According to Archimedes principle
Upthrust = Wt. of fluid displaced
$$\eqalign{ & {F_{{\text{bottom}}}} - {F_{{\text{top}}}} = V\rho g \cr & \therefore {F_{{\text{bottom}}}} = {F_{{\text{top}}}} + V\rho g \cr & = {P_1} \times A + V\rho g \cr & = \left( {h\rho g} \right) \times \left( {\pi {R^2}} \right) + V\rho g \cr & = \rho g\left[ {\pi {R^2}h + V} \right] \cr} $$

$$Y = \frac{F}{A} \times \frac{L}{{\Delta L}} = \frac{{100}}{{2 \times {{10}^{ - 4}}}} \times \frac{2}{{1 \times {{10}^{ - 2}}}} = {10^8}\,N/{m^2}$$

Oil will float on water so, (B) or (D) is the correct option. But density of ball is more than that of oil,, hence it will sink in oil.

Range is same for both holes
$$\therefore 2\sqrt {\left( {H - {h_1}} \right){h_1}} = 2\sqrt {\left( {H - {h_2}} \right){h_2}} $$
Squaring both sides,
$$\eqalign{ & 4\left( {H - {h_1}} \right){h_1} = 4\left( {H - {h_2}} \right){h_2} \cr & H{h_1} - h_1^2 = H{h_2} - h_2^2 \cr} $$
On solving we get,
$$H = {h_1} + {h_2}\left( {{\text{not}}\,{\text{possible}}} \right)\,{\text{and}}\,{h_1} - {h_2} = 0$$
Hence, the ratio of $$\frac{{{h_1}}}{{{h_2}}} = 1:1.$$

Volumetric strain, $$\frac{{\Delta V}}{V} = 3 \times \frac{{\Delta \ell }}{\ell }$$
$$ = 3 \times 0.01 = 0.03$$
Bulk modulus, $$B = \frac{{\Delta P}}{{\left( {\frac{{\Delta V}}{V}} \right)}} = \frac{{5 \times {{10}^5}}}{{0.03}}$$
$$ = 1.67 \times {10^7}\,N/{m^2}$$

Let $$L = PQ =$$   length of rod
$$\therefore SP = SQ = \frac{L}{2}$$
Weight of rod,
$$W = Al\rho g.$$   acting
At point $$S$$
And force of buoyancy, $${F_B} = Al{\rho _0}g.\left[ {l = PR} \right]$$
Which acts at mid-point of $$PR.$$
For rotational equilibrium.
$$\eqalign{ & Al{\rho _0}g \times \frac{\ell }{2}\cos \theta = AL\rho g \times \frac{L}{2}\cos \theta \cr & \Rightarrow \frac{{{l^2}}}{{{L^2}}} = \frac{\rho }{{{\rho _0}}} \Rightarrow \frac{l}{L} = \sqrt {\frac{\rho }{{{\rho _0}}}} \cr} $$$
From figure, $$\sin \theta = \frac{h}{l} = \frac{L}{{2l}} = \frac{1}{2}\sqrt {\frac{{{\rho _0}}}{\rho }} $$

According to Bernoulli’s theorem,
$$P + \frac{1}{2}\rho {v^2} = {\text{constant}}\,{\text{and}}\,Av = {\text{constant}}$$
If $$A$$ is minimum, $$v$$ is maximum, $$P$$ is minimum.

$$\eqalign{ & m = \frac{{Y \times {\text{strain}} \times A}}{2} \cr & = \frac{{2 \times {{10}^{11}} \times {{10}^{ - 3}} \times 3 \times {{10}^{ - 6}}}}{2} \cr & = 100\,kg \cr} $$

$${C_1} = \frac{{\pi \eta \left( {r_2^4 - r_1^4} \right)}}{{2\ell }},{C_2} = \frac{{\pi \eta {r^4}}}{{2\ell }}$$
Initial volume = Final volume
$$\eqalign{ & \therefore \pi \left[ {r_2^2 - r_1^2} \right]\ell \rho = \pi {r^2}\ell \rho \cr & \Rightarrow {r^2} = r_2^2 - r_1^2 \Rightarrow {r^2} = \left( {{r_2} + {r_1}} \right)\left( {{r_2} - {r_1}} \right) \cr & \Rightarrow {r^2} = \left( {8.02 + 7.98} \right)\left( {8.02 - 7.98} \right) \cr & \Rightarrow {r^2} = 16 \times 0.04 = 0.64\,cm \Rightarrow r = 0.8\,cm \cr & \therefore \frac{{{C_1}}}{{{C_2}}} = \frac{{r_2^4 - r_1^4}}{{{r^4}}} = \frac{{{{\left[ {8.02} \right]}^4} - {{\left[ {7.98} \right]}^4}}}{{{{\left[ {0.8} \right]}^4}}} \cr} $$

Since pressure is transmitted undiminished throughout the fluid (Pascal's law)
$$\eqalign{ & {F_1} = \frac{{{A_1}}}{{{A_2}}}{F_2} = \frac{{\pi \left( {5 \times 5} \right)}}{{\pi \left( {15 \times 15} \right)}}\left( {1350 \times 9.81} \right) \cr & \approx 1.5 \times {10^3}N \cr} $$
The air pressure that will produce this force is
$$P = \frac{{{F_1}}}{{{A_1}}} = \frac{{1.5 \times {{10}^3}}}{{\pi {{\left( {5 \times {{10}^{ - 2}}m} \right)}^2}}} = 1.9 \times {10^5}Pa$$