Pressure of two points lie in the same horizontal level should be same and $$p = hdg$$
Both ends of the $$U$$ tube are open, so the pressure on both the free surfaces must be equal.
i.e., $${p_1} = {p_2}$$
$$\eqalign{ & {h_{{\text{oil}}}} \cdot {S_{{\text{oil}}}}\,g = {h_{{\text{water}}}} \cdot {S_{{\text{water}}}} \cdot g \cr & {S_{{\text{oil}}}} = {\text{specific density of}}\,{\text{oil}} \cr & {S_{{\text{oil}}}} = \frac{{{h_{{\text{water}}}} \cdot {S_{{\text{water}}}} \cdot g}}{{{h_{{\text{oil}}}} \cdot g}} \cr} $$
From figure $${S_{{\text{oil}}}} = \frac{{\left( {65 + 65} \right) \times 1000}}{{\left( {65 + 65 + 10} \right)}}$$
$$ = 928\,kg{m^{ - 3}}$$

Newton’s law of viscosity, $$F = \eta A\frac{{dv}}{{dy}}$$
Stress $$ = \frac{F}{A} = \eta \left( {\frac{{dv}}{{dy}}} \right) = \eta k\left( {\frac{{4y}}{{{a^2}}} - \frac{{3{y^2}}}{{{a^3}}}} \right)$$
At $$y = a,$$  stress $$ = \eta k\left( {\frac{4}{a} - \frac{3}{a}} \right) = \frac{{\eta k}}{a}$$

Small amount of work done in extending the spring by $$dx$$  is
$$\eqalign{ & dW = k\,x\,dx \cr & \therefore W = k\int\limits_{0.05}^{0.15} {x\,dx} = \frac{{800}}{2} = \left[ {{{\left( {0.15} \right)}^2} - {{\left( {0.05} \right)}^2}} \right] = 8\,J \cr} $$

$$\eqalign{ & W = \frac{1}{2} \times F \times l = \frac{1}{2}mgl \cr & = \frac{1}{2} \times 10 \times 10 \times 1 \times {10^{ - 3}} = 0.05\,J \cr} $$

Using the usual expression for the Young’s modulus, the force constant for the wire can be written as $$k = \frac{F}{{\Delta l}} = \frac{{YA}}{L}$$
where the symbols have their usual meanings.
When the two wires are connected together in series, the effective force constant is given by
$${k_{eq}} = \frac{{{k_1}{k_2}}}{{{k_1} + {k_2}}}$$
Substituting the corresponding lengths, area of cross sections and the Young's moduli, we get
$${k_{eq}} = \frac{{\left( {\frac{{{Y_1}A}}{{{L_1}}}} \right)\left( {\frac{{{Y_2}A}}{{{L_2}}}} \right)}}{{\frac{{{Y_1}A}}{{{L_1}}} + \frac{{{Y_2}A}}{{{L_2}}}}} = \frac{{\left( {{Y_1}{Y_2}} \right)A}}{{{Y_1}{L_2} + {Y_2}{L_1}}}$$

From Stoke's law,
viscous force $$F = 6\pi \eta rv$$
hence $$F$$ is directly proportional to radius & velocity.

Total cross-sectional area of the thigh bones
$$A = 2\left( {10 \times {{10}^{ - 4}}} \right) = 2 \times {10^{ - 3}}{m^2}$$
Force acting on the bones $$ = mg = 40 \times 10 = 400\,N$$
$$\therefore {P_{av}} = \frac{F}{A} = \frac{{400}}{{2 \times {{10}^{ - 3}}}} = 2 \times {10^5}\,N/{m^2}$$

Total volume of rain drops, received $$100\,cm$$  in a year by area $$1\,{m^2}$$
$$ = 1\,{m^2} \times \frac{{100}}{{100}}m = 1\,{m^3}$$
As we know, density of water,
$$d = {10^3}kg/{m^3}$$
Therefore, mass of this volume of water
$$M = d \times v = {10^3} \times 1 = {10^3}kg$$
Average terminal velocity of rain drop
$$v = 9\,m/s\left( {{\text{given}}} \right)$$
Therefore, energy transferred by rain,
$$\eqalign{ & E = \frac{1}{2}m{v^2} = \frac{1}{2} \times {10^3} \times {\left( 9 \right)^2} = \frac{1}{2} \times {10^3} \times 81 \cr & = 4.05 \times {10^4}J \cr} $$

As we know,
Pressure $$P = Vdg$$
Here, $$LAdg = \left( {pL} \right)A\left( {n\rho } \right)g + \left( {1 - p} \right)LA\rho g$$
$$ \Rightarrow d = \left( {1 - p} \right)\rho + pn\rho = \left[ {1 + \left( {n - 1} \right)p} \right]\rho $$

We have, $$h = \frac{{2T\cos \theta }}{{r\rho g}}$$
Mass of the water in the capillary
$$\eqalign{ & m = \rho V = \rho \times \pi {r^2}h = \rho \times \pi {r^2} \times \frac{{2T\cos \theta }}{{r\rho g}} \cr & \Rightarrow m\,\alpha \,r \cr & \therefore \frac{{{m_1}}}{{{m_2}}} = \frac{r}{{2r}} \cr & {\text{or,}}\,\,{m_2} = 2{m_1} = 2m \cr} $$