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.

A straight line.

(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.

[(\(\frac{p}{γB}\))]

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 \(\infty \)

Given,
L=50 cm=50×10⁻² m
t=10 cm=10×10⁻² m
G=5.6×10⁹ Pa
F=9.0×10⁴ N
Area of the face on which force is applied,
A=50×10=500 cm²=0.05 m²
Let ΔL be the displacement of the upper edge of the slab, due to tangential force F applied.
\( \text { Then, } G=\frac{\left(\frac{F}{A}\right)}{\left(\frac{\Delta L}{L}\right)} \)
\( \Rightarrow \Delta L=\frac{F L}{G A}=\frac{9 \times 10^4 \times 50 \times 10^{-2}}{5.6 \times 10^9 \times 0.05} \)
\( \therefore \Delta L=1.6 \times 10^{-4} \mathrm{~m} \)

(a) Let Y = Young's modulus of the material of wire.
If a be its area of cross-section, then
\(\Upsilon =\frac { F/A }{ l/L } =\frac { WL }{ Al } \)    (∵ F = W)
or l=\(\frac { WL }{ A\Upsilon } \)             ...........(i)
(b) When the wire is passed over a pully, let I' be the increase in length of the each segment. Since \(\frac{L}{2}\) = length of each segment
∴ \(\quad \Upsilon =\frac { W(L/2) }{ Al^{ ' } } \)
or l' = \(\frac { 1 }{ 2 } \frac { WL }{ A\Upsilon } =\frac { 1 }{ 2 } \) [From eqn. (i)]         ......(ii)
∴ Total increase in the length of the wire
= l' + l' = 2l'= 2 x \(\frac{l}{2}\)= l

Let l and A be the original length and area of cross-section of the metal wire.
Change in length in the first case = (l₁ - l)
Change in length in the second case = (l₂ - l)
\(\Upsilon =\frac { { T }_{ 1 } }{ A } \times \frac { l }{ ({ l }_{ 1 }-l) } =\frac { { T }_{ 2 } }{ A } \times \frac { l }{ ({ l }_{ 2 }-l) } \)
or T₂l₂ - T₁l = T₂l₁-T₂l
or l(T₂-T₁) = T₂l₁-T₁l₂ ⇒ l = \(\frac { { T }_{ 2 }{ l }_{ 1 }-{ T }_{ 1 }{ l }_{ 2 } }{ ({ T }_{ 2 }-{ T }_{ 1 }) } \).

(i) The resolved part of F along the normal is the tensile force on this plane and the resolved part parallel to the plane is the shearing force on the plane. 
Area of MO plane section = A \(\sec { \theta } \)
Tensile stress =\(\frac { Force }{ Area } \)= \(\frac { F\cos { \theta } }{ F\sec { \theta } } \) 
=\(\frac { F }{ A } \cos ^{ 2 }{ \theta } \)     \(\left[ \because \quad \sec { \theta } =\frac { 1 }{ \cos { \theta } } \right] \quad \)
(ii) Shearing stress applied on the top face
So,      F = \(F\sin { \theta } \)
Shearing stress =\(\frac { Force }{ Area } \)=\(\frac { F\sin { \theta } }{ A\sec { \theta } } \)
= \(\frac { F }{ A } \sin { \theta } \cos { \theta } \)
= \(\frac { F }{ 2A } \sin { 2\theta } \)     \(\left[ \because \quad \sin { 2\theta } =2\sin { \theta } \cos { \theta } \right] \)
(iii) Tensile stress will be maximum when \(\cos ^{ 2 }{ \theta } \)i is maximum i.e. \(\cos { \theta } \) = 1 or \( \theta\) = 0°.
(iv) Shearing stress will be maximum when \(sin 2 \theta\) is maximum i.e. \(sin 2 \theta\) = 1 or \(2\theta\) = 90° or \( \theta\) = 45°.

\(Work\ done\ in\ strtching\ a\ wire\ is\ given\ by\)
\(\\ W=\frac { 1 }{ 2 } F\times \triangle l\)
\(\\ As\ springs\ of\ steel\ and\ copper\ are\ equally\ streched.Therefore,for\ same\ force(F).\)
\(\\ W\propto \triangle l\)
\(\\ Young's\ modulus\ (Y)=\frac { F }{ A } \times \frac { l }{ \triangle l } \)
\(\\ or\ \triangle l=\frac { F }{ A } \times \frac { l }{ Y } \)
\(\\ As\ both\ springs\ are\ identical\)
\( \triangle l\propto \frac { 1 }{ Y } \)
\(\\ From \ Eqs.(i) \ and \ (ii),\ we \ get \ W\propto \frac { 1 }{ Y }\)
\( \\ \therefore \frac { { W }_{ steel } }{ { W }_{ copper } } =\frac { { Y }_{ copper } }{ { Y }_{ steel } } <1\ [as\ { Y }_{ steel }>{ Y }_{ copper }]\)
\(\\ or\ { W }_{ steel }\ <\ { W }_{ copper }\)
\(\\ Therefore,\ more\ work\ will\ be\ done\ for\ stretching\ copper\ spring.\)

Due to extension produced in the cord, energy is stored in it which is converted into kinetic energy when the stone flies away. Assuming that there is no loss of energy in this process, the kinetic energy of the stone is given by
W = \(\frac{1}{2}\)mv² = 4J
This must be equal to the work done in stretching the cord.
Using the equation.
W = \(\frac{1}{2}\)FΔl = 4J
Where F is the stretching force. Since
Δl = 20 cm = 0.2 m
F= \(\frac { 4\times 2 }{ 0.2 } \)= 40 N
Stress = \(\frac { F }{ A } =\frac { 40 }{ { \pi r }^{ 2 } } \)
Now, r = 3 mm = 3 x 10^-3 m
Hence, Stress = \(\frac { 40 }{ \pi (3\times 10^{ - })^{ 2 } } \)=1.415 x 10⁶ Nm^-2
But Strain = \(\frac { 20 }{ 42 } \)=0.476
Young's modulus = \(\Upsilon =\frac { 1.415\times { 10 }^{ 6 } }{ 0.476 } \) 
= 2.97 x 10⁶ Nm^-2.

For Hooke's law and Young's modulus of elasticity, see fact that matter on pages 386-387 of this text book.
Let L be the original length of the wire, A be its .area of cross-section and W be the load attached to the wire. Then, Young's modulus of the wire is given by
Y = \(\frac { W\times L }{ A\times \triangle L } \)
Since ΔL = 90-L
∴ \(\Upsilon =\frac { W\times L }{ A(90-L) } \)      ......(1)
Volume of weight attached = \(\frac { W }{ density\quad of\quad weight } \)
= \(\frac { W }{ 7.6 } \)cm³
Weight of water displaced = \(\frac { W }{ 7.6 } \) x density of water
\(\frac { W }{ 7.6 } \times 1=\frac { W }{ 7.6 } \)
∴ Net weight after immersing in water is
W'=W-\(\frac { W }{ 7.6 } =\frac { 6.6W }{ 7.6 } \)
Length of wire after immersing in water
= (90 - 0.18) = 89.82 cm
∴ Change in length on immersing in water,
ΔL'=(89.82-L)cm
∴ \(\Upsilon \)=\(\frac { W'L }{ A\triangle L' } \)
= \(\frac { 6.6W\times L }{ 7.6\times A(89.82-L) } \)     ....(2)
Comparing eqns. (1) and (2), we get
\(\frac { W\times L }{ A(90-L) } =\frac { 6.6W\times L }{ 7.6\times A\times (89.82-L) } \)
7.6 x (89.82-L) = 6.6 (90-L)
682.632-7.6L = 594-6.6L
∴ L = 88.632 cm

Given, modulus of rigidity, \(\eta =2\times { 10 }^{ 7 }\) dyne/cm³
      
Side of the cube, l = 8 cm
Area, A = l² = 64 cm²
Force or load, F= 300 kgf = 300\(\times \) 1000\(\times \)981 dyne 
As,     \(\eta =\frac { F }{ A\theta } \)
\(\Rightarrow \quad \theta =\frac { F }{ A\eta } \) 
 \(\theta =\frac { 300\times 1000\times 981 }{ 64\times 2\times { 10 }^{ 7 } } \simeq \) 0.23 rad
As,     \(\eta =\frac { F }{ A } \frac { l }{ \Delta l } \)
\(\Rightarrow \quad \quad \quad \frac { \Delta l }{ l } =\theta\)
\( \Rightarrow \quad \Delta l=l\theta =8\times 0.23\)
\(\Delta l\)    = 1.84 cm

1.67 km