Multiple Choice Questions on Nuclear Fission

Multiple choice questions on nuclear fission at the level expected from you will be simple. Here are three such questions:

(1)The energy released per fission of a U²³⁵ nucleus is around

(a) 0.02 eV

(b) 2 eV

(c) 2 MeV

(d) 20 MeV

(e) 200 MeV

The mass of the products of fission will be less than the mass of the U²³⁵ nucleus. The mass difference gets converted into energy. Even though different fragments can be produced, the energy released per fission of a U²³⁵ nucleus is around 200 MeV.

(2) From the following, pick out the most suitable energy of neutrons which will produce nuclear fission in a reactor

(a) 0.04 eV

(b) 40 eV

(c) 400 eV

(d) 2 MeV

(e) 20 MeV

Nuclear fission is induced most effectively by neutrons of thermal energy and hence the correct option is 0.04 eV

(3) The function of the moderator in a nuclear reactor is

(a) to absorb fast neutrons

(b) to adjust the power output to moderate levels

(c) to slow down fast neutrons

(d) to absorb slow neutrons

(e) to cool the reactor core

This is a simple question which repeatedly appears in entrance test papers with minor changes in the wrong options. The moderator is used to slow down fast neutrons to thermal energies [Option (c)].

You can find all related posts on this site by clicking on the label ‘nuclear physics’ below this post.

You can find some useful multiple choice questions (with solution) from nuclear physics at physicsplus

Two Kerala Engineering Entrance 2007 Questions on Waves

The following question appeared in Kerala Engineering Entrance (2007) Examination question paper. It has appeared in modified forms in other question papers as well:

A transverse wave is described by the equation y = y₀sin2π(ft – x/λ). The maximum particle velocity is equal to four times the wave velocity if

(a) λ = πy₀/4

(b) λ = πy₀/2

(c) λ = πy₀

(d) λ = 2πy₀

(e) λ = πy₀/3

The particle velocity is dy/dt = 2πfy₀ cos2π(ft – x/λ) and the maximum particle velocity is 2πfy₀ ( when the cosine term has its maximum value equal to one).

The wave velocity is fλ since f is the frequency and λ is the wave length.

When the maximum particle velocity is equal to four times the wave velocity, we have

2πfy₀ = 4fλ so that λ = πy₀/2

An open pipe (or open organ pipe) in Acoustics means a pipe open at both ends where as a closed pipe (or closed organ pipe) means a pipe closed at one end. The following question on organ pipes which appeared in Kerala Engineering Entrance 2007 Examination question paper high lights the difference between the fundamental frequencies as well as the overtones possible in open and closed organ pipes:

An open organ pipe is closed suddenly with the result that the second overtone of the closed pipe is found to be higher in frequency by 100 than the first overtone of the original pipe. Then the fundamental frequency of the open pipe is

(a) 200 s^–1

(b) 100 s^–1

(c) 300 s^–1

(d) 250 s^–1

(e) 150 s^–1

The closed end of a pipe is always a node where as the open end is always an antinode. In the fundamental mode, the frequency of vibration of the air column in the open pipe is such that the open ends are antinodes as usual, but they are consecutive antinodes. This means that the length of the open pipe is equal to half the wave length of the fundamental note produced by the pipe. But in a closed pipe in the fundamental mode, the closed end is a node and the open end is the next antinode so that the length of the closed pipe is equal to a quarter of the wave length of the fundamental note produced by the pipe. The fundamental frequency (n) of the closed pipe is therefore equal to half the fundamental frequency (N) of the open pipe of the same length.

Or, n = N/2

All harmonics (odd as well as even) are possible in open pipes where as odd harmonics only are possible in closed pipes. The first overtone of the closed pipe is the 3^rd harmonic and the second overtone is the 5^th harmonic.

Therefore, frequency of the second overtone of the closed pipe = 5n

The first overtone of the open pipe is its second harmonic and is equal to 2N

Therefore, we have

5n = 2N + 100

Since n = N/2, the above equation becomes

5 N/2 = 2N + 100 from which N = 200 s^–1

^{You will find similar questions from different branches of Physics at physicsplus.blogspot.com}