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.

From a height of 'h' above the ground, a ball is projected up at an angle \( 30^\circ \) with the horizontal. If the ball strikes the ground with a speed of 1.25 times its initial speed of \( 40 \ ms^{-1} \), the value of 'h' is:

In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by: