Here are two multiple choice questions on thermodynamics which appeared in AIPMT 2009 question paper:
(i) In thermodynamic processes which of the following statements is not true?
(1) In an isochoric process pressure remains constant
(2) In an isothermal process the temperature remains constant
(3) In an adiabatic process PVγ = constant
(4) In an adiabatic process the system is insulated from the surroundings
The only statement which is not true is statement (1). A process in which the pressure remains constant is called isobaric process (and not isochoric process).
[Note that an isochoric process is one in which the volumeremains constant].
(ii) The internal energy change in a system that has absorbed 2 Kcals of heat and
done 500 J of work is:
(1) 6400 J
(2) 5400 J
(3) 7900 J
(4) 8900 J
When a quantity Q of heat is supplied to a system it is used to do an amount of work W by the system and to increase the internal energy of the system by ∆U:
Q = ∆U + W
Note that Q given in kilo calories is to be converted into joule which is the SI unit of heat. Therefore Q = 2×1000×4.2 J = 8400 J (remembering that 1 cal = 4.2 J very nearly).
Theincrease the internal energy, ∆U = Q – W =8400– 500 = 7900 J.
Find some more multiple choice questions (with solution) in this section hereas well ashere.
Today we will discuss the multiple choice questions on waves included in All India Pre-Medical / Pre-Dental Entrance Examination (AIPMT) 2009 and 2008 question papers. Here are the AIPMT 2009 questions:
(1) The electric field part of of an electromagnetic wave in a medium is represented by
Ex = 0;
Ey = 2.5 N/C cos[(2π×106 rad/s)t – (π×10–2 rad/m)x];
Ez = 0.
The wave is:
(1) moving along x-direction with frequency 106 Hz and wave length 100 m.
(2) moving along x-direction with frequency 106 Hz and wave length 200 m.
(3) moving along –x-direction with frequency 106 Hz and wave length 200 m.
(4) moving along y-direction with frequency 2π×106Hz and wave length 200 m.
The electric field variation is in accordance with the equation y = A sin (ωt – kx) which represents a progressive wave proceeding along the positive x-direction. Instead of the usual displacement y we have the electric field E. In place of the angular frequency ω we have 2π×106 (remember ω = 2πn) which means that the linear frequency n is 106 Hz.
Since the propagation constant k = 2π/λ where λ is the wave length, we have
π×10–2 = 2π/λ from which λ = 200 ms–1.
So the correct option is (2).
[The units of electric field E (N/C), angular frequency ω (rad/s) and the propagation constant k (rad/m) given in the wave equation in the question should not distract you.
You should note that the negative sign in the wave equation y = A sin (ωt – kx) or y = A sin (kx – ωt) indicates that the wave is propagatingalong the positive x-direction. A wave propagatingalong the negative x-direction is represented by y = A sin (ωt + kx).
You can make use of any other form of the wave equation, for instance, y = A sin [2π(t/T – x/ λ)], also to solve the problem].
(2) A wave in a string has an amplitude of 2 cm. The wave travels in the + ve direction of x axis with a speed of 128 m/sec. and it is noted that 5 complete waves fit in 4 m length of the string. The equation describing the wave is:
(1) y = (0.02) m sin (15.7x − 2010t)
(2) y = (0.02) m sin (15.7x + 2010t)
(3) y = (0.02) m sin (7.85x − 1005t)
(4) y = (0.02) m sin (7.85x + 1005t)
The unit of amplitude is metre, shown by its symbol ‘m’ in the equation (which should not distract you).
The equation describing the wave propagating in the +ve x-direction is
y = A sin (kx – ωt)
The amplitude A as given in the question is 2 cm = 0.02 m.
The velocity of the wave, v =ω/k = 128 ms–1.
Wave length λ = 4/5 m.
But k = 2π/λ = 2π×5/4 = 7.85 m and so ω = kv = 7.85×128 = 1005
So the equation of the wave is
y = (0.02) m sin (7.85x − 1005t)
The correct option is (3).
Here is the AIPMT 2008 question:
The wave described by y = 0.25 sin(10πx– 2πt) where x and y are in metres and t in seconds is a wave traveling along the
(1) negative x-direction with amplitude 0.25 m and wavelength λ= 0.2 m.
(2) negative x-direction with frequency 1 Hz.
(3) positive x-direction with frequency π Hz and wavelength λ= 0.2 m.
(4) positive x-direction with frequency 1 Hz and wavelength λ= 0.2 m.
The given wave equation is in the form y = A sin [2π(x/λ – t/T)].
The negative sign in the equation shows that the wave is propagating along the positive x-direction.
By comparison we obtain 2πx/λ = 10πx from which λ = 0.2 m. Also, 2πt/T =2πtfrom which the frequency 1/T = 1 Hz. [Option (4)].
The following three questions appeared in Karnataka CET 2006 question paper:
(1) When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. This is due to
(1) diffraction of signal received from the antenna
(2) interference of the direct signal received by the antenna with the weak signal reflected by the passing aircraft
(3) change of magnetic flux occurring due to the passing aircraft
(4) vibrations created by the passing aircraft.
The microwaves reaching the antenna directly and after reflection at the aircraft interfere. The interference pattern produced is unstable since the aircraft is moving. This results in a shaking picture [Option (2)].
(2) A beam of light of wave length 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between the first dark fringes on either side of the central bright fringe is
(1) 1.2 cm
(2) 1.2 mm
(3) 2.4 cm
(4) 2.4 mm
This is a popular question repeatedly asked in various entrance tests with changes in numerical values.
The diffraction bands are distributed symmetrically on either side of the central maximum and the angular separation of the centre of the first dark fringe from the centre of the central maximum is λ/a where λ is the wave length of light and a is width of the slit.
Therefore, the angular separation between the first dark fringes on either side of the central bright fringe is 2λ/a.
The distance (linear separation) between the first dark fringes on either side of the central bright fringe is D×2λ/a where D is the distance between the slit and the screen and we have
D×2λ/a = 2×(2×600×10–9/10–3) = 2.4×10–3 m = 2.4 mm.
(3) If white light is used in the Newton’s ring experiment, the colour observed in the reflected light is complementary to that observed in the transmitted light through the same point. This is due to
(1) 90º change of phase in one of the reflected waves
(2) 180º change of phase in one of the reflected waves
(3) 145º change of phase in one of the reflected waves
(4) 45º change of phase in one of the reflected waves
The mechanism of formation of Newton’s rings is the same as that of the production of colours in thin films and is shown in the figure. Usually the light beam falls normally on the air film included between the upper lens and the lower glass plate, but we have shown the rays slanting to make things clear.
Newton’s rings in the reflected system are formed by the interference of waves 1 and 2 (fig.) where as Newton’s rings in the transmitted system are formed by the interference of waves 3 and 4.
If the phase change of π introduced due to reflection at a denser medium is ignored, the difference between the path lengths of the rays 1 and 2 is the same as the difference between the path lengths of the rays 3 and 4.
In the case of waves 1 and 2 there is an additional path difference of λ/2 because of the phase differenceπ introduced due to reflection at B. In the case of waves 3 and 4 there is an additional path difference of λ because of the phase difference2π introduced due to reflections at B and C. If the condition for brightness is satisfied for one colour in the case of the interfering waves in the reflected system, the condition for darkness will be satisfied for the same colour in the case of the interfering waves in the transmitted system. So the colours present in the reflected system will be absent in the transmitted system and vice versa. The basic reason for this is the 180º phase change [Option (2)].
You will find a useful post in this section at physicsplus.
You will find many questions (with solution) involving magnetic fields on this blog, which can be accessed by trying a search for ‘magnetic field’ using the search box at the top of this page (or, you may click on the label ‘magnetic field’ below this post.
Today we will discuss two multiple choice questions on magnetic field produced by current carrying conductors:
(1) The adjoining figure shows a plane wire loop made of two semicircular portions of radii R1 and R2 and two straight portions. The wire loop contains a battery which drives a current I through the loop. What is the magnitude of the magnetic flux density at the common centre ‘O’ of the semicircular portions of this wire loop?
(a) (μ0I/2π)(1/R1 + 1/R2)
(b) (μ0I/4)(1/R1 + 1/R2)
(c) (μ0I/4)(1/R2 – 1/R1)
(d) (μ0I/4π)(1/R2– 1/R1)
(e) Zero
The magnetic field produced at O by the straight portions of the loop is zero. The semicircular portion of smaller radius R2 produces a magnetic field μ0I/4R2 which is directed normally into the plane of the figure (away from the reader).
[Note that a single turn plane circular coil of radius R produces a magnetic field μ0I/2R at its centre].
The semicircular portion of larger radius R1 produces a smaller magnetic field μ0I/4R1 which is directed normally outwards (towards the reader).
The resultant magnetic field at the centre O has magnitude (μ0I/4R2– μ0I/4R1)= (μ0I/4)(1/R2 – 1/R1), as given in option (c).
[The resultant field is directed normally into the plane of the figure (away from the reader). The above question can be modified to check your understanding of this fact also].
(2) A current I enters a circular coil of radius R, branches into two parts and then recombines as shown in the circuit diagram.
The resultant magnetic field at the centre of the coil is
(a) zero
(b) μ0I/2R
(c) (¾)(μ0I/2R)
(d) (¼ )(μ0I/2R)
(e) (½)(μ0I/2R)
This question appeared in KEAM 2009 (Engineering) question paper.
The magnetic field produced at the centre by the straight current leads is zero. So it is sufficient to consider the fields produced by the circular portions.
The currents through the branches are inversely proportional to the lengths of the branches while the magnetic fields for a given current are directly proportional to the lengths of the branches. The magnitudes of the fields due to the two branches are therefore equal. Since the fields due to the branches are perpendicular to the plane of the loop and oppositely directed, the net field at the centre is zero. This argument is enough for answering the above question.
If you want to make your argument more rigorous, you will proceed as follows:
Since the longer branch is made of three quarters of the circle and the shorter branch is made of one quarter of the circle, the resistance of the longer branch is 3 times that of the shorter branch. The currents through the shorter and longer branches are therefore given respectively by
I1 = 3I/4 and
I2 = I/4
The magnetic field produced at the centre by the shorter branch is directed normally into the plane of the coil (away from the reader) and has magnitude B1 given by
B1 = (¼)(μ0I1/2R) = 3μ0I/32R, on substituting for I1.
The magnetic field produced at the centre by the longer branch is directed normally outwards (towards the reader) and has magnitude B2 given by
B2 = (¾)(μ0I2/2R) = 3μ0I/32R, on substituting for I2.
The fields B1 and B2 get canceled and the resultant field at the centre is zero [Option (a)].
[Note that in all situations of branching of current at a circular coil, if the current leads are straight and point to the centre of the coil, the magnetic field at the centre will be zero (the angle shown need not necessarily be 90º)].
Now, find an interesting question in this section here.
(1) b + e = f
