Let ፒ be the torque applied at the free end and ф be the angle of twist at the junction. Then
\(\tau =\frac { \pi \eta { r }^{ 4 }(\Phi '-0) }{ 2l } =\frac { \pi \eta (2r)^{ 4 }(\Phi -\Phi ') }{ 2l } \)
⇒ ф'= 16(ф-ф')
or 17 ф' = 16 ф
or ф' = \(\\ \frac { 16 }{ 17 } \) ф.

Due to the repeated stress and strain, the material used in the bridges loses elastic strength and ultimately may collapse. Hence, bridges are declared unsafe after long use.

Bulk modulus of elasticity only.

Rubber, glass, copper and steel.

Twisting of cylinder produces pure shear.

Work done = elastic potential energy of stretched wire
= \(\frac { 1 }{ 2 } \) x \(\Upsilon \) x (strain)² x volume = \(\frac { 1 }{ 2 } \) x \(\Upsilon \) x \(\left( \frac { x }{ L } \right) ^{ 2 }\) x (A x L)
= \(\frac { \Upsilon A{ x }^{ 2 } }{ 2L } \).

A straight line.

The ratio of lateral strain to the longitudinal strain is called Poisson's ratio.

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.

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.

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.

The situation is shown in the diagram.
Now, Young's modulus (Y) = \(\frac { f }{ A } \times \frac { L }{ l } \)
  
For first wire, Y = \(\frac { f }{ \pi { r }^{ 2 } } \times \frac { L }{ l } \)      ....(i)
For second wire, Y = \(\frac { 2f }{ \pi { (2r) }^{ 2 } } \)\(\times \frac { 2L }{ l\prime } =\frac { f }{ \pi { r }^{ 2 } } \times \frac { L }{ l\prime } \)  ...(ii) 
For Eqs. (i) and (ii), \(\frac { f }{ \pi { r }^{ 2 } } \times \frac { L }{ l } \)= \(\frac { f }{ \pi { r }^{ 2 } } \times \frac { L }{ l\prime } \) 
\(​​\therefore\)    l = l'     [ \(​​\because\) both wires are of same material, hence, Young's modulus will be same].

(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.

Bulk modulus of rubber (b) = 9.8 x 10⁸ N/m²
Density of seawater (p) = 10³ kg/m³
Percentage decrease in volume
\(\left( \frac { \Delta V }{ V } \times 100 \right) =0.1\ or\ \frac { \Delta V }{ V } =\frac { 0.1 }{ 100 } \)
\(\\ or \frac { \Delta V }{ V } =\frac { 1 }{ 1000 } \)
Let the rubber ball be taken up to depth h.
Change in pressure (p) = hpg
Bulk modulus \((B)=\frac { p }{ (\Delta V/V) } =\frac { hpg }{ (\Delta V/V) } \)
\(or\ h=\frac { B\times (\Delta V/V) }{ pg } =\frac { 9.8\times 10^{ 8 }\times \frac { 1 }{ 1000 } }{ 10^{ 3 }\times 9.8 } =100m\) 

When mass M is placed on the piston, the excess pressure, p=Mg/A. As the pressure is equally applicable from all the direction on the sphere, hence there will be decrease in volume due to decrease in raius sphere. Volume of the sphere, \(V=\frac { 3 }{ 4 } \pi { R }^{ 3 }\)
Differentiating it , we get,
\(\Delta V=\frac { 4 }{ 3 } \pi (3{ R }^{ 2 })\Delta R=4\pi { R }^{ 2 }\Delta R\)
\(\\ \frac { \Delta V }{ V } =\frac { 4\pi { R }^{ 2 }\Delta R }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } =\frac { 3\Delta R }{ R } \)
We know that, \(B=\frac { P }{ dV/V } =\frac { Mg }{ A } \diagup \frac { 3\Delta R }{ R } \)
\(or\ \frac { \Delta R }{ R } =\frac { Mg }{ 3BA } \)