Kerala Engineering Entrance 2008 Questions on One Dimensional Motion

The following questions (on one dimensional kinematics) numbered 1, 2 and 3 appeared in KEAM (Engineering) 2008 question paper:

(1) A particle starts from rest at t = 0 and moves in a straight line with an acceleration as shown below. The velocity of the particle at t = 3 s is

(a) 2 ms^–1

(b) 4 ms^–1

(c) 6 ms^–1

(d) 8 ms^–1

(e) 1 ms^–1

According to the acceleration – time graph shown, the particle has an acceleration of 4 ms^–2 during the first two seconds. Therefore, its velocity (v₂) at the end of 2 seconds is given by

v₂ = v₀ + at = 0 + 4×2 = 8 ms^–1

From 2 second to 3 second the particle has a retardation of 4 ms^–2. Hence its velocity (v₃) at the end of 3 seconds is given by

v₃ = v₂ – at = 8 – 4×1 = 4 ms^–1 [Option (b)].

(2) Two cars A and B are moving with same speed of 45 km/hr along same direction. If a third car C coming from the opposite direction with a speed of 36 km/hr meets two cars in an interval of 5 minutes, the distance of separation of two cars A and B should be (in km)

(a) 6.75

(b) 7.25

(c) 5.55

(d) 8.35

(e) 4.75

The relative velocity of car A with respect to car B (and that of car B with respect to A) is zero since they have the same speed in the same direction. The relative velocity of car C with respect to A and B is 45 + 36 = 81 km/hr. Since the car C takes a time of 5 minutes to cover the distance between A and B, the separation between A and B is 81×(5/60) km = 6.75 km.

(3) An object is dropped from rest. Its v - t graph is

The correct option is (a) since the velocity ‘v’ is directly proportional to the time t in accordance with the equation v = v₀ + at where v₀ is the initial velocity which is zero and a is the acceleration which is the constant acceleration due to gravity. The graph should evidently pass through the origin.

Suppose you were asked to draw the v – t graph in the case of a ball projected vertically upwards with a velocity u. If you are asked to draw the graph from the instant the ball leaves your hand to the instant you catch it while returning, you can do it as shown, ignoring the air resistance and the time taken for the velocity to reduce to zero on hitting your hand.

Questions involving Viscosity and Buoyancy

The following simple question involving viscous force and force of buoyancy is meant for checking whether you have a good understanding of basic points:

A wooden ball of relative density 0.5 is released from the bottom of a still water reservoir. Its acceleration while moving up will

(a) go on decreasing initially

(b) go on increasing initially

(c) remain constant at g/2 (in magnitude)

(d) remain constant at g (in magnitude)

(e) be zero throughout

The net force acting on the ball at the moment of releasing is its apparent weight which is upwards. [The apparent weight = Real weight – up thrust = Vρg – Vσg. Since the density of wood (ρ) is less than that of water (σ), the apparent weight is negative which means it is directed upwards]

The sphere therefore moves up with an acceleration. But, when the velocity (v) increases from zero, the opposing viscous force (6πrηv) also increases thereby reducing the net upward force. The upward acceleration therefore goes on decreasing. So, the correct option is (a).

The following question which appeared in Kerala Engineering Entrance 2008 question paper also involves viscous force and force of buoyancy; but the latter is negligible:

Eight drops of a liquid of density ρ and each of radius ‘a’ are falling through air with a constant velocity of 3.75 cm s^–1. When the eight drops coalesce to form a single drop, the terminal velocity of the new drop will be

(a) 1.5×10^–2 ms^–1

(b) 2.4×10^–2 ms^–1

(c) 0.75×10^–2 ms^–1

(d) 25×10^–2 ms^–1

(e) 15×10^–2 ms^–1

The viscous force acting on a sphere of radius ‘a’ is 6πahv where h is the coefficient of viscosity of the fluid and ‘v’ is the velocity of the sphere. Terminal velocity is attained when the opposing viscous force is equal in magnitude to the apparent weight of the sphere. Therefore,

6πahv = (4/3)πa³(ρ – σ)g where ρ is the density of the sphere, σ is the density of the fluid and ‘g’ is the acceleration due to gravity.

The above equation shows that the terminal velocity ‘v’ is directly proportional to the square of the radius of the sphere.

In the above problem, eight identical drops coalesce to form a single drop. The new drop thus formed has eight times the volume of each small drop. The radius ‘R’ of the new drop is given by

(4/3)πR³ = 8×(4/3)πa³

Therefore, R =2a

Since the radius of the new drop is twice that of each small drop, the terminal velocity of the new drop must become four times. The correct option therefore is 15×10^–2 ms^–1.

Two Multiple Choice Questions involving Kinetic Energy

Check whether you can find the answers to the following two questions in a couple of minutes:

(1) When a running boy increases his speed by 2 ms^–1, his kinetic energy becomes three times the original value. His original speed is (in ms^–1)

(a) √3 +1

(b) √3 –1

(c) √3

(d) 2√3

(e) 2

If his original speed is ‘v’ we can write

(½)mv²×3 = (½)m(v+2)² where ‘m’ is his mass.

This gives v√3 = (v+2) so that v(√3 –1) =2.

Therefore, v = 2 /(√3 –1) = √3 +1 on multiplying the numerator and denominator by (√3 –1).

(2) A particle of mass ‘m’ at rest is acted on by a force ‘F’ for a time‘t’. Its kinetic energy after the time t is

(a) Ft²/2m

(b) F²t/2m

(c) F²t²/2m

(d) F²t²/3m

(e) F²t²/m

The impulse received by the particle during the time t is Ft. But impulse is equal to the change of momentum. Since the initial momentum of the particle is zero, Ft is the final momentum of the particle. The kinetic energy of the particle after the time t is therefore equal to F²t²/2m (remembering that KE = p²/2m where p is the momentum).

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}