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

$$F = Y \times A \times \frac{l}{L} \Rightarrow F \propto {r^2}\,\,\left( {Y,l\,{\text{and and }}L\,{\text{are constant}}} \right)$$
If diameter is made four times then force required will be 16 times, $$16 \times {10^3}N$$

It is given that
$$\eqalign{ & {v_{rms}} = 200\;m{s^{ - 1}},{T_1} = 300\;K,{P_1} = {10^5}\;N/{m^2} \cr & {T_2} = 400\;K,{P_2} = 0.05 \times {10^5}\;N/{m^2} \cr} $$
As, $$rms$$  velocity of gas molecules,
$$\because {V_{rms}} \propto \sqrt T \,\,\left( {\because {v_{rms}} = \sqrt {\frac{{3RT}}{m}} } \right)$$
For two different cases
$$\eqalign{ & \Rightarrow \quad \frac{{{{\left( {{v_{rms}}} \right)}_1}}}{{{{\left( {{v_{rms}}} \right)}_2}}} = \sqrt {\frac{{{T_1}}}{{{T_2}}}} \Rightarrow \frac{{200}}{{{{\left( {{v_{rms}}} \right)}_2}}} = \sqrt {\frac{{300}}{{400}}} = \sqrt {\frac{3}{4}} \cr & \Rightarrow {\left( {{v_{rms}}} \right)_2} = \frac{2}{{\sqrt 3 }} \times 200 = \frac{{400}}{{\sqrt 3 }}\;m{s^{ - 1}} \cr} $$

Pressure in limb I at $$B =$$   Pressure in limb II at $$A$$
$$\eqalign{ & h{\rho _1}g = h{\rho _2}g \cr & \Rightarrow {\rho _1} = {\rho _2} \cr} $$

Let $${\theta _1}$$ and $${\theta _2}$$ are the angle of twist produced in cylinders $$A$$ and $$B$$ respectively.
$${\text{Given,}}\,{\theta _1} + {\theta _2} = \theta \,......\left( {\text{i}} \right)$$
On being in series, the torque acts at their free ends are equal.
We have $$\tau = \frac{{\pi \eta {r^4}\theta }}{{2\ell }}$$
$$\therefore \frac{{\pi \eta {r^4}{\theta _1}}}{{2\ell }} = \frac{{\pi \eta {{\left( {2r} \right)}^4}{\theta _2}}}{{2\ell }} \Rightarrow {\theta _1} = 16{\theta _2}\,......\left( {{\text{ii}}} \right)$$
From (i) and (ii) , we have
$${\theta _1} = \frac{{16}}{{17}}\theta $$

$$\eqalign{ & \frac{{dV}}{V} = \left( {1 + 2\sigma } \right)\frac{{dL}}{L} \cr & \frac{{dV}}{V} = 2 \times 2 \times {10^{ - 3}} = 4 \times {10^{ - 3}}\,\,\left[ {\because \sigma = 0.5 = \frac{1}{2}} \right] \cr} $$
∴ Percentage change in volume $$ = 4 \times {10^{ - 1}}$$
$$ = 0.4\% $$

The condition for terminal speed $$\left( {{v_t}} \right)$$ is
Weight $$=$$ Buoyant force $$+$$ Viscous force
$$\eqalign{ & \therefore V{\rho _1}g = V{\rho _2}g + kv_t^2 \cr & \therefore {v_t} = \sqrt {\frac{{Vg\left( {{\rho _1} - {\rho _2}} \right)}}{k}} \cr} $$

Let $$\sigma $$ be the density of the gas, then that of the air is $$15\sigma .$$  Then the weight of the balloon = weight of the gas + weight of the envelope $$ = Vg\sigma + w.$$
If $$f$$ be the required acceleration of the balloon acting vertically upward and then from "mass $$ \times $$ acceleration = forces acting in the sense of acceleration" we get $$\frac{{\left( {Vg\sigma + w} \right)}}{g} \times a = $$     force of buoyance - wt. of the balloon with gas $$ = V15\sigma g - \left( {Vg\sigma + w} \right)$$
$${\text{or}}\,\,a = \left( {\frac{{14Vg\sigma - w}}{{Vg\sigma + w}}} \right) \times g$$

The object is spherical and the bulk modulus is represented by $$B.$$ It is the ratio of normal stress to the volumetric strain.
Hence, $$B = \frac{{\frac{F}{A}}}{{\frac{{\Delta V}}{V}}}$$
$$\eqalign{ & \Rightarrow \frac{{\Delta V}}{V} = \frac{p}{B} \cr & \Rightarrow \left| {\frac{{\Delta V}}{V}} \right| = \frac{p}{B} \cr} $$
Here $$p$$ is applied pressure on the object and $$\frac{{\Delta V}}{V}$$ is volume strain
Fractional decreases in volume
$$ \Rightarrow \frac{{\Delta V}}{V} = 3\frac{{\Delta R}}{R}\,\,\left[ {\because V = \frac{4}{3}\pi {R^3}} \right]$$
Volume of the sphere decreases due to the decrease in its radius.
Hence $$\frac{{\Delta V}}{V} = \frac{{3\Delta R}}{R} = \frac{p}{B}$$
$$ \Rightarrow \frac{{\Delta R}}{R} = \frac{p}{{3B}}$$

Water fills the tube entirely in gravity less condition i.e., $$20\,cm.$$