Steel is more elastic than Rubber . Why ?

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

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

 

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

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} $

(c) $\displaystyle \frac{m g}{ A } $

(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 .

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

(c) 3.6 × 109 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

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

(c) 1 is more malleable

(d) 2 is more malleable

Click to See Answer :
Ans: (b) , (c)