Elasticity

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

(c) 20

(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

(c) 2:1

(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 , ΔLr > ΔLs

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

Thus , Ys > Yr

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

(c) 0.99997 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