Questions (MCQ) on Tansmission of Heat

In this section you have to remember the following:

(1) The quantity (Q) of heat conducted through a rod of thermal conductivity ‘K’ and cross section area ‘A’ in time ‘t’ under a temperature gradient dθ/dx is given by

Q = KA(dθ/dx)t

(2) If ‘n’ conductors of the same area of cross section having lengths d₁, d₂, d₃….. d_n and thermal conductivities K₁, K₁, K₃,….K_n are joined in series, the same quantity of heat will flow through them in a given time. The thermal conductivity of such a compound rod is given by K = (d₁+ d₂+ d₃+….. +d_n)/[(d₁/K₁)+ (d₂/K₂)+ (d₃/K₃)+…...+(d_n/K_n)]

If two rods of the same length and the same area of cross section are joined in series, the equivalent thermal conductivity can be obtained from the above general expression as K = 2K₁K₂/ (K₁+K₂)

(3) Stefan’s law: Energy (E) radiated per second from a perfectly black body of surface area ‘A’ is given by E = AσT⁴ where ‘σ’ is Stefan’s constant equal to 5.57×10^–8Wm^–2K^–4.

If the body is not a perfect black body, E = AeσT⁴ where ‘e’ is a dimensionless fraction called emissivity.

In the case of a perfectly black body at temperature T, with surroundings at temperature T_0, the net rate of loss of radiant energy is E = Aσ(T⁴ – T₀⁴), since the body emits energy AσT⁴ and absorbs energy AσT₀⁴ per second in accordance with Kirchhoff's law.

(4) Wein’s displacement law: λ_mT = constant (= 0.29cmK) where λ_m is the wavelength for which the energy radiated is maximum at temperature T. If the constant is taken as 0.29 cmK, the value of λ_m is to be substituted in cm.

(5) Newton’s law of cooling: The rate of cooling is directly proportional to the excess of temperature of the body over the surroundings if the excess of temperature is small compared to the temperature of the body (T) and the temperature (T_S) of the surroundings: dT/dt α (T– T_S)

Now consider the following MCQ which appeared in IIT screening 2002 question paper:

An ideal black body at room temperature is thrown in to a furnace. It is observed that

(a) initially it is the darkest body and at later times the brightest

(b) it is the darkest body at all times

(c) it cannot be distinguished at all times

(d) initially it is the darkest body and at later times it cannot be distinguished

A black body is a good emitter as well as a good absorber. Initially it will absorb energy and hence will appear dark. Once it attains the temperature of the furnace, it emits better than the other parts of the furnace and hence appears the brightest. So the correct option is (a).

Consider the following question:

Two copper spheres A and B having the same emissivity have radii 12 cm and 3 cm respectively . If they are maintained at temperatures 727°C and 1727°C respectively, the ratio of energy radiated by A and B is

(a) 0.032 (b) 0.12 (c) 0.48 (d) 1 (e) 2

Since the energy radiated per second is given by E = AeσT⁴ and the emissivities are equal, we have E₁/E₂ = (A₁/A₂)(T₁/ T₂ )⁴.

Since the radii of A and B are in the ratio 4:1, the surface areas are in the ratio 16:1. Since the temperatures are 1000K and 2000K, the ratio (T₁/ T₂ )⁴ = 1/16. Therefore, E₁/E₂ = 1.

Here is a simple question involving Wein’s displacement law:
On examining the spectrum of a star, it is found that mximum energy is emitted at a wave length of 5800 Ǻ. The surface temperature of the star is
(a) 4500 K (b) 5000 K (c) 5500 K (d) 6000 K (e) 6500 K
This question indicates how one can determine the temperature of a distant star from spectroscopic data. According to Wien’s law, we have λ_mT = 0.29cmK, from which T =0.29/(5800×10^–8) = 5000K. [Note that we have substituted the wave length in cm].

Consider the following question involving heat conduction:

Three identical iron rods are welded together to form the shape of Y. The top ends of the ‘Y’ are maintained at 0°C and the bottom end is maintained at 600°C. The temperature of the junction of the three rods is

(a) 100°C (b) 200°C (c) 250°C (d) 300°C (e) 400°C

The quantity of heat conducted per second through the bottom rod making the ‘Y’ gets divided equally at the junction of the three rods. If ‘θ’ is the temperature of the junction, we have

KA(θ–0)/L = 2KA(600 – θ)/L where K is the thermal conductivity, A is the area of cross section and L is the length of the identical rods.

[ Note that the L.H.S. is the quantity of heat conducted through the lower single rod making the ‘Y’ and the R.H.S. is the sum of the quantities of heat conducted through the upper two rods].

The above equation yields θ = 400°C

Heat engine and Refrigerator

From the section Thermodynamics, you will usually get questions on heat engines or refrigerators. You will definitely remember the expression for the efficiency (η) of a Carnot engine: η = (Q₁ –Q₂)/Q₁ = (T₁–T₂)/T₁ where Q₁ is the heat absorbed from the source at temperature T₁ and T₂ is the heat rejected to the sink at the lower temperature T₂.
The coefficient of performance (β) of a refrigerator (heat pump) is given by
β = (Heat removed from the cold body) / (Work done by the pump)
= Q₂/W = Q₂/(Q₁– Q₂) = T₂/(T₁–T₂)Now consider the following simple question which appeared in AIEEE 2002 question paper:
Even Carnot engine cannot give 100% efficiency because we cannot
(a) prevent radiation (b) find ideal source
(c) reach absolute zero temperature (d) eliminate frictionThe correct option is (c). The efficiency is given by the expression, η = (T₁–T₂)/T₁. The percentage efficiency is [(T₁–T₂)/T₁] ×100. This shows that the efficiency is 100% only if either the source temperature T₁ is infinite or the sink temperature T₂ is zero. Both are impossibilities.
Now see the following MCQ:
In a Carnot engine 800 J of heat is absorbed from a source at 400 K and 640 J of heat is rejected to the sink. The temperature of the sink is
(a) 320 K (b) 100 K (c) 273 K (d)250 K (e) 200 KIn a Carnot engine, Q₁/T₁ = Q₂/T₂ so that the temperature of the sink, T₂ = T₁Q₂/Q₁ = 400×640/800 = 320 K.
The following question appeared in Kerala Engineering Entrance 2000 question paper:
A heat engine undergoes a process in which its internal energy decreases by 400 J and it gives out 150 J of heat. During the process
(a) it does 250 J of work and its temperature rises
(b) it does 250 J of work and its temperature falls
(c) it does 550 J of work and its temperature rises
(d) it does 550 J of work and its temperature falls
(e) 250 J of work is done on the systemThe internal energy of the system will decrease when the system does work and/or gives off heat. Since the heat given out is 150 J and the reduction in internal energy is 400 J, the work done by the engine is 400– 150 = 250 J.
When the internal energy is reduced, the system is cooled. So, the correct option is (b).
Now, consider the following question:
The temperature inside a refrigerator is 4°C and the room temperature is 27°C. How many joules of heat will be delivered to the room for each joule of electricity consumed by the refrigerator?( Treat the refrigerator as ideal).
(a) 1 J (b) 12 J (c) 8.3 J (d) 13 J (e) 6 JThe coefficient of performance of the refrigerator, β = Q₂/W = Q₂/(Q₁– Q₂) = T₂/(T₁–T₂) = 277/(300–277) = 12. [Note that we have converted the temperature to the Kelvin scale].
Therefore, Q₂ =12 W. Heat delivered to the room is Q1 = Q₂+W = 12W+W = 13W. Here W is the work done by the pump. So for each joule of work done (for each joule of electricity consumed), the quantity of heat pumped out in to the room will be 13 joule.

Given below is a question of the type which often finds a place in Entrance test papers:

An ideal gas is taken through a cycle of operations shown by the indicator diagram. The net work done by the gas at the end of the cycle is

(a) 6P0V0 (b) 4P0V0 (c) 15P0V0 (d) 10P0V0 (e) 3P0V0

The work done in a cyclic process indicated by a PV diagram is the area enclosed by the closed curve. The area under the slanting curve showing the expansion of the gas from volume 2V0 to volume 5V0 gives the work done by the gas. This is greater than the area under the (horizontal) curve showing the compression of the gas (from volume 5V0 to volume 2V0), which gives the work done on the gas. The vertical portion of the curve is an isochoric (volume constant) change which involves no work since the area under it is zero. The area enclosed by the closed curve gives the net work done by the gas. The triangular area enclosed is ½ ×3V0×2P0 = 3P0V0 [Option (e)].
[ Note that in problems of the above type, the work is done by the gas if the arrow showing the cycle is clockwise. If the arrow is anticlockwise, work is done on the gas. In either case, the work done is the area enclosed by the curve].

Questions on Isothermal and Adiabatic Changes

You can expect questions involving isothermal and adiabatic processes in most entrance examinations. Here is a typical question:
A gas at a temperature of 27°C inside a container is suddenly compressed to one sixteenths of its initial volume. The temperature of the gas immediately after the compression is (Ratio of specific heats of the gas, γ = 1.5)
(a) 19200 K (b) 1200°C (c) 927°C (d) 108°C (e) 19200°C
This is an adiabatic change since the compression is sudden so that the volume(V) and the temperature (T) are related as TV^γ–1 = constant.
Therefore we have 300 V^0.5 = T(V/16)^0.5. Note that the temperature is to be substituted in Kelvin. The final temperature is given by T = 300×16^0.5 = 1200 K = 927°C.
The following MCQ is meant for testing your understanding of the work done in thermodynamic processes:

Starting from the same initial conditions an ideal gas expands from volume V₁ to volume V₂ in three different ways: (i) Adiabatically, doing work W₁. (ii) Isothermally, doing work W₂. (iii) Isobarically, doing work W₃. Then,

(a) W₁ = W₂ = W₃ (b) W₁ > W₂ > W₃ (c) W₃ > W₂ > W₁ (d) W₂ > W₁ > W₃ (e) W₁ > W₃ > W₂

Since the work done is the area under the corresponding curve in a PV diagram, you can easily verify that the work done is the largest in the isobaric case since it is a straight line parallel to the volume axis. The adiabatic curve is the steepest one so that the area under it is the smallest. The correct option therefore is (c).
You can easily show that the slope of the adiabatic curve is γ times the slope ofthe isothermal curve as follows:
In the case of an adiabatic change, the pressure and volume are related as PV^γ = constant.
Differentiating, P γV^γ–1 dV + V^γdP = 0
Slope of adiabatic curve = dP/dV = (– γPV^γ–1)/V^γ = – γP/V
In the case of an isothermal change, the pressure and volume are related as PV= constant.
Differentiating, PdV + VdP = 0.
Slope of isothermal curve = dP/dV = – P/V.
This show that the slope of the adiabatic curve is γ times the slope ofthe isothermal curve.
The following MCQ also pertains to adiabatic compression:
A gas having a volume of 800 cm³ is suddenly compressed to 100cm³. If the initial pressure is P, the final pressure is (γ = 5/3)
(a) P/32 (b) 24P (c) 32P (d) 8P (e) 16P
We have PV^γ = constant so that P×800^5/3 = P'×100^5/3 from which P' = P×8^5/3 = P×2⁵ = 32P.

Now, see this simple question:
An ideal gas expands isothermally from volume V₁ to volume V₂. It is then compressed to the original volume V₁ adiabatically. The initial pressure is P₁, final pressure is P₂ and the net work done by the gas during the entire process is W. Then
(a) P₁ = P₂, W>0 (b) P₁> P₂, W>0 (c) P₂ > P₁, W>0 (d) P₂ > P₁, W=0 (e) P₂ > P₁, W<0

The adiabatic compression will increase the temperature of the gas so that the final pressure (P₂) when the volume is restored to the value V₁ is greater than the initial pressure P₁. Since the pressure is greater during the adiabatic compression, more work has to be done on the gas. The work done on the gas is thus greater than the work done by the gas. In other words, the net work done by the gas during the entire process is negative. So, the correct option is (e).

Questions on Optical Instruments

The essential points you have to remember to work out questions on optical instruments are the following:
(1) Myopia (short sightedness) is corrected by a concave lens of focal length ‘d’ where ‘d’ is the distance of the far point in the case of the defective eye. [Note that for a normal eye the far point will be at infinity where as for the defective eye, it will be at a finite distance‘d’].

Therefore, f = –d
(2) Hypermetropia (long sightedness) is corrected by a convex lens of focal length ‘f’ given by f = dD/(d – D) where ‘d’ is the distance of the near point for the defective eye and D is the least distance of distinct vision (25cm), [Note that for the normal eye, the near point will be at D and for the defective eye it will be at a greater distance ‘d’].
(3) Presbyopia is corrected by bifocal lens with the upper portion concave and the lower portion convex. [Note that for the defective eye in this case, the far point is nearer (and not at infinity) while the near point is farther away (and not at D)].
(4) Astigmatism is corrected by cylindrical lens.
(5) Magnifying power of a simple microscope M = 1+ D/f if the image is formed at the least distance of distinct vision ‘D’.
If the image is formed at infinity (normal setting or setting for relaxed eye), M= D/f .
(6) Magnifying power of compound microscope (M):
(i) M = (v_o/u_o)(1 + D/f_e) if the final image produced by the eye piece is at the least distance of distinct vision ‘D’. Here u_o is the distance of object from the objective, v_o is the distance of the image produced by the objective and f_e is the focal length of the eye piece.
(ii) If the final image is at infinity (normal setting or setting for relaxed eye),
M = (v_o/u_o)(1 + D/f_e)
Approximate expressions for the magnifying power of a compound microscope in the two cases are M = (L/f_o)(1 + D/f_e) and M = (L/f_o)(D/f_e) in the two cases respectively. Here L is the tube length of the microscope, which is the distance between the objective and the eye piece.
(7) (a)Limit of resolution (minimum separation between two point objects which can be resolved) of a microscope, dmin = λ /2n sinθ where λ is the wave length (in vacuum) of the light used for illuminating the object, ‘n’ is the refractive index of the medium between the object and the objective and θ is the semi angle of the cone of light proceeding from the object to the objective. This is Abbe’s expression for the resolving power when the object is not luminous and is therefore illuminated, as is usually the case. [Note that nsinθ is the numerical aperture]
(b) Resolving power of microscope = 1/d_min
(8) Magnifying power of telescope(M)
(i) M = β/α = fo/fe for normal adjustment (image at infinity) where β is the angle subtended at the eye by the image, α is the angle subtended at the eye by the object, fo is the focal length of the objective and fe is the focal length of the eye piece.
(ii) M = (f_o/f_e)(1 + f_e/D) if the final image is formed at the least distance of distinct vision ‘D’.
(9) Length of a telescope (in normal setting) = f_o + f_e
(10) Limit of resolution of telescope (minimum angular separation between two point objects that can be resolved) dθ = 1.22λ/a where λ is the wave length of light proceeding from the object and ‘a’ is the aperture (diameter) of the objective.
(11) Resolving power of a telescope = 1/dθ = a/1.22λ
Now let us now consider the following MCQ:
The length of a microscope is 12 cm and its magnifying power is 25 for relaxed eye.The focal length of the eye piece is 4 cm. The distance of the object from the objective is
(a) 2 cm (b) 2.5 cm (c) 3 cm (d) 3.5 cm (e) 4 cm
The image produced by the objective is formed at the focus of the eye piece since the final image is formed at infinity (for relaxed eye). If v_o is the distance of the image formed by the objective, the tube length of the microscope (distance between the objective and eye piece) is v_o+ f_e = v_o+ 4 = 12 cm from which v_o = 8 cm.
The magnifying power (for relaxed eye) is given by M = (v_o/u_o)(D/f_e). Substituting the known values, 25 = (8/u_o)(25/4) from which u_o = 2 cm. [Option (a)].
The following question appeared in MPPMT 2000 question paper:
The length of the tube of a microscope is 10 cm. The focal lengths of the objective and eye lenses are 0.5 cm and 1 cm. The magnifying power of the microscope is about
(a) 5 (b) 166 (c) 23 (d) 500
The magnifying power, M = (L/f_o)( D/f_e) = (10/0.5)( 25/1) = 500.
Consider now the following MCQ which appeared in EAMCET 2000 question paper:
In a compound microscope, the focal lengths of two lenses are 1.5 cm and 6.25 cm. If an object is placed at 2 cm from the objective and the final image is formed at 25 cm from the eye lens, the distance between the two lenses is
(a) 6 cm (b) 7.75 cm (c) 9.25 cm (d) 11 cm
You should note that the focal length of the objective in a microscope is less than that of the eye piece. (In a telescope it is the other way round). Therefore, f_o = 1.5 cm and f_e = 6.25 cm.
The distance between the objective and the eye piece is the sum of the image distance(vo) for the objective and the object distance (ue) for the eye piece.
From the equation, 1/f = 1/v – 1/u as applied to the objective, we have
1/1.5 = 1/v_o – 1/(–2).
Note that we have substituted the object distance as –2 in accordance with the Cartesian sign convention discussed in the post dated 22-11-06 (Questions (MCQ) on Refraction at Spherical Surfaces). This yields the image distance in the case of the objective as v_o = 6 cm.The image distance for the eye piece is similarly given by
1/6.25 = 1/(–25) – 1/u_e.
Note that the sign of the object distance is negative in accordance with the Cartesian sign convention. This equation yields u_e = –5 cm. The negative sign just shows that the object distance for the eye piece is measured opposite to the direction of the incoming rays.In fact, we know that the real image (of the object) formed by the objective serves as the object for the eye piece and therefore its distance is negative. However, since u_e is an unknown quantity, we did not bother about its sign. If we had substituted its sign as negative in the law of distances, we would have obtained the value as +5 cm.
The distance between the objective and the eye piece is 6+5 =11 cm.
Here is a simple question involving myopia:
A man who cannot see clearly beyond 10m wants to see stars clearly. He should use a lens of power (in dioptre)
(a) 10 (b) – 10 (c) –1 (d) 0.1 (e) – 0.1
His defect is myopia and hence he should use a concave lens of focal length equal to the distance of the far point, which is given as 10m. The power of the lens is 1/(–10) = – 0.1 dioptre.

Now consider the following question:
The near point of a person with defective eye is at 65 cm. To correct his defect, he should use spectacle lenses of focal length
(a) 65 cm (b) 55.6 cm (c) 50.6 cm (d) 45,5 cm (e) 40.6 cm
As his defect is hypermetropia (long sightedness), he should use convex lenses of focal length ‘f’ given by f = dD/(d – D) where ‘d’ is the distance of his far point and ‘D’ is the least distance of distinct vision. Therefore, f = 65×25/(65 – 25) = 40.6 cm.
Here is a typical simple question on telescopes:
The magnifying power of a small telescope is 25 and the separation between its objective and eye piece is 52 cm in normal setting. The focal lengths of its objective and eye piece are respectively
(a) 50 cm and 2 cm (b) 27 cm and 25 cm (c) 45 cm and 7 cm (d) 50.5 cm and 1.5 cm (e) 47 cm and 5 cm
We have magnifying power, M = f_o/f_e so that 25 = f_o/f_e, from which f_o = 25 fe.
Sincethe length of the telescope = f_o + f_e, we have 52 = 25f_e + f_e = 26 f_e so that f_e = 2 cm and hence fo = 50 cm.
The following question pertains to the resolving power of a telescope:
The distance between the earth and the moon is nearly 3.8×10⁵ km. What is the separation of two points on the moon that can be just resolved using a 400 cm telescope, using light of wave length 6000 Ǻ?
(a) 46.5 m (b) 56 m (c) 69.5 m (d) 78.6 m (e) 85.5 m
The limit of resolution dθ = 1.22λ/a. This is the minimum angular separation between objects that can be just resolved.
The linear separation between the objects that can be just resolved is rdθ = 3.8×10⁸×1.22×6000×10^–10/4 = 69.5 m. [Note that we have converted all distances in to metre].
The following m.c.q. appeared in the MPPMT 2000 question paper:
The focal lengths of the eye piece and objective of a telescope are respectively 100 cm and 2 cm. The moon subtends an angle of 0.5º at the eye. If it is looked through the telescope, the angle subtended by the moon’s image will be
(a) 100º (b) 25º (c) 50º (d) 10º
the magnifying power of a telescope is given by M = β/α = f_o/f_e with usual notations. Therefore, β = α f_o/f_e = 0.5×100/2 = 25º [Option (b)].

Now Consider the following MCQ:
A telescope has an objective lens of focal length 1.5m and an eye piece of focal length 5 cm. If this telescope is used in normal setting to view a tower of height 100m located 3km away, what will be the height of the image of the tower?
(a) 10 cm (b) 15 cm (c) 20 cm (d) 25 cm (e) 50 cm
The angle subtended at the objective by the tower will be the same as the angle subtended (at the objective) by the image produced by the objective so that we have
100/3000 = h/1.5 where ‘h’ is the height of the image produced by the objective. [Note that this image is at the focus of the objective].
From the above, h = 0.05m = 5 cm.
The magnifying power of the eye piece in normal adjustment is D/f_e = 25cm/5cm = 5.
The height of the final image = 5h = 25 cm.

Kerala Medical and Engineering Entrance Examinations- 2007

The Commissioner of Entrance Examinations (Govt. of Kerala) has invited applications for the Entrance Examinations for admission to the following Professional Degree courses in Kerala for 2007-08:
(a) Medical (i) MBBS (ii) BDS (iii) BPharm (iv) BSc (Nursing) (v) BSc (MLT) (vi) BAMS (vii) BHMS (viii) BSMS (Siddha) (ix) BSc–Nursing (Ayurveda) and (x) BPharm (Ayurveda) .
(b) Agriculture (i) BSc (Agriculture) (ii) BFSc (Fisheries) (iii) BSc (Forestry)
(c) Veterinary BVSc & AH
(d) Engineering B.Tech [including BTech (Agricultural Engg. / BTech (Dairy Sc. & Tech) courses under the Kerala Agr iculture University]
(e) Architecture B.Arch
Time Table for the Entrance Examinations:
The Entrance Examinations will be held in all the District Centres in Kerala, New Delhi and Dubai (UAE), on the dates mentioned below as per Indian Standard Time.
Engineering Entrance Examination (For Engineering courses except Architecture):
23-04-2007 Monday 10.00 A.M. to 12.30 PM Paper-I: Physics & Chemistry.
24-04-2007 Tuesday 10.00 A.M. to 12.30 PM Paper-II: Mathematics.
Medical Entrance Examination [For Medical (including B.Pharm), Agriculture and Veterinary Courses]:
25-04-2007 Wednesday 10.00 A.M. to 12.30 PM Paper-I: Chemistry & Physics.
26-04-2007 Thursday 10.00 A.M. to 12.30 PM Paper-II: Biology.
Appearance in the two papers of the concerned Entrance Examination is compulsory for being considered for inclusion in the Engineering/Medical rank lists (except Architecture)
Admission to the BPharm course will be based on a separate rank list prepared on the basis of the performance of candidates in the ‘Chemistry & Physics’ Paper of the Medical Entrance Examination. Those who wish to be considered for admission to the BPharm course should therefore appear for the Chemistry & Physics Paper of the Medical Entrance Examination.
Application Forms:
The application form and Prospectus will be sold from 07-02-2007 to 05-03-2007 through selected Canara Bank branches in Kerala and outside the State.
Cost of Application form: General candidates: Rs. 700/- ; SC/ST candidates : Rs. 350/- ( Candidates opting Dubai centre should enclose a bank draft for Rs 7000/- along with the application)
Last date and time for receipt of filled in Application Forms: The filled in Application Form along with the OMR DATA SHEET and other relevant documents to be submitted with the Application form is to be sent in the printed envelope bearing the address of the Commissioner for Entrance Examinations supplied along with the application form so as to reach him before 5 p.m. on 07.03 200 7, by Hand Delivery / Registered Post/ Speed Post.
Visit the site www.cee-kerala.org for complete details about the eligibility for applying for the examinations, sale centres of application forms, provision for downloading the application form in the case of certain categories etc. Visit the site to be informed of the latest developments in this regard.

New Template, New Look

After waiting for some time, I have switched over to the new version of Blogger today. The switch over was quite easy. I will have to spend some time on effecting a few changes in the lay out as well as the look of my blog. The priority will be in posting new questions and solution since examinations are fast approaching.
Thank you Blogger Staff for effecting the transition out of beta in a relatively short time span!

Multiple Choice Questions on Waves

The section on waves may appear to be somewhat boring and difficult to some of you. But you should not ignore this section because you will usually get a couple of questions from this. Remember the following important relations:
(1) Speed of transverse waves in a stretched string, v = √(T/m) where T is the tension and ‘m’ is the linear density (mass per unit length) of the string
(2) Frequency of vibration of a string, n = (1/2l)√(T/m) where ‘l’ is the length of the string.
Note that this is the frequency in the fundamental mode. Generally, the frequency is given by n = (s/2l)√(T/m) where s = 1,2,3,….etc. In the fundamental mode, s = 1.
(3) Speed of sound (v) in a medium is generally given by v = √(E/ρ) where E is the modulus of elasticity and ρ is the density of the medium.
(a) Newton-Laplace equation for the velocity (v) of sound in a gas:
v = √(γP/ρ) where γ is the ratio of specific heats and P is the pressure of the gas.
(b) Velocity of sound in a solid rod, v =√(Y/ρ) where Y is the Young’s modulus.
(4) Equation of a plane harmonic wave:
You will encounter the wave equation in various forms. For a progressive wave proceeding along the positive X-direction, the wave equation is
y = A sin [(2π/λ)(vt–x) +φ]
where A is the amplitude, λ is the wave length, v is the velocity (of the wave) and φ is the initial phase of the particle of the medium at the origin.
If the initial phase of the particle at the origin (φ) is taken as zero, the above equation has the following forms:
(i) y = A sin [(2π/λ)(vt–x)]
(ii) Since λ = vT and 2π/T = ω, where T is the period and ω is the angular frequency of the wave motion, y = A sin [(2π/T)(t – x/v)] and
(iii) y = A sin ω(t–x/v)
(iv) A common form of the wave equation (obtained from the above) is
y = A sin [2π(t/T – x/ λ)]
(v) Another form of the wave equation is y = A sin (ωt – kx), which is evident from the form shown at (iii), where k = ω/v = 2π/λ.
Note that unlike in the case of the equation of a simple harmonic motion, the wave equation contains ‘x’ in addition to ‘t’ since the equation basically shows the variation of the displacement ‘y’ of any particle of the medium with space and time.
It will be useful to remember that the velocity of the wave,

v = Coefficient of t /Coefficient of x
(6) Equation of a plane wave proceeding in the negative X-direction is
y = A sin [2π(t/T + x/ λ)] or
y = A sin [(2π/λ)(vt + x)] or
y = A sin ω(t + x/v) or
y = A sin (ωt + kx).
Note that the negative sign in the case of the equation for a wave proceeding along the positive X-direction is replaced with positive sign.
(7) Equation of a stationary wave is y = 2A cos(2πx/λ) sin(2πvt/λ) if the stationary wave is formed by the superposition of a wave with the same wave reflected at a free boundary of the medium (such as the free end of a string or the open end of a pipe).
If the reflection is at a rigid boundary (such as the fixed end of a string or the closed end of a pipe), the equation for the stationary wave formed is
y = – 2A sin(2πx/λ) cos(2πvt/λ).
Don’t worry about the negative sign and the inter change of the sine term and the cosine term. These occurred because of the phase change of π suffered due to the reflection at the rigid boundary.The important thing to note is that the amplitude has a space variation between the zero value (at nodes) and a maximum vlue 2A (at the anti nodes). Further, the distance between consecutive nodes or consecutive anti nodes is λ/2.
Now consider the following MCQ:
The equation, y = A sin [2π/λ(vt – x)] represent a plane progressive harmonic wave proceeding along the positive X-direction. The equation, y = A sin[2π/λ(x – vt)] represents
(a) a plane progressive harmonic wave proceeding along the negative X-direction (b) a plane progressive harmonic wave with a phase difference of π proceeding along the negative X-direction
(c) a plane progressive harmonic wave with a phase difference of π proceeding along the positive X-direction
(d) a periodic motion which is not necessarily a wave motion
(e) a similar wave generated by reflection at a rigid boundary.
The equation, y = A sin[2π/λ(x – vt)] can be written as y = – A sin [2π/λ(vt – x)] and hence it represents a plane progressive harmonic wave proceeding along the positive X-direction itself, but with a phase difference of π (indicated by the negative sign).
The following question appeared in IIT 1997 entrance test paper:
A traveling wave in a stretched string is described by the equation, y = A sin(kx– ωt). The maximum particle velocity is
(a) Aω (b) ω/k (c) dω/dk (d) x/t
This is a very simple question. The particle velocity is v = dy/dt = –Aω cos(kx– ωt) and its maximum value is Aω.
Consider the following MCQ:
The displacement y of a wave traveling in the X-direction is given by y = 10^–4 sin (600t–2x + π/3) metres. where x is expressed in metres and t in seconds. The speed of the wave motion in ms^–1 is
(a)200 (b) 300 (c) 600 (d) 1200
This MCQ appeared in AIEEE 2003 question paper. The wave equation given here contains an initial phase π/3 but that does not matter at all. You can compare this equation to one of the standard forms given at the beginning of this post and find out the speed ‘v’ of the wave. If you remember that velocity of the wave, v = Coefficient of t /Coefficient of x, you get the answer in notime: v = 600/2 =300 ms^–1.
Here is a question of the type often found in Medical and Engineering entrance test papers:
Velocity of sound in a diatomic gas is 330m/s. What is the r.m.s. speed of the molecules of the gas?
(a) 330m/s (b) 420m/s (c) 483m/s (d) 526m/s (e) 765m/s
At a temperature T, the velocity of sound in a gas is given by v = √(γP/ρ) where γ is the ratio of specific heats and P is the pressure of the gas. The r.m.s. velocity of the molecules of the gas is given by c = √(3P/ρ). Therefore, c/v = √(3/γ). For a diatomic gas, γ = 1.4 so that c/v = √(3/1.4). On substituting v=330m/s, c = 483m/s.
Now consider the following MCQ which appeared in EAMCET 1990 question paper:
If two waves of length 50 cm and 51 cm produced 12 beats per second, the velocity of sound is
(a) 360 m/s (b) 306 m/s (c) 331 m/s (d) 340 m/s
If n₁ and n₂ are the frequencies of the sound waves, n₁– n₂ =12 or, v/λ₁– v/λ₂ =12. Substituting the given wave lengths (0.5m and 0.51m), the velocity v works out to 306 m/s.
Here is a typical simple question on stationary waves:
A stationary wave is represented by the equation, y = 3 cos(πx/8) sin(15πt) where x and y are in cm and t is in seconds. The distance between the consecutive nodes is (in cm)
(a) 8 (b) 12 (c) 14 (d) 16 (e) 20
This stationary wave is in the form y = 2A cos(2πx/λ) sin(2πvt/λ) so that 2π/λ = π/8 from which λ = 16cm. The distance between consecutive nodes is λ/2 = 8cm.
You can find all posts on waves in this site by clicking on the label 'waves' below this post or on the side of this page.