 7 × 10^9 Nm-2 has a breaking strain of 0.2 % . The minimum cross-sectional area of the rod ….

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Q: An aluminium rod having Young’s modulus 7 × 10⁹ Nm^-2 has a breaking strain of 0.2 % . The minimum cross-sectional area of the rod (in m²) in order to support a load of 10⁴ N is

(a) 1 × 10^-2

(b) 1 × 10^-8

(d) 7.1 × 10^-4

Click to See Answer :

Ans: (d)

Sol: Breaking Strain = Δl/l = 0.2/100

Stress = F/AY = Stress /Strain

7 × 10⁹ = (10⁴/A)/(0.2/100)A = 7.1 × 10^-4