Added Luis's English solutions of 4.2.11 and 4.2.12

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+ <header style="text-align:center;">
+ <h2>Solutions of Savchenko Physics Textbook</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+
+ <p>
+ $4.2.12$
+ Determine the force of the tension of the thread connecting two balls of volume 10 cm$^3$ each, if the upper ball floats half-immersed in water. The lower ball is three times heavier than the upper one.
+ </p>
+ <center>
+ <figure>
+ <img src="statement.png"
+ loading="lazy" alt="4.2.12" width="200" />
+ <figcaption>
+ For problem 4.2.12
+ </figcaption>
+ </figure>
+ </center>
+
+ <h3>Solution</h3>
+ <p>
+ Applying Newton's Second Law...
+ </p>
+ <center>
+ <figure>
+ <img src="draw.png"
+ loading="lazy" alt="4.2.12" width="200" />
+ <figcaption>
+ Force analysis
+ </figcaption>
+ </figure>
+ </center>
+ <p>
+ For upper ball
+ $$m_{\uparrow}~g+T = F_{\uparrow}$$
+ $$m_{\uparrow}~g+T = \rho_0~g~\frac{V}{2} \;(1)$$
+ For lower ball
+ $$m_{\downarrow}~g = T+F_{\downarrow}$$
+ $$m_{\downarrow}~g = T+\rho_0~g~V \;(2)$$
+ Separating the gravity's force term from $(1)$
+ $$m_{\uparrow}~g = \rho_0~g~\frac{V}{2}-T \;(3)$$
+ Dividing $(2)$ by $(3)$, and taking account that $m_{\downarrow}=3~m_{\uparrow}$
+ $$\frac{\rho_0~g~V+T}{\rho_0~g~\frac{V}{2}-T}=3$$
+ and separating $T$
+ $$T = \frac{\rho_0~g~V}{8}$$
+ Taking $\rho_0 = 1~\rm{g/cm^3} = 10^{-3}~\rm{kg/cm^3}$ (water's density) and $g=9.8~\rm{m/s^2}$ (acceleration due to gravity at Earth's surface),
+ $$\boxed{T \simeq 1.2 \cdot 10^{-2}~\rm{N}}$$
+ </p>
+ <p style="text-align: right; font-style: italic; font-size: 14;">
+ BSc. Luis Daniel Fernández Quintana<br>
+ Physics Department (FCNE)<br>
+ Universidad de Oriente, Cuba<br>
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