$$\eqalign{ & P.E.{\text{ }} = \frac{1}{2} \times {\text{stress}} \times {\text{strain}} \times {\text{volume}} \cr & {\text{or}}\,\,\left( {\frac{{P.E.}}{{{\text{volume}}}}} \right){\text{ }} = \frac{1}{2} \times \left( {Y \times {\text{strain}}} \right)\left( {{\text{strain}}} \right) \cr & = \frac{1}{2}Y{\left( {{\text{strain}}} \right)^2} = \frac{1}{2}Y{X^2} \cr} $$

$$W=T × $$   change in surface area
$$\eqalign{ & W = 2T4\pi \left[ {\left( {{5^2}} \right) - {{\left( 3 \right)}^2}} \right] \times {10^{ - 4}} \cr & = 2 \times 0.03 \times 4\pi \left[ {25 - 9} \right] \times {10^{ - 4}}\,J \cr & = 0.4\pi \times {10^{ - 3}}\,J \cr & = \,0.4\pi \,mJ \cr} $$

Let us consider a small dotted segment of thickness $$dx$$ for observation. Since, this segment is accelerated towards right, a net force is acting in this segment towards right from the liquid towards the left of $$ABCD.$$  According to Newton's third law, the segment $$ABCD$$  will also apply a force on the previous section creating a pressure on it which makes the liquid rise.

The square of the velocity of efflux
$$\eqalign{ & {v^2} = \frac{{2gh}}{{\sqrt {1 - {{\left( {\frac{a}{A}} \right)}^2}} }}\,\,or,\,{v^2} = \frac{{2 \times 10 \times 2.475}}{{\sqrt {1 - {{\left( {0.1} \right)}^2}} }} = 50\,{m^2}/{s^2} \cr & h = 3 - 0.525 = 2.475\,m \cr} $$

We have, $$U = \frac{{{F^2}}}{{2k}}$$
where $$k = \frac{{Yl}}{A} = \frac{{Yl}}{{\frac{1}{4}\pi {d^2}}}$$
$$\therefore \quad U \propto \frac{{{d^2}}}{l}$$

For stress to be equal, $$\frac{{{T_1}}}{{{A_1}}} = \frac{{{T_2}}}{{{A_2}}}$$
$$\therefore \frac{{{T_1}}}{{{T_2}}} = \frac{{{A_1}}}{{{A_2}}} = \frac{1}{2}$$

As shown in the figure, the wires will have the same Young’s modulus (same material) and the length of the wire of area of cross-section $$3A$$  will be $$\frac{\ell }{3}$$ (same volume as wire 1).
For wire 1,
$$Y = \frac{{\frac{F}{A}}}{{\frac{{\Delta x}}{\ell }}}\,.....(i)$$
For wire 2,
$$Y = \frac{{\frac{{F'}}{{3A}}}}{{\frac{{\Delta x}}{{\left( {\frac{\ell }{3}} \right)}}}}\,.....(ii)$$
From (i) and (ii),
$$\frac{F}{A} \times \frac{\ell }{{\Delta x}} = \frac{{F'}}{{3A}} \times \frac{\ell }{{3\Delta x}}\,\,\, \Rightarrow F' = 9F$$

$$Y = \frac{{\frac{F}{A}}}{{\frac{{\Delta \ell }}{\ell }}} = \frac{F}{A}.\frac{\ell }{{\Delta \ell }} = \frac{{20 \times 1}}{{{{10}^{ - 6}} \times {{10}^{ - 4}}}} = 2 \times {10^{11}}\,{\text{N/}}{{\text{m}}^2}$$

According to questions,
$$\eqalign{ & \frac{{{\ell _s}}}{{{\ell _b}}} = a,\frac{{{r_s}}}{{{r_b}}} = b,\frac{{{y_s}}}{{{y_b}}} = c,\frac{{\Delta {\ell _s}}}{{\Delta {\ell _b}}} = ? \cr & {\text{As,}}\,\,y = \frac{{F\ell }}{{A\Delta \ell }} \Rightarrow \Delta \ell = \frac{{F\ell }}{{Ay}} \cr & \Delta {\ell _s} = \frac{{3mg{\ell _s}}}{{\pi r_s^2.{y_s}}}\,\,\left[ {\because {F_s} = \left( {M + 2M} \right)g} \right] \cr & \Delta {\ell _b} = \frac{{2Mg{\ell _b}}}{{\pi r_b^2.{y_b}}}\,\,\left[ {\because {F_b} = 2Mg} \right] \cr & \therefore \frac{{\Delta {\ell _s}}}{{\Delta {\ell _b}}} = \frac{{\frac{{3Mg{\ell _s}}}{{\pi r_s^2.{y_s}}}}}{{\frac{{2Mg.{\ell _b}}}{{\pi r_b^2.{y_b}}}}} = \frac{{3a}}{{2{b^2}c}} \cr} $$

As the block moves up with the fall of coil, $$l$$ decreases, similarly $$h$$ will also decrease because when the coin is in water, it displaces water equal to its own volume only.