Binding energy gain indicates increased stability after nuclear reactions
Step 1: Calculate initial and final binding energies - Initial binding energy \[ B_{\text{initial}} = 242 \times 7.6 = 1839.2 \, \text{MeV}. \] Final binding energy \[ B_{\text{final}} = 2 \times (121 \times 8.1) = 2 \times 980.1 = 1960.2 \, \text{MeV}. \]
Step 2: Calculate the gain in binding energy - Gain in binding energy is given by: \[ \text{Gain} = B_{\text{final}} - B_{\text{initial}} = 1960.2 - 1839.2 = 121 \, \text{MeV}. \]
Final Answer: The total gain in binding energy is 121 MeV.
A bullet of mass \(10^{-2}\) kg and velocity \(200\) m/s gets embedded inside the bob of mass \(1\) kg of a simple pendulum. The maximum height that the system rises by is_____ cm.

Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
Consider the following two reactions A and B: 
The numerical value of [molar mass of $x$ + molar mass of $y$] is ___.
Consider an A.P. $a_1,a_2,\ldots,a_n$; $a_1>0$. If $a_2-a_1=-\dfrac{3}{4}$, $a_n=\dfrac{1}{4}a_1$, and \[ \sum_{i=1}^{n} a_i=\frac{525}{2}, \] then $\sum_{i=1}^{17} a_i$ is equal to