To solve this problem, we will use the kinematic equations for uniformly accelerated motion.$\\$Given:$\\$• Initial velocity of the bullet $u$ when it enters the block.$\\$• Final velocity after traveling 24 cm is $\frac{u}{3}. \\$Let's denote:$\\$• $s_1 = 24$ cm as the distance traveled by the bullet to reach velocity $\frac{u}{3}. \\$• $v_1 = \frac{u}{3}$ as the velocity after traveling $s_1. \\$• $a$ as the acceleration (deceleration in this case) of the bullet.$\\$Using the kinematic equation:$\\ v_1^2 = u^2 + 2a s_1 \\$Substitute the known values:$\\ \left(\frac{u}{3}\right)^2 = u^2 + 2a \cdot 24 \\$Simplify:$\\ \frac{u^2}{9} = u^2 + 48a \\$Rearrange to solve for $a: \\ 48a = \frac{u^2}{9} - u^2 \\$Combine the terms:$\\ 48a = \frac{u^2 - 9u^2}{9} \\ 48a = \frac{-8u^2}{9} \\ a = \frac{-8u^2}{432} \\ a = \frac{-u^2}{54} \\$Now, consider the bullet coming to rest at the other end of the block.$\\$Let $s_2$ be the additional distance traveled by the bullet before coming to rest.$\\$Final velocity $v_2 = 0$ when the bullet stops.$\\$Using the kinematic equation:$\\ v_2^2 = v_1^2 + 2a s_2 \\$Substitute the known values:$\\ 0 = \left(\frac{u}{3}\right)^2 + 2\left(\frac{-u^2}{54}\right) s_2 \\$Simplify:$\\ 0 = \frac{u^2}{9} - \frac{u^2}{27} s_2 \\$Rearrange to solve for $s_2: \\ \frac{u^2}{27} s_2 = \frac{u^2}{9} \\ s_2 = \frac{u^2}{9} \cdot \frac{27}{u^2} \\ s_2 = 3 \\$The total length of the block $s_{total}$ is:$\\ s_{total} = s_1 + s_2 \\ s_{total} = 24 + 3 \\ s_{total} = 27 \\$Therefore, the total length of the block is $27$ cm. This corresponds to Option 1.