To solve this problem, we need to understand how the time period of a simple pendulum is affected by changes in mass and length.$\\$The time period $T$ of a simple pendulum is given by the formula:$\\ T = 2\pi \sqrt{\frac{L}{g}} \\$where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity.$\\$Note that the time period is independent of the mass of the bob.$\\$Let's denote:$\\$• $L_0$ as the original length of the pendulum.$\\$• $T_0$ as the original time period of the pendulum.$\\$The original time period $T_0$ is:$\\ T_0 = 2\pi \sqrt{\frac{L_0}{g}} \\$According to the problem, the mass is increased to thrice its original mass,$\\$but this does not affect the time period.$\\$The length is made half its original length, so the new length $L$ is:$\\ L = \frac{L_0}{2} \\$The new time period $T$ is:$\\ T = 2\pi \sqrt{\frac{L}{g}} = 2\pi \sqrt{\frac{L_0/2}{g}} \\ T = 2\pi \sqrt{\frac{L_0}{2g}} \\ T = 2\pi \frac{1}{\sqrt{2}} \sqrt{\frac{L_0}{g}} \\ T = \frac{2\pi}{\sqrt{2}} \sqrt{\frac{L_0}{g}} \\ T = \frac{1}{\sqrt{2}} \cdot 2\pi \sqrt{\frac{L_0}{g}} \\ T = \frac{1}{\sqrt{2}} \cdot T_0 \\$The new time period is $\frac{1}{\sqrt{2}}$ times the original time period.$\\$According to the problem, the new time period is $\frac{x}{2}$ times the original time period.$\\$Therefore, we equate:$\\ \frac{1}{\sqrt{2}} = \frac{x}{2} \\$Solving for $x: \\ x = 2 \cdot \frac{1}{\sqrt{2}} \\ x = \frac{2}{\sqrt{2}} \\ x = \sqrt{2} \\$Thus, the value of $x$ is $\sqrt{2}. \\$This corresponds to Option 4.