Two Questions (MCQ) on Doppler Effect in Sound

You will find the formulae to be remembered in connection with Doppler effect and some useful multiple choice questions (with solution) on Doppler effect at this location on this site.

[If you want all questions on Doppler effect on this site, you may click on the label ‘Doppler effect’ below this post].

Today I give you two more multiple choice questions on Doppler effect:

(1) A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationary person with speed v ms^–1. The velocity of sound in air is 300 ms^–1. If the person can hear frequencies up to 10000 Hz, the maximum value of v up to which he can hear the whistle is

(1) 30 ms^–1

(2) 15√2 ms^–1

(3) 15/√2 ms^–1

(4) 15 ms^–1

The apparent frequency (n’) in terms of all possible variables is given by

n’ = n(v+w–v_L)/(v+w–v_S) where ‘n’ is the real frequency of sound, ‘v’ is the velocity of sound, ‘w’ is the velocity of wind, ‘v_L’ is the velocity of listener and ‘v_S’ is the velocity of the source of sound. Note that in the above expression all velocities are in the same direction and the source is behind the listener and is therefore approaching the listener.

Since the wind velocity is zero and the listener is stationary in the first case, the above expression reduces to

n’ = nv/(v–v_S)

[The apparent frequency is therefore greater than the real frequency].

As the person can hear frequencies up to 10000 Hz only, the maximum value of v_S upto which he can hear the whistle is given by

10000 = 9500×300/(300 – v_S)

Therefore, 300 – v_S = 285 so that v_S = 15 ms^–1

[The above question appeared in AIEEE 2006 question paper]

(2) A person moves away with constant velocity ‘v₀’ from a stationary train which blows its whistle. The ratio of the real frequency of the whistle to the apparent frequency as measured by the person is 1.25. If the person is stationary and the whistle is moving away from the person with the same velocity ‘v₀’, the ratio of the real frequency of the whistle to the apparent frequency as measured by the person will be

(a) 1.2

(b) 1.25

(c) 1.4

(d) 1.45

(e) 1.5

The apparent frequency (n₁) as measured by the person moving away from the whistle with velocity v_L is given by

n₁ = n(v–v_L)/v where ‘n’ is the real frequency of the whistle and ‘v’ is the velocity of sound.

Therefore, n/n₁ = v/(v–v_L) = v/(v–v₀).

Since n/n₁ = 1.25,we have v/(v–v₀) =1.25 so that 1– (v₀/v) = 0.8 from which v₀ = 0.2v.

If the person is stationary and the whistle is moving away from the person with the same velocity ‘v₀’, the apparent frequency (n₂) as measured by the person is given by

n₂ = nv/(v +v_S) where v_S = v₀

Therefore, n/n₂ = (v+v₀)/v.

Substituting for v₀ (= 0.2v) we obtain n/n₂ = (v+0.2v)/v = 1.2

Geometric Optics- Multiple Choice Questions involving Refraction at Plane Surfaces

I am neither especially clever nor especially gifted. I am only very, very curious.

– Albert Einstein

The following question at the first glance may appear to be a difficult one to many of you; but, you will realise how easy it is when you apply basic points you studied in geometric optics:

A glass jar has the plane inner surface PQ of its bottom silvered and contains water (of refractive index n = 4/3) column of height t = 6 cm. A small light emitting diode (LED) is arranged at O at a height d = 8 cm from the water surface AB (Fig.). The silvered bottom of the jar acts as a plane mirror. At what distance from the free surface (AB) of water will this plane mirror form the image of the light emitting diode?

(a) 11 cm

(b) 14 cm

(c) 17 cm

(d) 18 cm

(e) 20 cm

When you look into the plane mirror (silvered surface) PQ from the position O of the LED, the plane mirror will appear to be located at P₁Q₁ (fig.) at a distance t/n from the free surface of water (because of normal refraction at the water surface). The distance of the LED from this refracted image P₁Q₁ of the plane mirror is therefore equal to (d + t/n) as shown in the adjoining figure.

The image of the LED must be formed at O₁ which is at the same distance (d + t/n) from the effective plane mirror P₁Q₁.

As is clear from the adjoining figure, the distance of the image O₁ from the free surface (AB) of water is (d + t/n) + t/n which is equal to (d + 2t/n) = 8 + 2×6/(4/3) = 17 cm.

The following question appeared in EAMCET (Engineering) 2003 question paper:

One of the refracting surfaces of a prism of refractive index √2 is silvered. The angle of the prism is equal to the critical angle of a medium of refractive index 2. A ray of light incident on the unsilvered surface passes through the prism and retraces its path after reflection at the silvered face. Then the angle of incidence on the unsilvered surface is

(a) 0º

(b) 30º

(c) 45º

(d) 60º

The angle A of the prism (as mentioned in the question) is given by n = 1/sin A where n = 2.

[Remember n = 1/sin C where n is the refractive index and C is the critical angle].

Therefore, sin A = ½ so that A = 30º

Since the ray retraces its path after reflection at the silvered face, it is incident normally at the silvered face (at the point N in the figure). With reference to the figure, angle QNA in the triangle QNA is 90º.

Since the angle A is 30º it follows that angle AQN = 60º so that the angle of refraction (r) at Q is 30º.

The angle of incidence (i) at the unsilvered face is given by

n = sin i/sin r from which sin i = n sin r = √2 sin 30º.

This gives sin i =1/√2 so that i = 45º.

You may search for ‘optics’ using the ‘search blog’ facility at the top left of this page to find all related posts on this site.

A useful post on the equations to be remembered in Geometric Optics can be found here.

Kerala Engineering Entrance 2008 and other Multiple Choice Questions (MCQ) on Work and Energy

Setting an example is not the main means of influencing others; it is the only means.

– Albert Einstein

Let us discuss a few multiple choice questions on work and energy.

(1) A small sphere of mass 20 g is projected vertically up with a velocity of 10 ms^–1. If air resistance is negligible, what is the total work done by gravity during the upward trip of the sphere?

(a) 10 J

(b) –10 J

(c) – 1 J

(d) 1 J

(e) 9.8 J

When the sphere rises up its kinetic energy goes on decreasing because of the work done by the gravitational force against the motion of the sphere. (The gravitational potential energy of the sphere goes on increasing by an equal amount). The work done by gravity is evidently negative. When the sphere reaches the maximum height the entire kinetic energy gets converted into gravitational potential energy. The total work done by gravity during the the upward trip of the sphere is numerically equal to the initial kinetic energy (½ mv²) of the sphere but its sign is negative.

Therefore, the answer is – ½ mv² = – ½ ×0.020×10² = –1 J.

(2) A world class athlete covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

(a) 2 × 10⁵ J to 5 × 10⁵ J

(b) 2 × 10⁴ J to 5 × 10⁴ J

(c) 2 × 10³ J to 5 × 10³ J

(d) 200 J to 500 J

(e) 20 J to 50 J

This question has appeared in various entrance tests with slight differences in the wording and in the options. Last year it appeared in the All India Engineering Entrance Examination.

In the 100 m dash the velocity v is almost constant so that we have

v = 100 m/10 s = 10 ms^–1

We may take the mass of the athlete to be 70 kg to 80 kg. Let us use 80 kg.

His kinetic energy will be ½ mv² = ½ ×80×10² = 4000 J so that the correct option is (c).

Questions (3) and (4) appeared in Kerala Engineering Entrance Examination 2008 question paper.

(3) Two bodies A and B have masses 20 kg and 5 kg respectively. Each one is acted upon by a force of 4 kg wt. If they acquire the same kinetic energy in times t_A and t_B, then the ratio t_A/ t_B is

(a) ½

(b) 2

(c) 2/5

(d) 5/6

(e) 1/5

The concept of impulse will be very useful here. The impulse received by A and B in the times t_A and t_B are respectively F t_A and F t_B where F is the force acting on them (4 kg wt. here). But impulse is the change in momentum so that the momenta acquired by A and B are F t_A and F t_B.

Since the kinetic energy is p²/2m where p is the momentum, we have

(F t_A)²/2m_A = (F t_B)²/2m_B

Therefore, t_A/ t_B = √( m_A/ m_B) = √(20/5) = 2.

(4) A particle acted upon by constant forces 4i + j – 3k and 3i + j – k is displaced from the point i + 2j + 3k to the point 5i + 4j + k. The total work done by the forces in SI units is

(a) 20

(b) 40

(c) 50

(d) 30

(e) 35

The resutant force (F) on the particle is the sum of the forces given by

F = (4i + j – 3k) + (3i + j – k) = 7i + 2j – 4k.

The displacement (s) of the particle is given by

s = (5i + 4j + k) – (i + 2j + 3k) = (4i + 2j – 2k)

The work done (W) is given by

W = F.s = (7i + 2j – 4k) . (4i + 2j – 2k) = 28 + 4 + 8 = 40.

You will find similar useful multiple choice questions (with solution) at AP Physics Resources.

KEAM 2009 Rescheduled

The Commissioner for Entrance Examinations, Govt. of Kerala, has notified that the dates of the Entrance Examinations for Admission to Medical/ Agriculture/ Veterinary/ Engineering Degree Courses 2009 (KEAM 2009), Kerala have been changed (for the second time) as follows:

Modified Dates of Exam:

Engineering Stream:

25.05.2009 Monday 10.00 A.M. to 12.30 P.M. Paper-I : Physics & Chemistry.

26.05.2009 Tuesday 10.00 A.M. to 12.30 P.M. Paper-II: Mathematics.

Medical Stream:

27.05.2009 Wednesday 10.00 A.M. to 12.30 P.M. Paper-I : Chemistry & Physics.

28.05.2009 Thursday 10.00 A.M. to 12.30 P.M. Paper-II: Biology.

For more details visit the site http://www.cee-kerala.org/ where you will find information regarding NATA (National Aptitude Test in Architecture) required for those who apply for Architecture