Kerala Medical Entrance (KEAM) 2008 Questions on Nuclear Physics

Try not to be a person of success, but rather a person of virtue.

– Albert Einstein

Here are the two questions from nuclear physics which were included in KEAM (Medical) 2008 question paper:

(1) If the mass defect of ₈O¹⁶ nucleus is 0.128 amu, then the binding energy per nucleon of oxygen is

(a) 8.2 MeV

(b) 7.45 MeV

(c) 7.3 MeV

(d) 7.1 MeV

(e) 8.15 MeV

One atomic mass unit (amu) is equivalent to 931 MeV. Therefore, the total binding energy of the ₈O¹⁶ nucleus is 0.128×931 MeV.

Since there are 16 nucleons in the ₈O¹⁶ nucleus, the binding energy per nucleon of oxygen is (0.128×931)/16 = 7.45 MeV, very nearly.

(2) Two radioactive samples have decay constants 15x and 3x. If they have the same number of nuclei initially, the ratio of number of nuclei after a time 1/6x is

(a) 1/e

(b) e/2

(c) 1/e⁴

(d) 2e/3

(e) 1/e²

The number N of nuclei at time t is given by

N = N₀e^–λt where N₀ is the initial number, e is the base of natural logarithms, and λ is the decay constant.

The required ratio is (N₀e^–15x/6x)/ (N₀e^–3x/6x) = e^–2.5/ e^–0.5 = e^–2 = 1/e²

Three questions from nuclear physics were included in the physics question paper of Kerala Engineering Entrance (KEAM) 2008 examination. You will find those questions with solution here

Two Kerala Engineering Entrance 2005 Questions on Electrostatics

I never did a day’s work in my life. It was all fun.

– Thomas Alva Edison

Here are two multiple choice questions which appeared in Kerala Engineering Entrance 2005 question paper:

(1) A soap bubble is charged to a potential of 16 V. Its radius is then doubled. The potential of the bubble now will be

(a) 16 V

(b) 8 V

(c) 4 V

(d) 2 V

(e) zero

The potential of a bubble of radius R carrying charge Q is Q/4πε₀R so that for a given charge the potential is inversely proportional to the radius. Therefore when the radius is doubled, the potential is halved. The answer is 8 V [Option (b)].

(2) A parallel plate capacitor of capacitance 10 μF is charged to 1μC. The charging battery is removed and then the separation between the plates is doubled. Work done during the process is

(a) 5 μJ

(b) 0.05 μJ

(c) 1 μJ

(d) 10 μJ

(e) 50 μJ

The work done is equal to the increase in the energy of the capacitor.

The initial energy is Q²/2C where C is the initial capacitance and Q is the charge.

Therefore initial energy, Q²/2C = 10^–12/(20×10^–6) = 5×10^–8 J = 0.05 μJ.

When the separation between the plates is doubled, the capacitance is halved (since the capacitance is Kε₀A/d with usual notations) and hence the energy is doubled. The final energy is thus 0.1 μJ.

The increase in energy, which is equal to the work done, is 0.05 μJ.

You will find all posts related to electrostatics on this site by clicking on the label ‘electrostatics’ below this post. More useful questions with solution on electrostatics can be found here as well as here.