The deformation of a body on application of a force depends on the nature of the material and the magnitude of the applied force.

$$\eqalign{ & {\text{In }}\Delta PCM\cos \left( {\theta + \frac{\alpha }{2}} \right) = \frac{b}{R}\,.....(1) \cr & {\text{Also }}\,\,\left( {{P_0} - \frac{{2S}}{R}} \right) + h\rho g = {P_0} \cr & h = \frac{{2S}}{{R\rho g}} = \frac{{2S}}{{b\rho g}}\cos \left( {\theta + \frac{\alpha }{2}} \right) \cr} $$

$$\ell $$ decreases as the block moves up $$h$$ will also decreases because when the coin is in water it will displace a volume of water, equal to its own volume, whereas when it is on the block it displaces more volume than to own volume (because density of coin is greater than density of water).

$$P = h\rho g$$   i.e. pressure does not depend upon the area of bottom surface.

The pressure outside the submarine is
$$P = {P_a} + \rho gh$$
Pressure inside the submarine is $${P_a}.$$
Net pressure acting on the window is
$$\eqalign{ & {P_g} = P - {P_a} = \rho gh \cr & = {10^{ - 3}}kg{m^{ - 3}} \times 10\,m{s^{ - 2}} \times 2000\,m \cr & = 2 \times {10^7}Pa \cr} $$
Area of window is
$$A = 50\,cm \times 50\,cm = 2500 \times {10^{ - 4}}{m^2}$$
Force on the window is
$$\eqalign{ & F = {P_g}A = 2 \times {10^7}Pa \times 2500 \times {10^{ - 4}}{m^2} \cr & = 5 \times {10^6}N \cr} $$

Shear stress $$ = \frac{F}{A}$$
$$\eqalign{ & \tan \theta = \frac{{{\text{shear}}\,{\text{stress}}}}{{{\text{shear}}\,{\text{modulus}}}} \cr & \Leftarrow \tan \theta = \frac{{4000}}{{{{\left( {0.25} \right)}^2} \times 8 \times {{10}^{10}}}} \cr & \tan \theta = 8 \times {10^{ - 7}} \cr} $$
for small angle $$\tan \theta = \theta = 8 \times {10^{ - 7}}rad$$

For equilibrium $$4dA\ell g = \left( {dAL + m} \right)g$$
$$\eqalign{ & {\text{and}}\,\frac{\ell }{2} > {Y_{cm}}\,\left( {{\text{for}}\,{\text{rotational}}\,{\text{equilibrium}}} \right) \cr & {Y_{cm}} = \frac{{m \times 0 + dAL\left( {\frac{L}{2}} \right)}}{{m + dAL}} = \frac{{dA{L^2}}}{{2\left( {m + dAL} \right)}} \cr & \ell > \frac{{A{L^2}d}}{{\left( {ALd + m} \right)}} \cr & \Rightarrow \frac{{m + dAL}}{{4dA}} > \frac{{A{L^2}d}}{{\left( {ALd + m} \right)}}m > ALd \cr} $$

KEY CONCEPT:
According to Archimedes principle Upthrust = Wt. of fluid displaced

$$\eqalign{ & {F_{{\text{bottom}}}} - {F_{{\text{top}}}} = V\rho g \cr & \therefore {F_{{\text{bottom}}}} = {F_{{\text{top}}}} + V\rho g \cr & = {P_1} \times A + V\rho g \cr & = \left( {h\rho g} \right) \times \left( {\pi {R^2}} \right) + V\rho g \cr & = \rho g\left[ {\pi {R^2}h + V} \right] \cr} $$

For a beam, the depression at the centre is given by, $$\delta = \left( {\frac{{fL}}{{4Yb{d^3}}}} \right)\,\,\left[ {f,L,b,d\,{\text{are constants for a particular beam}}} \right]$$
$${\text{i}}{\text{.e}}{\text{.}}\,\delta \propto \frac{1}{Y}$$

Let $$V$$ be the volume of iceberg and let $$x$$ be the fraction of volume above water.
Using law of floatation, weight of floating body = weight of liquid displaced by part of the floating body inside the liquid.
Therefore, $$V{\rho _{{\text{ice}}}}g = \left( {1 - x} \right)V{\rho _{{\text{water}}}}g.$$
Using the value of $${\rho _{{\text{ice}}}}$$  and $${\rho _{{\text{water'}}}}$$  We get $$x = \left( {\frac{{13}}{{103}}} \right)$$