Volume of wire
V = \(\pi\)r²L
∴ Density = \(\frac { mass }{ V } \)
9 x 10³ = \(\frac { 31.4 }{ \pi r^{ 2 }\times L } \)
L = \(\frac { 3.14 }{ { \pi r }^{ 2 }\times 9\times 10^{ 3 } } \)
L = \(\frac { 31.4 }{ 3.14\times 10^{ -6 }\times 9\times 10^{ 3 } } \)
∴ L = \(\frac { 10 }{ 9 } \) x 10³m
=\(\frac { { 10 }^{ 4 } }{ 9 } \)m
Now \(\Upsilon =\frac { mg\quad L }{ { \pi r }^{ 2 }l } \)
l = \(\frac { mg\quad L }{ { \pi r }^{ 2 }.\Upsilon } \)
=\(\frac { 3.14\times 9.8\times { 10 }^{ 4 } }{ 3.14\times 10^{ -6 }\times 9\times 9.8\times 10^{ 10 } } \)
= \(\frac { 10 }{ 9 } \)m
Now the work done =\(\frac { 1 }{ 2 } \)F.l
=\(\frac { 1 }{ 2 } \) x 3.14 x 9.8 x \(\frac { 10 }{ 9 } \)
75%of the work done is converted into heat energy
∴ Heat energy =\(\frac { 1 }{ 2 } \) x 31.4 x 9.8 x \(\frac { 10 }{ 9 } \times \frac { 75 }{ 100 } \)
But heat energy  =mass x S.P. heat x temp difference
= 1.4 x 490 x t
∴ 31.4 x 490 x t = \(\frac { 1 }{ 2 } \) x31.4 x 9.8 x \(\frac { 10 }{ 9 } \times \frac { 75 }{ 100 } \)
∴ t = \(\frac { \frac { 1 }{ 2 } \times 3.14\times 9.8\times \frac { 10 }{ 9 } \times \frac { 75 }{ 100 } }{ 3.14\times 490 } \)
t = \(\frac { 1 }{ 120 } \)K or 0.0083⁰C

For steel wire A, l₁ =I; A_z = 1 mm²: \(\Upsilon \)₁ = 2 x 10¹¹ Nm^-2
For aluminium wire B, l₂= I; A₂ = 2mm²; \(\Upsilon \)₂ = 7 x 10¹⁰ Nm^-2
(a) Let mass m be suspended from the rod at distance x from the end where wire A is connected. Let F₁ and F₂ be the tensions in two wires and there is equal stress in two wires, then
\(\frac { { F }_{ 1 } }{ { A }_{ 1 } } =\frac { { F }_{ 2 } }{ { A }_{ 2 } } \Rightarrow \frac { { F }_{ 1 } }{ { F }_{ 2 } } =\frac { { A }_{ 1 } }{ { A }_{ 2 } } =\frac { 1 }{ 2 } \)           ...(i)
Taking moment of forces about the point of suspension of mass from the rod, we have
F₁x = F₂(1.05 - x) or \(\frac { 1.05-x }{ x } =\frac { { F }_{ 1 } }{ F_{ 2 } } =\frac { 1 }{ 2 } \)
or 2.10-2x = x ⇒ x = 0.70 m = 70 cm
(b) Let mass m be suspended from the rod at distance x from the end where wire A is connected. Let F₁ and F₂ be the tension in the wires and there is equal strain in the two wires i.e.,
\(\frac { { F }_{ 1 } }{ { A }_{ 1 }{ \Upsilon }_{ 1 } } =\frac { F_{ 2 } }{ { A }_{ 2 }{ \Upsilon }_{ 2 } } \Rightarrow \frac { { F }_{ 1 } }{ { F }_{ 2 } } =\frac { { A }_{ 1 }{ \Upsilon }_{ 1 } }{ { A }_{ 2 }{ \Upsilon }_{ 2 } } =\frac { 1 }{ 2 } \times \frac { 2\times 10^{ 11 } }{ 7\times 10^{ 10 } } =\frac { 10 }{ 7 } \)
As the rod is stationary, so F1x = F2 (1.05 - x) or \(\frac { 1.05-x }{ x } =\frac { { F }_{ 1 } }{ { F }_{ 2 } } =\frac { 10 }{ 7 } \)
⇒ 10x = 7.35 - 7x or x = 0.4324 m
= 43.2 cm.

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.

Quick reaction of locals and their help without waiting for police to arrive helped in saving lives.

The amount by which the cable streches depends on the elascity of the steel cable. Young's Modulus for steel is given as Y = 20 x 10⁸ N/m²
Given Diameter D = 3.0mm = 3.0 x 10^-3m
Length, L = 8.0 m; Mass m = 130 kg
Radius \(r=\frac { D }{ 2 } =\frac { 3.0\times { 10 }^{ -3 } }{ 2 } =1.5\times { 10 }^{ -3 }m\)
The area of cross-section of the cable
 \(A=\pi { r }^{ 2 }=\pi \times (1.5\times { 10 }^{ -3 })^{ 2 }=7.065\times { 10 }^{ -6 }{ m }^{ 2 }\)
Thus,     F = w = mg =130 x 98
F = 1274 N
The change in length
\(Y=\frac { Stress }{ Strain } =\frac { F/A }{ \Delta L/L } \Rightarrow \Delta L=\frac { LF }{ AY }\) 
\( \\ \Delta L=\frac { 8.0\times 1274 }{ 7.065\times { 10 }^{ -6 }\times 20\times { 10 }^{ 8 } } =0.72m=720mm\) 

Let L be the initial length of the each side of the cube.
Volume, V = L x L x L = L³
= initial volume (V_i say)
If the each side of the cube is shortened by 2 %, then final length of the cube
= L - 2 % of L
= \(\left( L-\frac { 2L }{ 100 } \right) \) 
= L \(\left( 1-\frac { 2 }{ 100 } \right) \) 
\(\therefore\)   Final volume, V_f = L³ \(\left( 1-\frac { 2 }{ 100 } \right) ^{ 3 }\) 
= V \(\left( 1-\frac { 2 }{ 100 } \right) ^{ 3 }\) 
Change in volume, \(\Delta V\)  = V_f - V_i = V\(\left( 1-\frac { 2 }{ 100 } \right) ^{ 3 }\) - V
= V \(\left[ \left( 1-\frac { 2 }{ 100 } \right) ^{ 3 }-1 \right] \) 
 \(\frac { \Delta V }{ V } \) = \(\left( 1-\frac { 2 }{ 100 } \right) ^{ 3 }\) - 1 = \(\left[ 1-\frac { 2\times 3 }{ 100 } \right] \) -1
( \(\because\) ( 1 - x )ⁿ \(\simeq \) 1 - nx for x <<1]
\(\therefore\)  Volumetric strain = \(\frac { \Delta V }{ V } \) = 1 - 0.06 - 1 = 0.06          (take positive sign)                                     

V =1 litre = 10^-3 m³; ΔV/V  = 0.10/100 = 10^-3
K = \(\frac { pV }{ \triangle V } \)
p = K\(\frac { pV }{ \triangle V } \) = (2.2 x 10⁹) x 10^-3 = 2.2 x 10⁶ Pa.

(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

0.01m

2 x 10^-2J

Bulk modulus \((B)=\frac { p }{ \Delta V/V } =\frac { pV }{ \Delta V } \)
For perfectly rigid body, change in volume Δ V=0
\(B=\frac { pV }{ 0 } =\infty \)
Therefore, Bulk modulus for a perfectly rigd body is \(\infty \)

The slope of stress (on y-axis) and strain (on x-acis) gives modulus of elasticity.
The Slope of stress (on x-axis) and strain (on y-axis) gives the reciprocal of modulus of elasticity.

Longitudinal strain and shear strain.

Pressure is an external force per unit area, while stress is the internal restoring force which comes into play in  a deformed body acting transversely per unit area of  body.

(i) Energy density = \(\frac { 1 }{ 2 } \left( \frac { \Upsilon l }{ L } \right) .\frac { 1 }{ L } \)
=\(\frac { 1 }{ 2 } \left[ \frac { 2\times 10^{ 11 }\times 2\times { 10 }^{ -3 } }{ 4 } \right] \times \left[ \frac { 2\times 10^{ -3 } }{ 4 } \right] \)
=\(\frac { 1 }{ 2 } \) x 10⁸ x \(\frac { 1 }{ 2 } \) x 10^-3
= 0.25 x 10⁵
= 2.5 x 10⁴ J/m³.
(ii) Potential energy stored in the wire
U = \(\frac { 1 }{ 2 } \left( \frac { \Upsilon Al }{ L } \right) .l\)
= \(\frac { 1 }{ 2 } \left[ \frac { 2\times { 10 }^{ 11 }\times 2\times 10^{ -6 }\times 2\times 10^{ -3 } }{ 4 } \right] \) x 2 x 10^-3
= 10² x 2 x 10^-3
= 0.2 J.

(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.

As shown in figure, consider a small element of thickness dx at distance x from the fixed support. Force acting on the element dx is                                      
F = Weight of length (l- x) of the rod
= \(\frac { W }{ l } \)(l-x)
Elongation of the element = Original length \(\times \)\(\frac { Stress }{ Y } \)
= \(dx\times \frac { F/A }{ Y } =\frac { W }{ l\quad Ay } (l-x)dx\)               
Total elongation produced in the rod =  \(\frac { W }{ l\quad Ay } \int _{ 0 }^{ l }{ (l-x)dx } =\frac { W }{ l\quad Ay } { \left[ lx-\frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ l }\)
= \(\frac { W }{ l\quad Ay } \left[ { l }^{ 2 }-\frac { { l }^{ 2 } }{ 2 } \right] =\frac { W }{ 2Ay } \)

Given, compressional force, F = 50000 N
Diameter, D = 0.5 mm = 5\(\times \)10^-4 m
Radius, r = \(\frac { D }{ 2 } \) = 2.5 \(\times \)10^-4m
Pressure at the tip of the anvil (p) = \(\frac { Force }{ Area } \)
\( p=\frac { F }{ \pi { r }^{ 2 } } =\frac { 50000 }{ { \left( 2.5\times { 10 }^{ -4 } \right) }^{ 2 } } \) 
=  2.5 \(\times \)10¹¹ Pa.