From Bernoulli's theorem $${p_1} + \frac{1}{2}pv_1^2 = {p_2} + \frac{1}{2}pv_2^2$$
where, $${p_1},{p_2}$$  are pressure inside and outside the roof and $${v_1},{v_2}$$  are velocities of wind inside and outside the roof. Neglect the width of the roof. Pressure difference is
$$\eqalign{ & {p_1} - {p_2} = \frac{1}{2}\rho \left( {v_2^2 - v_1^2} \right) = \frac{1}{2} \times 1.2\left( {{{40}^2} - 0} \right) \cr & = 960\,N/{m^2} \cr} $$
Force acting on the roof is given by
$$\eqalign{ & F = \left( {{p_1} - {p_2}} \right)A = 960 \times 250 \cr & = 24 \times {10^4}N = 2.4 \times {10^5}N \cr} $$
As the pressure inside the roof is more than outside to it. So the force will act in the upward direction,
i.e. $$F = 2.4 \times {10^5}N{\text{ - upwards}}{\text{.}}$$

$$dv = 8\,cm/s\,{\text{and}}\,dx = 0.1\,cm$$
Velocity gradient $$ = \frac{{dv}}{{dx}} = \frac{8}{{0.1}} = 80/s.$$

As linear dimension increases by a factor of 9
$$\therefore \frac{{{v_f}}}{{{v_i}}} = {9^3}$$
$$\because $$ Density remain same
So, mass $$ \propto $$ volume
$$\eqalign{ & \frac{{{m_f}}}{{{m_i}}} = {9^3}\,\,\,\,\,\,\,\,\,\, \Rightarrow \frac{{{{\left( {Area} \right)}_f}}}{{{{\left( {Area} \right)}_i}}} = {9^2} \cr & {\text{Strees }}\left( \sigma \right) = \frac{{force}}{{area}} = \frac{{\left( {mass} \right) \times g}}{{area}} \cr & \frac{{{\sigma _2}}}{{{\sigma _1}}} = \left( {\frac{{{m_f}}}{{{m_i}}}} \right)\left( {\frac{{{A_i}}}{{{A_f}}}} \right) = \frac{{{9^3}}}{{{9^2}}} = 9 \cr} $$

$$\eqalign{ & {p_0} - {p_i} = \frac{{2T}}{R} \cr & \therefore {P_1} > {P_2} \cr} $$
hence air moves from smaller bubble to bigger bubble.

Fluid resistance is given by $$R = \frac{{8\eta L}}{4}$$
When two capillary tubes of same size are joined in series, then equivalent fluid resistance is
$${R_S} = \frac{{8\eta L}}{{\pi {R^4}}} + \frac{{8\eta \times 2L}}{{\pi {{\left( {2R} \right)}^4}}} = \left( {\frac{{8\eta L}}{{\pi {R^4}}}} \right) \times \frac{9}{8}$$
Rate of flow = $$ = \frac{P}{{{R_S}}} = \frac{{\pi P{R^4}}}{{8\eta L}} \times \frac{8}{9} = \frac{8}{9}X\,\,\left[ {{\text{as}}\,X = \frac{{\pi P{R^4}}}{{8\eta L}}} \right]$$

According to Toricelli’s theorem,
Velocity of efflex,
$${V_{{\text{eff}}}} = \sqrt {2gh} = \sqrt {2 \times 9.8 \times 5} \cong 9.8\,m{s^{ - 1}}$$

Given, the breaking strength of cable $${f_u} = 7 \times {10^7}\,N/{m^2}$$
The force carried by the cable,
$$F = m\left( {g + a} \right) = 2000\left( {9.8 + 1.5} \right) = 22600\,N$$
The area of cross-section, $$A = \frac{F}{{{f_u}}} = \frac{{22600}}{{7 \times {{10}^7}}}$$
$$ = 3.28 \times {10^{ - 4}}\,{m^2}.$$

$$Y = \frac{{\frac{T}{A}}}{{\frac{{\Delta \ell }}{\ell }}}\,\,\, \Rightarrow \Delta \ell = \frac{{T \times \ell }}{{A \times Y}} = \frac{T}{Y} \times \frac{\ell }{A}$$
Here, $$\frac{T}{Y}$$  is constant. Therefore, $$\Delta \ell \propto \frac{\ell }{A}.$$
$$\frac{\ell }{A}$$  is largest in the first case.

Bulk modulus,
$$\eqalign{ & B = - {V_0}\frac{{\Delta p}}{{\Delta V}} \Rightarrow \Delta V = - {V_0} = - {V_0}\frac{{\Delta p}}{B} \cr & \Rightarrow V = {V_0}\left[ {1 - \frac{{\Delta p}}{B}} \right] \cr & \therefore {\text{Density,}}\,\rho = {\rho _0}{\left[ {1 - \frac{{\Delta p}}{B}} \right]^{ - 1}} = {\rho _0}\left[ {1 + \frac{{\Delta p}}{B}} \right] \cr & {\text{where,}}\,\Delta p = p - {p_0} = h{\rho _0}g \cr} $$
= pressure difference between depth and surface of ocean
$$\therefore \rho = {\rho _0}\left[ {1 + \frac{{{\rho _0}gy}}{B}} \right]\,\,\left( {{\text{As}}\,h = y} \right)$$

$$\frac{U}{{{\text{volume}}}} = \frac{1}{2}Y \times {\text{strai}}{{\text{n}}^2} = 3600\,J\,{m^{ - 3}}\,\left[ {{\text{Strain}} = 0.06 \times {{10}^{ - 2}}} \right]$$