Kerala Engineering Entrance test-2006 – Two Questions

The following two questions were omitted by a fairly bright student who appeared for Kerala Engineering Entrance test (I wonder what is so unattractive about these questions!):
(1) The momentum of a body is increased by 25%. The kinetic energy is increased by about
(a) 25% (b) 5% (c) 56% (d) 38% (e) 65%
The kinetic energy of a body is given by E = p²/2m. Therefore, when the momentum ‘p’ is increased by 25%, the kinetic energy E is increased by about 56%. (Note that 1.25² = 1.5625).
(2) If the potential energy of a gas molecule is U = M/r⁶ – N/r¹², M and N being positive constants, then the potential energy at equilibrium must be
(a) zero (b) M²/4N (c) N²/4M (d) MN²/4 (e) NM²/4
You know that the molecule will be in equilibrium if its potential energy U is a minimum. For this, dU/dr = 0. Therefore, -6Mr^-7 + 12Nr^-13 = 0 from which r⁶ = 2N/M. Therefore, r¹² = 4N²/M². Substituting these values in the expression (given in the question) for U, the potential energy at equilibrium is M/(2N/M) – N/(4N²/M²) = M²/2N – M²/4N = M²/4N [Option (b)].