The pressure exerted by a 3000 m column of water on the bottom layer
p = hρ g = 3000 m x 1000 kg m^–3 x 10 m s^–2
= 3 x 10⁷ kg m^–1 s^-2
= 3 x 10⁷ N m^–2
Fractional compression ∆V/V, is
∆V/V = stress/B = (3 x 10⁷ N m^-2) / (2.2 x 10⁹ N m^–2 )
= 1.36 x 10^-2 or 1.36 %

Here, P =10 atm = 10 x 1.013 x 10⁵ Pa; k = 37 x 10⁹ Nm^-2
Volumetric strain = \(\frac { \triangle V }{ V } =\frac { P }{ K } =\frac { 10\times 1.013\times 10^{ 5 } }{ 37\times 10^{ 9 } } \) = 2.74 x 10^-5
∴ Fractional change in volume = \(\frac { \triangle V }{ V } \) = 2.74 x 10^-5.

Since each wire is to have same tension therefore, each wire has same extension. Moreover, each wire has the same initial length. So, strain is same for each wire.
Now, \(\Upsilon =\frac { Stress }{ Strain } =\frac { F/\pi D^{ 2 }/4) }{ Strain } \)
or \(\Upsilon \propto \frac { 1 }{ { D }^{ 2 } } \Rightarrow D\propto \frac { 1 }{ \sqrt { \Upsilon } } \)
\(\frac { D_{ copper } }{ D_{ iron } } =\sqrt { \frac { { \Upsilon }_{ iron } }{ { \Upsilon }_{ copper } } } =\sqrt { \frac { 190\times 10^{ 9 } }{ 110\times 10^{ 9 } } } =\sqrt { \frac { 19 }{ 11 } } \)
= 1.14.

Here, A =15.2 x 19.2 x 10^-6 m²; F = 44500 N; η = 42 x 10⁹ Nm^-2
Strain = \(\frac { Stress }{ modulus\ of\ elasticity } =\frac { F/A }{ \eta } \)
=\(\frac { F }{ A\eta } =\frac { 44500 }{ (15.2\times 19.2\times 10^{ -6 })\times 42\times 10^{ 9 } } \)
= 3.65 x 10^-3.

(i) Young's modulus of the given material (Y) = Slope of strain-stress curve
Y = \(\frac { 150\times { 10 }^{ 6 } }{ 0.002 } =75\times { 10 }\)  N/m²
= 7.5\(\times \)10¹⁰ N/m²
(ii) Yield strength of the given material = Maximum stress, the material can sustain 
= 300 \(\times \) 10⁶ N/m² = 3 \(\times \) 10⁸ N/m²

(i) In the two graphs, the slope of graph in Fig. (a) is greater than the slope of graph in Fig. (b), so material A has greater Young's modulus. 
(ii) Material A is stronger than material B because it can withstand more load without breaking. For material A, the break even point (D) is higher.

Given, radius of steel cable (r) = 1.5 cm = 1.5 x 10^-2m
Maximum stress = 10⁸ N/m²
Area of cross-section of steel cable (A) = \(\pi { r }^{ 2 }\) 
= 3.14 x (1.5 x 10^-2)² m²
= 3.14 x 2.25 x 10^-4 m²
Maximum stress = \(\frac { Maximum\ force }{ Area\ of\ cross-section } \) 
or Maximum force = Maximum stress x Area of cross-section
= 10⁸ x (3.14 x 2.25 x 10^-4 ) N
= 7.065 x 10⁴ N
= 7.1 x 10⁴ N

Here  \(V=1 \text { litre }=10^{-3} m^{3} ;(\Delta \mathrm{V} / \mathrm{V})=\frac{0.10}{100}=10^{-3}\)
\(K=\frac{p v}{\Delta V}\)
\(\text { or } p=k \frac{\triangle V}{V}=\left(2.2 \times 10^{9}\right) \times 10^{-3}=2.2 \times 10^{6} \mathrm{~Pa}\)

The lead slab is fixed and the force is applied parallel to the narrow face as shown in Fig. 

The area of the face parallel to which this force is applied is
A = 50 cm x 10 cm
= 0.5 m x 0.1 m
= 0.05 m².
Therefore, the stress applied is
= (9.4 x 104 N/0.05 m² )
= 1.80 x 106 N.m^–2
We know that shearing strain = (∆x/L)= Stress /G.
Therefore the displacement ∆x = (Stress x L)/G
= (1.8 x 10⁶ N m^–2 x 0.5m)/(5.6 x 10⁹ N m^–2)
= 1.6 x 10^–4 m = 0.16 mm

The copper and steel wires are under a tensile stress because they have the same tension (equal to the load W) and the same area of cross-section A. From Eq. we have stress = strain x Young’s modulus. Therefore
W/A = Y_c x (∆L_c /L_c ) = Y_s x (∆L_s /L_s )
where the subscripts c and s refer to copper and stainless steel respectively. Or,
∆L_c /∆L_s = (Y_s /Y_c ) x (L_c /L_s )
Given L_c = 2.2 m, L_s = 1.6 m
Y_c = 1.1 x 10¹¹ N.m^–2, and
Y_s = 2.0 x 10¹¹ N.m^–2
∆L_c /∆L_s = (2.0 x 10¹¹/1.1 x 10¹¹) x (2.2/1.6) = 2.5
The total elongation is given to be
∆L_c + ∆L_s = 7.0 x 10^-4 m
Solving the above equations
∆L_c = 5.0 x 10^-4 m, and ∆L_s = 2.0 x 10^-4 m
Therefore
W = (A x Y_c x ∆L_c )/L_c.
= π (1.5 x 10^-3) ² x  [(5.0 × 10^-4 x 1.1 x 10¹¹)/2.2]
= 1.8 x 10² N

We assume that the rod is held by a clamp at one end, and the force F is applied at the other end, parallel to the length of the rod. Then the stress on the rod is given by
\(\text { Stress }=\frac{F}{A}=\frac{F}{\pi r^{2}}\)
\(=\frac{100 \times 10^{3} \mathrm{~N}}{3.14 \times\left(10^{-2} \mathrm{~m}\right)^{2}}\)
= 3.18 x 108 N m^–2
The elongation
\(\Delta L=\frac{(F / A) L}{Y}\)
\(=\frac{\left(3.18 \times 10^{8} \mathrm{~N} \mathrm{~m}^{-2}\right)(1 \mathrm{~m})}{2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}}\)
= 1.59 x 10^–3 m
= 1.59 mm
The strain is given by
Strain = ∆L/L
= (1.59 x 10^–3 m)/(1m)
= 1.59 x 10^–3
= 0.16 %

(a) False. The Young's modulus is defined as the ratio of stress to the strain within elastic limit. For a given stretching force elongation is more in rubber and quite less in steel. Hence, rubber is less elastic than steel.
(b) True. Stretching of a coil is determined by its shear modulus. When equal and opposite forces are applied at opposite ends of a coil, the distance as well as shape of helicals of the coil change and it involves shear modulus.

Given, each side of cube()=10cm=0.1m
Hydralic pressure (p)=7×10⁶ Pa
Bulk modulus for copper(B)=140×10⁹ Pa
Volume contraction(△V)=?|
Volume of the cube(V)=I³=(0.1)³=1×10⁻³m³
Bulk modulus for copper(B) \(=\frac { p }{ \triangle V/V } \)
\(\\ \triangle V=\frac { pV }{ B }\)
\( \\ \triangle V=\frac { 7\times 10^{ 6 }\times 1\times 10^{ -3 } }{ 140\times 10^{ 9 } } =\frac { 1 }{ 20 } \times 10^{ -6 }m^{ 3 }\)
\(\\ =0.05\times 10^{ -6 }m^{ 3 }=5\times 10^{ -8 }m^{ 3 }\)

Given for steel wireLength (l₁) = 4.7 m
Area of cross-section (A₁) = 3.0 x 10^-5m²
For copper wire
Length (l₂) =3.5 m
Area of crosssection (A₂) = 4.0 x 10^-5m²
Let F be the given load under which steel and copper wires be streched by the same amount \(\Delta l\)
Young's modulus \((Y)=\frac { F/A }{ \Delta l/l } =\frac { F\times l }{ A\times \Delta l } \)
For steel   \({ Y }_{ s }=\frac { F\times l_{ 1 } }{ { A }_{ 1 }\times { \Delta l } } \)
For copper \({ Y }_{ c }=\frac { F\times l_{ 2 } }{ { A }_{ 2 }\times { \Delta l } } \)
Dividing Eq (i)by Eq(ii) we get
\(\frac { { Y }_{ s } }{ { Y }_{ c } } =\frac { F\times { l }_{ 1 } }{ { A }_{ 1 }\times \Delta l } \times \frac { { A }_{ 2 }\times \Delta l }{ F\times { l }_{ 2 } }\)
\( \\=\frac { { l }_{ 1 } }{ { l }_{ 2 } } \times \frac { { A }_{ 2 } }{ { A }_{ 1 } } =\frac { 4.7 }{ 3.5 } \times \frac { 4.0\times { 10 }^{ -5 } }{ 3.0\times { 10 }^{ -5 } } \)
\( \frac { { Y }_{ s } }{ { Y }_{ c } } =\frac { 18.8 }{ 10.5 } =1.79=1.8\)

Total mass of all the performers, tables, plaques etc. = 280 kg
Mass of the performer = 60 kg
Mass supported by the legs of the performer at the bottom of the pyramid
= 280 – 60 = 220 kg
Weight of this supported mass = 220 kg wt. = 220 x 9.8 N = 2156 N
Weight supported by each thighbone of the performer = ½ (2156) N = 1078 N.
From Table the Young’s modulus for bone is given by
Y = 9.4 x 10⁹ N m^–2 .
Length of each thighbone L = 0.5 m
the radius of thighbone = 2.0 cm
Thus the cross-sectional area of the thighbone A = π x (2 x 10^-2) ² m² = 1.26 x 10^-3 m² .
Using Eq. = (F x L) /(A x ∆L) the compression in each thighbone (∆L) can be computed as
∆L = [(F x L)/(Y x A)]
= [(1078 x 0.5)/(9.4 x 10⁹ x 1.26 x 10^-3)]
= 4.55 x 10^-5 m or 4.55 x 10^-3 cm.
This is a very small change! The fractional decrease in the thighbone is ∆L/L = 0.000091 or 0.0091%.

Compressibility of water,
k =\(\frac{1}{B}\) = 45.8 x 10^-11 Pa^-1
Change in pressure,
Δp = 80 atm - 1 atm
= 79 atm = 79 x 1.013 x 10⁵ Pa.
\(\rho \)=1.03 x 10³ kg m^-3
As B = \(\frac { \triangle p.V }{ \triangle V } \) or \(\frac { \triangle V }{ V } =\frac { \triangle p }{ B } =\triangle p\times \frac { 1 }{ B } \)=Δp x k
or \(\\ \frac { \triangle V }{ V } \)=\(\frac { \triangle V }{ V } =\frac { (M/\rho )-(M/\rho ') }{ (M/\rho ) } =1-\frac { \rho }{ { \rho }^{ ' } } \)
or \(\frac { \rho }{ { \rho }^{ ' } } =1-\frac { \triangle V }{ V } \)
or \({ \rho }^{ ' }=\frac { \rho }{ 1-(\triangle V/V) } \)
or \({ \rho }^{ ' }=\frac { 1.03\times 10^{ 3 } }{ 1-3.665\times 10^{ -3 } } =\frac { 1.03\times 10^{ 3 } }{ 0.996 } \)
= 1.034 x 10³ kg/m³.

P =100 atmosphere
= 100 x 1.013 x 10⁵ Pa     (∵ 1 atm = 1.103 x 10⁵ Pa)
Initial volume, V₁ = 100 litre = 100 x 10^-3 m³
Final volume, V₂ = 100.5 litre = 100.5 x 10^-3 m³
∴ Change in volume = ΔV= V₃ - V₁
= (100.5 - 100) x 10^-3 m³
= 0.5 x 10^-3 m³.
Using formula of bulk modulus,
B =\(\frac { P }{ \frac { \triangle V }{ V } } =\frac { PV }{ \triangle V } \)
= \(\frac { 100\times 1.013\times 10^{ 5 }\times 100\times 10^{ -3 } }{ 0.5\times 10^{ -3 } } \)
B = 2.026 x 10⁹ Pa
Also we know that the bulk modulus of air = 1.0 x 10⁵ Pa
Now, \(\frac { Bulk\ modulus\ of\ water }{ Bulk\ modulus\ of\ air } =\frac { 2.206\times 10^{ 9 } }{ 1.0\times 10^{ 5 } } \)
= 2.026 x 10⁴
= 20260.
The ratio is too large. This is due to the fact that the strain for air is much larger than for water at the same temperature. In other words, the intermolecular distances in case of liquids are very small as compared to the corresponding distances in the case of gases. Hence there are larger interatomic forces in liquids than in gases.

Here total mass to be supported, M = 50,000 kg
∴ Total weight of the structure to be supported = Mg
= 50,000 x 9.8 N
Since this weight is to be supported by 4 columns,
∴ Compressional force on each column (F) is given by
F = \(\frac { Mg }{ 4 } =\frac { 50,000\times 9.8 }{ 4 } \)N
Inner radius of a column, r₁= 30 cm = 0.3 m
Outer radius of a column, r₂ = 60 cm = 0.6 m.
∴ Area of cross-section of each column is given by
A = \(\pi ({ r }_{ 2 }^{ 2 }-{ r }_{ 1 }^{ 2 })\)
= \(\pi \)[(0.6)²-(0.3)²] = 0.27\(\pi \) m².
Young's modulus, \(\Upsilon \) = 2 x 10¹¹ Pa.
Compressional strain of each column =?
\(\Upsilon \)= \(\\ \frac { Compressional\ force \ area }{ Compressional\ Strain } \)
= \(\frac { F/A }{ Compressional\ Strain } \)
or Compressional strain of each column
= \(\frac { F }{ A\gamma } =\frac { 50,000\times 9.8\times 7 }{ 4\times 0.27\times 22\times 2\times 10^{ 11 } } \)
= 0.722 x 10^-6
∴ Compressional strain of all columns is given by
= 0.722 x 10^-6 x 4 = 2.88 x 10^-6
= 2.88 x 10^-6.

Given, diameter of wires (2r) = 0.25 cm
\(\therefore \)   r = 0.125 cm
= 1.25 \(\times\) 10^-3 m
For steel wire,
Load (F₁) = (4 + 6) kgf
= 10 \(\times\) 9.8N = 98N
Length of steel wire (l₁) = 1.5 m
Young's modulus (Y₁ ) = 2.0 \(\times\) 10¹¹ Pa
Young's modulus (Y) = \(\frac { { F }_{ 1 }\times { l }_{ 1 } }{ { A }_{ 1 }\times { \Delta l }_{ 1 } } \)
\(\therefore \)    Change in length (\({ \Delta l }_{ 1 }\)) = \(\frac { { F }_{ 1 }\times { l }_{ 1 } }{ { A }_{ 1 }\times { Y }_{ 1 } } =\frac { { F }_{ 1 }\times { l }_{ 1 } }{ { \pi }{ r }_{ 1 }^{ 2 }\times { Y }_{ 1 } } \)
= \(\frac { 98\times 1.5 }{ 3.14\times { \left( 1.25\times { 10 }^{ -3 } \right) }^{ 2 }\times 2.0\times { 10 }^{ 11 } } \)
= 1.5\(\times\)10^-4 m
For brass wire,
Load (F₂) = 6 kgf = 6\(\times\) 9.8N = 58.8 N
Length of brass wire ( l₂) = 1.0m
Young's modulus (Y₂) = 0.91 \(\times\) 10¹¹Pa
Change in length(\({ \Delta l }_{ 2 }\)) =  \(\frac { { F }_{ 2 }\times { l }_{ 2 } }{ { \pi }{ r }_{ 2 }^{ 2 }\times { Y }_{ 2 } } \)
= \(\frac { 58.8\times 1.0 }{ 3.14\times { \left( 1.25\times { 10 }^{ -3 } \right) }^{ 2 }\times 0.91\times { 10 }^{ 11 } } \)
= 1.3\(\times\)10^-4 m

Given, mass(m) = 14 .5kg
Length of wire (l ) = 1 m
Angular frequency (v) = 2 revls
Angular velocity (\(\omega \)) = 2\(\pi \)v
= 2\(\pi \)\(\times \)2 rad/s = 4\(\pi \) rad/s
 
Area of cross-section of wire (A) = 0.065 cm²
= 6.5 \(\times \)10^-6 m²
Young's modulus for steel (Y) = 2 \(\times \) 10¹¹ N/m².
At lowest point of the vertical circle, T - mg = ml\({ \omega }^{ 2 }\)
or  T= mg +  m\({ \omega }^{ 2 }\)
= (14.5\(\times \)9.8)+14.5\(\times \)1\(\times \)\(​​{ (4\pi ) }^{ 2 }\)
= 14.5(9.8+16\({ \pi }^{ 2 }\))
= 14.5(9.8\(\times \)16\(\times \)9.87)          [\(\because \) \({ \pi }^{ 2 }\)=9.87]
= 14.5\(\times \) 167.72N=2431.94 N
Young's modulus (Y) =\(\frac { Stress }{ Strain } =\frac { (T/A) }{ \Delta l/l } =\frac { Tl }{ A.\Delta l } \)
\(\therefore \Delta l=\frac { T.l }{ A.Y } =\frac { 2431.94\times 1 }{ 6.5\times { 10 }^{ -6 }\times 2\times { 10 }^{ 11 } } \)
= 1.87\(\times \)10^-3 m =1.87 mm

Given, side of a cube (l) = 10 cm = 0.1 m
Area of its each face (A) = l² = (0.1)² = 0.01 m²
Load(m) = 100 kg
Tangential force acting on one face of the cube, F = mg = 100\(\times \) 9.8 =980 N
Shear stress acting on this face = \(\frac { F }{ A } \) = \(\frac { 980 }{ 0.01 } \) N/m²
= 9.8 \(\times \)10⁴ N/m²
Shear modulus of aluminium (\(\eta \)) = 25 GPa
= 25\(\times \) 10⁹ N/m²
Shear modulus (\(\eta \))=\(\frac { Shearing\ stress }{ Shearing\ strain } \)
or    shearing strain \(\left( \frac { \Delta L }{ L } \right) \) = \(\frac { Shearing\ stress }{ Shear\ modulus } \)
or      \(\Delta L\)   = \(\frac { Shearing\ stress }{ Shear\ modulus } \) \(\times \) L = \(\frac { 9.8\times { 10 }^{ 4 } }{ 25\times { 10 }^{ 9 } } \)\(\times \) 0.1
= 0.0392 \(\times \)10^-5m
= 3.92 \(\times \) 10^-7m