In case of water, the meniscus shape is concave upwards. Also according to ascent formula $$h = \frac{{2T\cos \theta }}{{r\rho g}}$$
The surface tension ($$T$$) of soap solution is less than water. Therefore rise of soap solution in the capillary tube is less as compared to water. As in the case of water, the meniscus shape of soap solution is also concave upwards.

$$\delta = \frac{{W{\ell ^3}}}{{3YI}},$$   where $$W =$$  load, $$\ell =$$  length of beam and $$I$$ is geometrical moment of inertia for rectangular beam,
$$I = \frac{{b{d^3}}}{{12}}$$   where $$b =$$  breadth and $$d =$$  depth
For square beam $$b = d$$
$$\therefore {I_1} = \frac{{{b^4}}}{{12}}$$
For a beam of circular cross-section,
$$\eqalign{ & {I_2} = \left( {\frac{{\pi {r^4}}}{4}} \right) \cr & \therefore {\delta _1} = \frac{{W{\ell ^3} \times 12}}{{3Y{b^4}}} = \frac{{4W{\ell ^3}}}{{Y{b^4}}}\,\left( {{\text{for sq}}{\text{. cross section}}} \right) \cr & {\text{and}}\,{\delta _2} = \frac{{W{\ell ^3}}}{{3Y\left( {\frac{{\pi {r^4}}}{4}} \right)}} = \frac{{4W{\ell ^3}}}{{3Y\left( {\pi {r^4}} \right)}}\,\left( {{\text{for circular cross section}}} \right) \cr & {\text{Now}}\,\frac{{{\delta _1}}}{{{\delta _2}}} = \frac{{3\pi {r^4}}}{{{b^4}}} = \frac{{3\pi {r^4}}}{{{{\left( {\pi {r^2}} \right)}^2}}} = \frac{3}{\pi }\,\,\left( {\because {b^2} = \pi {r^2}\,{\text{i}}{\text{.e}}{\text{.,they have same cross - sectional area}}} \right) \cr} $$

The bulk modulus is defined as
$$B = - \frac{{\Delta p}}{{\frac{{\Delta V}}{V}}},$$   where the minus sign is inserted because $${\Delta V}$$ is negative when $${\Delta p}$$  is positive.
$$100\left| {\frac{{\Delta V}}{V}} \right| = 100\frac{{\Delta p}}{B} = 100\frac{{345 \times {{10}^6}}}{{138 \times {{10}^9}}} = 0.25\% $$

$$\eqalign{ & {T_1} = {T_2},Mg + \sqrt 2 {T_1} \cr & = \frac{4}{3}\pi {r^3}{\rho _w}g{T_1} \cr & = 250\sqrt 2 N \cr} $$

Weight of the bowl = $$mg$$
$$ = V\rho g = \frac{4}{3}\pi \left[ {{{\left( {\frac{D}{2}} \right)}^3} - {{\left( {\frac{d}{2}} \right)}^3}} \right]\rho g$$
Where $$D$$ = Outer diameter
$$d$$ = Inner diameter, $$\rho $$ = Density of bowl
Weight of the liquid displaced by the bowl
$$ = V\sigma g = \frac{4}{3}\pi {\left( {\frac{D}{2}} \right)^3}\sigma g$$
where $$\sigma $$ is the density of the liquid
For the floation
$$\eqalign{ & \frac{4}{3}\pi {\left( {\frac{D}{2}} \right)^3}\sigma g = \frac{4}{3}\pi \left[ {{{\left( {\frac{D}{2}} \right)}^3} - {{\left( {\frac{d}{2}} \right)}^3}} \right]pg \cr & \Rightarrow {\left( {\frac{1}{2}} \right)^3} \times 1.2 \times {10^3} = \left[ {{{\left( {\frac{1}{2}} \right)}^3} - {{\left( {\frac{d}{2}} \right)}^3}} \right]2 \times {10^4} \cr} $$
By solving we get $$d = 0.98\,m.$$

To find the minimum diameter, and hence minimum cross-sectional area, we assume that the force $$F = 400\,N$$   brings us to the elastic limit. Then from the stress,
$$\eqalign{ & \frac{F}{A} = 379 \times {10^6}\,Pa,{\text{we}}\,{\text{get}} \cr & A = \frac{{400N}}{{379 \times {{10}^6}Pa}} = 1.0554 \times {10^{ - 6}}{m^2}. \cr & {\text{Then,}}\,A = \frac{{\pi {D^2}}}{4} \cr & {D^2} = \frac{{4A}}{\pi } = \frac{{4\left( {1.0554 \times {{10}^{ - 6}}{m^2}} \right)}}{\pi } \cr & = 1.344 \times {10^{ - 6}} \cr & {\text{and}}\,D = \sqrt {1.344 \times {{10}^{ - 6}}{m^2}} \cr & = 1.16 \times {10^{ - 3}}m \cr & = 1.16\,mm \cr} $$

From the figure it is clear that liquid 1 floats on liquid 2. The lighter liquid floats over heavier liquid. Therefore we can conclude that $${\rho _1} < {\rho _2}$$
Also $${\rho _3} < {\rho _2}$$   otherwise the ball would have sink to the bottom of the jar.
Also $${\rho _3} < {\rho _1}$$   otherwise the ball would have floated in liquid 1. From the above discussion we conclude that $${\rho _1} < {\rho _3} < {\rho _2}.$$

According to Hooke’s law
Stress $$ \propto $$ strain i.e., $$\frac{F}{A} \propto \frac{{\Delta l}}{l}$$
⇒ For same $$F\,\& \,l,\Delta l \propto \frac{l}{A}$$

According to question,
Coefficient of linear expression of brass = $${\alpha _1}$$
Coefficient of linear expression of steel = $${\alpha _2}$$
Length of brass and steel rods are $${l_1}$$ and $${l_2}$$ respectively.
As given difference increase in length $$\left( {{{l'}_2} - {{l'}_1}} \right)$$  is same for all temperature.
So, $${{l'}_2} - {{l'}_1} = {l_2} - {l_1}$$
$$\eqalign{ & \Rightarrow {l_2}\left( {1 + {\alpha _2}\Delta t} \right) - {l_1}\left( {1 + {\alpha _1}\Delta t} \right) = {l_2} - {l_1} \cr & \Rightarrow {l_2}{\alpha _2} = {l_1}{\alpha _1} \cr} $$

As the sphere floats in the liquid. Therefore its weight will be equal to the upthrust force on it
Weight of sphere $$ = \frac{4}{3}\pi {R^3}\rho g\,......\left( {\text{i}} \right)$$
Upthrust due to oil and mercury
$$ = \frac{2}{3}\pi {R^3} \times {\sigma _{{\text{oil}}}}g + \frac{2}{3}\pi {R^3}{\sigma _{Hg}}g\,......\left( {{\text{ii}}} \right)$$
Equation (i) and (ii)
$$\eqalign{ & \frac{4}{3}\pi {R^3}\rho g = \frac{2}{3}\pi {R^3}0.8g + \frac{2}{3}\pi {R^3} + 13.6g \cr & \Rightarrow 2\rho = 0.8 + 13.6 = 14.4 \Rightarrow \rho = 7.2 \cr} $$