Let $$h$$ be the depth at which the rubber ball be taken, Then
$$P=h\rho g...(i)$$
By definition of bulk modulus
$$B=-\cfrac { P }{ \Delta V/V } $$
The negative sign shows that with increase in pressure, a decrease in volume occurs.
$$\quad \therefore P=B\cfrac { \Delta V }{ V } $$
Using (i)
$$h\rho g=B\cfrac { \Delta V }{ V } \quad or\quad h=\cfrac { B }{ \rho g } \cfrac { \Delta V }{ V } $$
Substituting the given values, we get
$$h=\left( \cfrac { 9\times { 10 }^{ 8 }N\quad { m }^{ -2 } }{ { 10 }^{ 3 }kg\quad { m }^{ -3 }\times { 10 }^{  }m{ s }^{ -2 } }  \right) \left( \cfrac { 0.1 }{ 100 }  \right) =90m$$

Given,

Here $$L=10cm=10\times { 10 }^{ -2 }m;P=7MPa=7\times { 10 }^{ 6 }Pa;B=140GPa=140\times { 10 }^{ 9 }Pa$$
As $$B=\cfrac { P }{ \Delta V/V } $$
$$\therefore \Delta V=\cfrac { PV }{ B } =\cfrac { P{ L }^{ 3 } }{ B } =\cfrac { \left( 7\times { 10 }^{ 6 }Pa \right) { \left( 10\times { 10 }^{ -2 }m \right)  }^{ 3 } }{ 140 } =5\times { 10 }^{ -8 }{ m }^{ 3 }=5\times { 10 }^{ -2 }{ cm }^{ 3 }$$

Normally, we use Searle's method to measure Young's modulus of some material. Young's modulus is independent of the shape of the material, so we can utilize any shape for its calculation. Searle’s apparatus is used for the measurement of Young’s modulus experimentally. This apparatus consists of two equal length wires that are attached to a rigid support.

Sln :
We know Y = $$\dfrac{stress}{strain} \,= \,\dfrac{Wl}{2Ax} \, \times \, \dfrac{2l^2}{x^2}$$

Hint:

The axial force $$25kN$$ is transmitted to each of the three bars.

Given,

Volumetric strain in tooth cavity $$ =\dfrac { \Delta V }{ V }  $$ 
Let $$ \gamma  $$ be the coefficient of volume expansion with the change in temperature $$ \Delta T $$ .
Change in volume is 
$$ \Delta V=\gamma V\Delta T\quad or\quad \dfrac { \Delta V }{ V } =\gamma \Delta T $$ 
Thermal stress in tooth cavity
     $$ =\beta \times volumetric \ strain=\beta \gamma \Delta T $$
    $$ =\beta \times 3\alpha \Delta T $$               $$  \left( \because \gamma =3\alpha  \right)  $$
$$ =140\times { 10 }^{ 9 }\times 3\times 1.7\times { 10 }^{ -5 }\times \left( 57-37 \right)  $$
$$ =1.43\times { 10 }^{ 8 }N{ m }^{ -2 } $$

Y=F/AΔl/l=FAlΔl

Consider the expression for Young’s modulus.

$$ Y=\dfrac{FL}{A\Delta L} $$

$$ \Delta L=\dfrac{FL}{AY} $$

Where A is the area, F is the applied force, $$\Delta L$$ is the increase in length.

So,

$$ \Delta L=\dfrac{FL}{\pi {{r}^{2}}Y} $$

$$ \Delta L\propto \dfrac{1}{{{r}^{2}}} $$

$$ \dfrac{{{l}_{2}}}{{{l}_{1}}}={{\left( \dfrac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}} $$

$$ ={{\left( \dfrac{{{d}_{1}}/2}{{{d}_{2}}/2} \right)}^{2}} $$

$$ ={{(n)}^{2}} $$

$$ {{l}_{2}}=n^2\,{{l}_{1}} $$

Hence, on applying the same load, the length is increased by a factor of $${{n}^{2}}$$.

Sln :
$$Y \, = \, \dfrac{Fla}{\Delta lll} \, = \, \dfrac{Fl}{a \Delta l} \, or \, \Delta l \, \, or \, \dfrac{1}{D^2}$$

$$\dfrac{\delta V}{V} \, = \, -\dfrac{P}{B}$$
Where $$P \, = \, \rho gh$$ 
Then $$-\dfrac{\delta V}{V}\, = \, \dfrac{\rho gh}{B}$$
Since, the volume of the sphere is V = $$\dfrac{4}{3}\pi r^{3}$$, we have 
$$\dfrac{\delta V}{V}\, = \, \dfrac{3\delta r}{r}$$
Using Eqs, (i) and (ii), we have 

$$Y \, = \, \dfrac{Fl}{a\Delta l}$$
Y, l and F are constants.
$$\therefore \, \Delta l \, \, or \, \dfrac{1}{D^2}$$

Youngs modulus is the property of material; it does not depend on shape and size of body.

Two prominent mechanisms of plastic deformation, namely slip and twinning occur . 
1. Slip involves sliding of blocks of crystal over one other along definite crystallographic planes, called slip planes. During slip each atom usually moves same integral number of atomic distances along the slip plane producing a step, but the orientation of the crystal remains the same.

2. Portion of crystal takes up an orientation that is related to the orientation of the rest of the untwined lattice in a definite, symmetrical way. The twinned portion of the crystal is a mirror image of the parent crystal.

$$Vp^2 = C$$
$$\beta = -V \left(\dfrac{dp}{dv}\right)$$

$$p^2 + 2vp \dfrac{dp}{dv} = 0$$

$$\dfrac{dp}{dv} = \dfrac{-p}{2v}$$

$$\beta = -v \times \left(\dfrac{-p}{2v}\right) = \dfrac{P}{2}$$

The tensile strength = stress = force/area

Thus, $$area = Force/ Stress =100/ (70 \times  10^6) = 1.42 \times 10^{-6} m^2$$

The correct option is option (b)

The yield point is the point on a stressstrain curve that indicates the limit of elastic behavior and the beginning of plastic behavior. Yielding means the start of breaking of fibers.

The correct option is option(a)

Stress is defined by Stress = $$1000/0.2=5000 Nm^{-2}$$

The correct option is option (b)

Plasticity is defined as the property which enables a material to be deformed continuously and permanently without rupture during the application of force. This deformation determines the viscous behavior of the material and is irrecoverable.

Substances that elongate considerably and undergo plastic deformation before they break are known as ductile substances

The yield point is the point on a stressstrain curve that indicates the limit of elastic behavior and the beginning of plastic behavior. Yielding means the start of breaking of fibers.

Thus, Hooke's law dosent hold good in this region, as stress is not proportional to strain in this region

Elasticity is defined as the ability of a body to  resist a distorting influence and to return to its original size and shape when that influence or force is removed.

Bulk modulus is pressure/volumetric strain. The strain dosen't have any units and hence bulk's modulus takes the units of pressure. Thus, dimensions of bulk's modulus is $$[B]=[MLT^{-2}]/[L^2] =[ML^{-1}T^{-2}]$$

The correct option is (b)

Longitudinal strain is calculated using the formula Change in length/ original length

The correct option is option (a)

Poisson' ratio is defined as the ratio of lateral stress and longitudinal stress

The correct option is (c)

Pressure is applied on an object to compress it, while stress is induced inside the body and acts opposite to the direction of the pressure applied

The correct option is option (d)

Bulk Modulus k is related to the compression of a liquid and the decrease in volume per unit volume. It is the ratio of compressive stress to the volumetric strain.

The correct option is (a)

The applied force can be resolved in to two components namely, 
F cos 30 along the horizontal and F sin 30 along the vertical

Thus, the normal stress can be determined as F sin 30 x Area of cross section
The areal stress can be determined using F cos 30 x Area perpendicular to horizontal axis

Thus, the correct option is (a)

The plot is between extension and the force (mg), which will be a straight line

The correct option is option (c)

We know that $$Y=\dfrac{F/A}{\delta L/L}=\dfrac{FL}{A \delta L}$$

Breaking force area of cross-section of wire i.e. load hold by the wire does not depend upon the length of the wire.   

The correct option is (b)

$$Stress = Weight/area = mg/A = (\rho A g)/A=\rho g$$ and
$$strain=\delta L/L \implies $$ change in length = $$\rho g L/Y$$
Also the length to be considered here is from the centre of gravity to the point of suspension, since the entire mass is assumed to be concentrated at the centre

Substituting the values, we get, extension = $$(1500 \times 10 \times 5)/(5 \times 10^8) =0.15 mm$$

The correct option is (b)

Bulk Modulus, $$B=\dfrac{\triangle P}{\triangle V/ V}$$
$$\dfrac{\triangle V}{V}\times 100$$% $$=10$$%
$$\dfrac{\triangle V}{V}=\dfrac{10}{100}=\dfrac{1}{10}$$
$$\triangle P=(1.165-1.01)\times10^5\,Pa$$
$$=0.155\times10^5 \,Pa$$
$$B=\dfrac{0.155\times 10^5}{1/10}\,Pa$$
$$=1.55\times 10^5\,Pa$$

Poisson's ratio = change in area /  change in length. If poisson's ratio >1, then change in area > change in length. Thus area expands when length increases

The option (b) is the correct option

Let Y, K, n and $$\sigma$$ be the Young's Modulus, Bulk modulus, Modulus of Rigidity and Poisson's Ratio, respectively. 
Y = 3K (1 - 2$$\sigma$$) [Standard formula] 
Y = 2n (1 + $$\sigma$$) [Standard formula] 
Hence, 3K (1 - 2$$\sigma$$) = 2n (1 + $$\sigma$$) 
Now K and n are always positive, so 
i) If $$\sigma$$ be +ve, then RHS is always +ve. So LHS must also be +ve. Therefore, 2$$\sigma$$ < 1 or $$\sigma$$ <1/2 
ii) If $$\sigma$$ be -ve, then LHS will always be +ve. Therefore, 1+$$\sigma$$ > 0 or $$\sigma$$ > -1 
Thus the limiting values of Poisson's ratio are -1 < $$\sigma$$ < 1/2

The correct option is (a)

$$\dfrac{\triangle  V}{V}=\dfrac{\triangle l}{l}+ \dfrac{2\triangle r}{r}, \dfrac{\triangle l}{l}= \dfrac{stress}{Y} = 2.5 \times 10^{-4}$$

We know that,

When there is no change in liquid level in vessel then,

$$\gamma _{real}^{'}=\gamma _{vessel}^{'}$$

Change in volume in liquid relative to vessel

$$ \Delta {{V}_{app}}=V\gamma _{app}^{'}\Delta \theta  $$

$$ \Delta {{V}_{app}}=V\left( \gamma _{real}^{'}-\gamma _{vessel}^{'} \right) $$

Hence, the fraction is $$1:4$$  

Poisson's ratio can lie only between 0 to 1. It cannot be greater than 1 for any material

The correct option is (d)

Proof resilience is defined as the maximum energy that can be absorbed up to the elastic limit, without creating a permanent distortion. 

Pressure at $$1\ km$$ depth inside water,

$$P=1000\times 10\times 1000=10^7\ Nm^{_2}$$

say,  V=volume of ball

$$\dfrac{0.05}{100}\times V$$  = Reduction in volume  (i.e) $$\Delta V$$

Bulk Modulus, $$\beta=\dfrac{-PV}{\Delta V}=\dfrac{-10^7\times V}{\dfrac{-0.05V}{100}}$$
Hence Option $$\textbf A $$ is correct.