As the balloon rises, the density of air decreases, leading to a decrease in buoyant force until it equals the weight of the balloon, so it stops rising. Thus, Assertion is true. However, atmospheric pressure does not become zero at this height, making Reason false.

The excess pressure inside a soap bubble is \(P = \frac{4T}{R}\). Thus, pressure is greater inside the smaller bubble (lesser radius). When in contact, the common interface bulges towards the larger bubble due to this pressure difference. Both are true and Reason explains Assertion.

For a perfectly plastic body, there is no tendency to regain its shape, meaning the restoring force (and thus stress) is zero. Since Young's modulus \(Y = \frac{\text{stress}}{\text{strain}}\), it is also zero. Both statements are true and Reason is the correct explanation.

Applying the work-energy theorem, work done by the buoyant force over depth \(h\) equals total work done against gravity: \(V \sigma g h = V \rho g (h + H)\). Solving for height \(H\) gives \(H = \left(\frac{\sigma}{\rho} - 1\right)h\).

At terminal velocity, the upward forces balance the downward force: \(F_v + F_b = mg\). Since buoyant force is \(F_b = V \rho_l g = \left(\frac{m}{\rho}\right)\left(\frac{\rho}{4}\right)g = \frac{mg}{4}\), we get \(F_v = mg - \frac{mg}{4} = \frac{3mg}{4}\).

Volume conservation gives \(R = 5r\). Initial surface area is \(A_i = 125 \times 4\pi r^2 = 500\pi r^2\) and final is \(A_f = 4\pi R^2 = 100\pi r^2\). The decrease in area is \(400\pi r^2\), so loss in surface energy is \(T \times 400\pi r^2\).

Using capillary rise formula \(h = \frac{2T \cos\theta}{r \rho g}\) with \(\theta = 0^\circ\) in CGS: \(4 = \frac{2 \times 60 \times 1}{r \times 1 \times 1000} ⇒ r = 0.03\text{ cm}\).

Soap reduces surface tension, so \(h\) decreases (A-p). In a free fall, effective gravity \(g_{eff} = 0\), so water rises to full height (B-s). Upward acceleration increases \(g_{eff}\) hence \(h\) decreases (C-p). Downward acceleration decreases \(g_{eff}\) hence \(h\) increases (D-q).

Converting Young's modulus to CGS units gives \(Y = 10^5\text{ dyne/cm}^2\). Stress is \(F/A = 2 \times 10^5 / 2 = 10^5\text{ dyne/cm}^2\). Since \(\text{Strain} = \text{Stress}/Y = 1\), we have \(\Delta L = L\). Thus, the final length becomes \(L + \Delta L = 2L\).

The bulk modulus is defined as \(B = \frac{\Delta P}{\Delta V/V}\). Moving to vacuum causes a pressure change of \(\Delta P = 10^5\text{ N/m}^2\). Thus, the fractional volume change is \(\frac{\Delta V}{V} = \frac{\Delta P}{B} = \frac{10^5}{1.25 \times 10^{11}} = 8 \times 10^{-7}\).