Questions on Satellites

The following questions on satellites are simple, but slightly different from the common direct questions. See whether you can solve them yourself. Go through the solution given, only after your trial.
(1)A satellite is moving in an orbit of radius ‘r’ around the earth. A second satellite is moving in an orbit of radius (1.02)r. Then the orbital period of the second satellite is greater than that of the first by approximately
(a) 0.2% (b) 2% (c) 3% (d) 10.2% (e) 6%
We have T² α r³ (Kepler’s law) from which T α r^3/2. Since the increment in the orbital radius ‘r’ is 0.02r, the fractional increment in the orbital period ‘T’ is (∆T)/T = (3/2)(∆r)/r = (3/2) × 0.02r/r = 0.03. The percentage increase is 0.03×100 = 3% [option (c)].
(2) If the orbital period of a satellite is ‘T’, its kinetic energy is directly proportional to
(a) T (b) T^-1 (c) T^3/2 (d) T^2/3 (e) T^-2/3

We have, kinetic energy, E = GMm/2r where G is the gravitational constant, M is the mass of the earth ‘m’ is the mass of the satellite and ‘r’ is the radius of the orbit. Therefore, E α 1/r. Since T² α r³ in accordance with Kepler’s law, we can write r α T^2/3. Therefore, E α T^-2/3. The correct option is (e).

Questions on Bohr Atom Model

You will usually find a couple of questions pertaining to Bohr model of hydrogen atom in most entrance test papers. Questions in this section are interesting and often simple. Consider the following M.C.Q.:
Which state of the triply ionised Be+++ has the same orbital radius (for the electron) as that of the hydrogen atom in the ground state?
(a) First excited state (b) Second excited state (c) Third excited state (d) Fourth excited state (e) Ground state
The Be atom has lost 3 electrons since it is triply ionised and it behaves as a hydrogen like atom. Therefore, the radius of the orbit of quantum number ‘n’ is r n²/z where ‘r’ is the Bohr radius (radius of the electron orbit in the hydrogen atom in the ground state) and ‘z’ is the atomic number of the hydrogen like atom. Since z = 4 for Be, we have r = r n²/4 from which n = 2, which means the first excited state. So, the correct option is (a).
Now consider the following questions which appeared in the Kerala Engineering Entrance Test of 2002:
(1) The energy of an electron in excited hydrogen atom is -3.4 eV. Then, according to Bohr’s theory, the angular momentum of the electron in Js is
(a) 2.11×10^-34 (b) 3×10^-34(c) 3×10^-34 (d) 0.5×10^-34
Since the energy of the electron is -3.4 eV, the electron is in the second orbit. This is checked using the energy (in eV) expression E = -13.6/n² = -3.4 eV, when n = 2.
Since the angular momentum is nh/2π, on substituting for h (=6.63×10^-34 Js) and n (=2) we obtain the first option as the correct answer.
(2) Consider the spectral line resulting from the transition from n = 2 to n = 1, in atoms and ions given below. The shortest wave length is produced by
(a) hydrogen atom (b) deuterium atom (c) singly ionized helium (d) doubly ionized helium (e) doubly ionized lithium

The correct option is (e). The expression for the energy of an electron in a hydrogen like atom is E = (- 13.6 z²)/n². The energy for a given value of ‘n’ is maximum for lithium in the present case (since z = 3). The energy difference also is therefore maximum (for the given transition) in the case of lithium, giving rise to the shortest wavelength.

You will find more questions with solution at AP Physics Resources: AP Physics B– Multiple Choice Questions (for practice) on Atomic Physics and Quantum Effects

Questions on Electromagnetic Waves

When an electromagnetic wave is propagating through any medium including empty space, the direction of propagation of the wave is related to the directions of the electric and magnetic field vectors (E and B) in a definite manner. Many students are found to commit mistake in marking the directions of these vectors. If you remember that the vector product E×B always points along the direction of propagation of the electromagnetic wave, you wont commit mistake in specifying the directions of these vectors. Now consider the following M.C.Q.:
In a plane electromagnetic wave, the magnetic field vector is along the negative X-direction and the direction of propagation is along the positive Y-direction. Then the direction of the electric field vector must be along
(a) positive Z-direction (b) negative X-direction (c) negative Z-direction
(d) positive Y-direction (e) negative Y-direction
Since the direction of the E has to be perpendicular to both the direction of B and the direction of propagation, it has to be in the positive or negative Z-direction. Since there are two options [(a) and (c)] you have a 50-50 chance to hit the bull’s eye. But don’t take such chances. The magnetic field vector B is along the negative X-direction and so you can get the vector product E×B along the positive Y-direction only if E is along the negative Z-direction [option(c)].
You might have noted that our eyes are most sensitive to light of wave length about 550 nm because the sun radiates maximum energy in the form of light at this wavelength. This wavelength is related to the surface temperature T of the sun in accordance with Wein’s law which states that λm T = constant where λm is the wave length at which maximum energy is radiated. If you express λm in cm and T in Kelvin the constant is 0.29 cmK. Now consider the following question:
If the temperature of the sun were twice the present value, the radiation of the sun would be mostly in
(a) microwave region (b) infra red region (c) visible region (d) X-ray region (e) ultraviolet region.
The correct option is (e) because the wave length of maximum energy radiation has to be reduced to half the present value when the temperature is doubled (in accordance with Wien’s law). The present value of λm is about 550 nm and so when the temperature is doubled, the value of λm should become 275 nm, which is in the ultra violet region.

Questions on Elasticity

Here is a question involving Young’s modulus. It’s a simple question, but there is a good chance of committing a mistake if you are not careful enough.
A wooden plank, used as a bridge over a canal sags by 10mm due to its own weight. If the thickness of the wooden plank is doubled, the sag will be
(a) 5mm (b) 2.5mm (c) 1.25mm (d) 10mm (e) 20mm
The expression for the depression (sag) ‘δ’ of a centrally loaded beam is,
δ = (WL³)/4Ybd³ where W is the load (W=mg), L is the length of the beam, Y is the Young’s modulus, ‘b’ is the breadth of the beam and ‘d’ is its thickness(depth).
In the above problem, no separate load is suspended from the centre of the beam, but the weight of the beam acts through its centre to depress it. When a beam of twice the thickness is used, its weight appearing in the numerator is doubled and the value of d³ appearing in the denominator becomes 8 times. Therefore, the depression is one-fourth, equal to 2.5 mm [option (b)].
Let us consider another question which also may baffle you slightly:
A thick rope of length L is hanging from the ceiling of a room. If Y and ρ are respectively the Young’s modulus and density of its material, its increase in length due to its own weight is
(a) (ρgL²)/2Y (b) (ρgL²)/Y (c) (ρgL)/Y (d) Y/ρgL² (e) data insufficient.
Since the Young’s modulus is given by the usual expression,Y = FL/Al where L is the original length, ‘l’ is the increase in length, F is the elongating force and A is the area of cross section, we have,
l = FL/YA. Here F = ALρg, which is the weight of the rope. You have to substitute the length L as it is, in the expression for F. But the weight of the rope acts through its centre of gravity and hence you have to substitute L/2 instead of L in the expression for the increase in length. The answer therefore is (ρgL²)/2Y [option (a)].

Friction Moves the Car and Friction Stops the Car

It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong.

– Richard Feynman

 
When you think of friction in detail you will find that it is an ordinary force with extraordinary features. Its direction is always self adjusting and is opposite to the direction of the force which tries to move a body. The magnitude of frictional force is self-adjusting up to the limiting value.
Have you ever thought how exactly a car moves forward? The engine of the car does not exert any forward force. It just spins the wheels of the car. The portion of the wheel (tyre) in contact with the ground moves backwards. The frictional force between the wheel and the ground is therefore directed forwards. It is this frictional force which moves the car forward.
When you apply the brakes, the braking mechanism does not apply any backward force. It just stops the spin of the wheels. The forward frictional force ceases. But the car will continue to move forward due to inertia. The wheel will then slide forward. Immediately, a backward frictional force is called into play and the car stops after traveling some distance. Now, consider the following questions:(1) If μ is the coefficient of friction between the road and the tyre of a car, the minimum time in which the car can cover a distance ‘s’, starting from rest is
(a) directly proportional to μ
(b) inversely proportional to μ
(c) inversely proportional to √μ
(d) directly proportional to √μ (e) independent of μThe correct option is not (b) but is (c). We have s = 0 + ½ at². Since the driving force is the frictional force μmg, acceleration a = μg. . Therefore s = ½ μg t² and hence t α 1/√μ. The correct option is (c).(2) If μ is the coefficient of friction between the road and the tyre of a car moving with a velocity ‘v’, the minimum distance in which it can be stopped is
(a) directly proportional to μ
(b) inversely proportional to μ
(c) inversely proportional to √μ
(d) directly proportional to √μ
(e) independent of μThe correct option is (b) which you can obtain using the equation, 0 = v² – 2as, on substituting for a = μg. Thus , stopping distance, s = v²/2μg.
Note this equation. It says that the stopping distance of any vehicle is directly proportional to the square of the speed of the vehicle.(3) A cube of mass 2kg is kept pressed against a vertical wall by applying a horizontal force of 60N. The coefficient of friction between the cube and the wall is 0.5. What is the frictional force between the cube and the wall?
(a) 60N
(b) 30N
(c) 19.6N
(d) 9.8N
(e) zeroDon’t be tempted to chose option (b) as the answer. Option (b) gives the maximum possible frictional force μR, where R is the normal reaction. Here friction is just sufficient to balance the weight of the cube (mg). So the correct option is (c).Certain simple questions may confuse you occasionally. Consider the following question:A wooden block of mass 3kg is resting on an inclined plane of inclination 30˚. If the coefficient of static friction between the block and the plane is 0.8, the frictional force between the block and the plane is (g=10m/s²)
(a) 20.8 N
(b) 24N
(c) 30N
(d) 12.5N
(e) 15NYou may be inclined to calculate the force as μmg cosθ, which yields the value20.8N, which is not the correct answer. Remember, μmg cosθ is the maximum possible frictional force (limiting friction). Since the block is resting on the inclined plane, the frictional force required is just enough to balance the component of the weight along the plane, which is mg sinθ = 15N. The correct option therefore is (e). Since frictional force is self adjusting (up to the limiting value), the value gets automatically adjusted to 15N in the present case.
Now consider the following MCQ:A heavy uniform chain is lying on a horizontal table with a certain portion hanging over one edge. If the coefficient of static friction between the chain and the table is 0.25, what is the maximum fraction of the length of the chain that can hang over the edge of the table?
(a) 10% (b) 20% (c) 33% (d) 45% (e) 50%If ‘L’ is the full length of the chain and ‘l’ the maximum length that can overhang, we have, mlg = μm(L-l)g where m is the mass per unit length (linear density) of the chain. The weight of the hanging portion tries to displace the chain while the frictional force between the table and the remaining portion tries to restore the chain. We have equated their magnitudes in the limiting case. From this we get l = 0.25(L-l) so that l/L = 0.2. The fraction written as percentage is 20%.An insect inside a hemispherical bowlHere is an interesting problem:An insect is trapped in a hemispherical bowl of radius R kept on a horizontal table. Up to what height can it crawl if the angle of friction is θ?
(a) Rcosθ
(b) Rsinθ
(c) R(1-cosθ)
(d) R(1-sinθ)
(e) RtanθAs the insect crawls upwards, the component of its weight along the surface of the bowl increases and the opposing frictional force also increases. When the insect reaches the maximum possible height, the frictional force is maximum and is equal to mg sinθ where θ is the angle of friction. This means that the inclination (with the horizontal) of the tangent to the hemispherical surface at the limiting position of the insect is equal to the angle of friction,θ. The angle between the vertical radius (of the hemispherical surface) and the radius drawn from the limiting position of the insect also is equal to θ, as ascertained from simple geometry.
Therefore, the maximum height attained by the insect is R-Rcosθ = R(1-cosθ).

Millikan’s oil drop experiment

Millikan’s famous oil drop experiment which brought him a Nobel Prize in Physics will always shine brightly in the memory of any one who loves Physics. Let us consider the following M.C.Q. related to Millikan’s experiment:
In an oil drop experiment (Millikan’s), an oil drop carrying a charge Q is held stationary between the plates by applying a potential difference of 400V. To keep another drop of half the radius stationary, the potential difference had to be increased to 600V. The charge on the second drop is
(a) Q/24
(b) Q/12
(c) Q/6
(d) 3Q/2
(e) 2Q/3
In the case of the first drop, we have
w = (400/d)×Q ……….. (1)
where ‘w’ is the apparent weight of the first drop and ‘d’ is the separation between the plates.
In the case of the second drop, the apparent weight is w/8 since the volume (and hence the mass) of the drop is reduced to one-eighth 
[Note that the volume is directly proportional to the cube of the radius]
Therefore we have,
w/8 = (600/d)×q ………... (2)
where ‘q’ is the charge on the second drop. Dividing eq(1) by eq(2), we obtain q = Q/12.
Consider now the following simple question which is meant for high lighting the quantum nature of electric charge:
An experimenter obtained the following values for the charge (in coulomb) on five different drops in Millikan’s oil drop experiment. Which one is the most unlikely value?
(a) 3.2×10^-19
(b) 4.8×10^-18
(c) 1.6×10^-17
(d) 8×10^-18
(e) 5.6×10^-19.
Remember that the minimum electric charge in nature is the electronic charge, which is equal to 1.6×10^-19 coulomb. All charges will be integral multiples of this minimum value. So, the unlikely value is 5.6×10^-19 coulomb [option (e)].
The number of electrons passing per second through any section of a conductor or a region of space to produce a given current is a constant and is independent of the voltage. 
Consider the following M.C.Q. which appeared in the I.I.T. screening test paper of 2002:
The potential difference applied to an X-ray tube is 15kV and the current through it is 3.2mA. Then the number of electrons striking the target per second is
(a) 2×10¹⁶
(b) 5×10⁶
(c) 1×10¹⁷
(d) 4×10¹⁵
As the current is 3.2mA, the charge reaching the target per second is 3.2 millicoulomb. Since the electronic charge is 1.6×10^-19 coulomb, the number of electrons striking the target per second is (3.2×10^-3)/(1.6×10^-19) = 2×10¹⁶.
The accelerating voltage of 15kV is just a distraction in the question.

Questions on Surface Tension

God used beautiful mathematics in creating the world.

– P. A. M. Dirac

 
You might be remembering the expression for the excess of pressure ‘p’ inside a liquid drop: p = 2T/r where ‘T’ is the surface tension of the liquid and ‘r’ is the radius of the drop. The excess pressure is due to surface tension by which the free surface of a liquid behaves like a stretched membrane. The excess of pressure inside a bubble is 4T/r, which is twice that inside a drop because a bubble has two free surfaces (inner and outer).
Consider the following question which often finds a place in Medical and Engineering Entrance test papers:
Two soap bubbles A and B are obtained at the ends of a glass tube with a stop cock at its middle. The stop cock is closed initially and the bubble B is larger in size. If the stop cock is opened what happens to the bubbles?
(a) Nothing
(b) The size of A decreases and the size of B increases
(c) The size of A increases and the size of B decreases
(d) Both A and B will decrease in size
(e) Both A and B will increase in size.The excess pressure inside the smaller bubble A is greater and hence air from it will rush to B through the tube. The size of A will decrease further while the size of B will increase. The correct option therefore is (b).
You should note that the pressure on the concave side of a liquid surface is always greater by 2T/r. Consider the following question:
A tall metallic jar has a small hole of radius 0.07mm at its bottom. What is the approximate depth up to which the jar can be lowered vertically in water before any water penetrates in to it through the hole? (Surface tension of water = 0.073N/m)
(a) 7cm
(b) 11cm
(c) 16cm
(d) 21cm
(e) 27cmWater will begin to enter the jar through the hole when the hydrostatic pressure exceeds the allowed excess pressure of 2T/r on the concave side of the water surface at the hole. Therefore, the depth ‘h’ up to which the jar can be lowered safely is given by
hdg = 2T/r so that h = 2T/dgr = 2×0.073/(1000×9.8×0.00007) = 0.21m =21cm.

Let us consider one more question involving surface tension:
If T is the surface tension of soap solution, the amount of work done in increasing the radius of a soap bubble from r to 2r is
(a) (64πr²)T
(b) (32πr²)T
(c) (24πr²)T
(d) (16πr²)T
(e) (8πr²)T

Work done = Increase in surface area ×T = 2 [4π(2r)² - 4πr² )]T = (24πr²)T. So, the correct option is (c). Note that the bubble has two surfaces.

The following MCQ appeared in Kerala Engineering Entrance 2006 test paper:Water rises in a capillary tube to a height ‘h’. Choose FALSE statement regarding capillary rise from the following:
(a) On the surface of Jupiter, height will be less than h.
(b) In a lift moving up with constant acceleration, height is less than h.
(c) On the surface of moon, height is more than h.
(d) In a lift moving down with constant acceleration, height is less than h.
(e) At the poles, height is less than that at equatorThe expression for surface tension (T) in terms of capillary rise in a capillary tube is
T = hrdg/2cosθ where ‘h’ is the capillary rise (or fall), ‘r’ is the radius of the tube, ‘d’ is the density of the liquid, ‘g’ is the acceleration due to gravity and ‘θ’ is the angle of contact. When ‘g’ is less, the capillary rise ‘h’ is more and vice versa. The value of ‘g’ is smaller in options (c) and (d) only so that the value of ‘h’ is greater in these two cases. So, option (d) is false. [Note that when a lift moves down with acceleration ‘a’, the effective value of acceleration due to gravity is (g-a)]

Introductory Remarks

This site is intended to deal with topics in Physics for helping students aspiring to become Engineering and Medical Professionals. I will discuss typical multiple choice questions (M.C.Q.) of the type you can expect to find in Medical and Engineering Entrance test papers of various institutions. Depending on the feedback from the viewers, other types of questions also will be included in due course. As Physics is not limited by regional boundaries, you will find that the topics discussed will be useful to you irrespective of your geographical location. An important case in point is the usefulness of the posts for students preparing for Graduate Record Examinations (GRE) and AP Physics B and C Examinations
The minimum I expect from the users of this site is some genuine interest in the subject. If you remember that the scientific and technological progress you find all around today is, to a very large extent, thanks to the hard work of physicists, you will find pleasure in learning the beautiful principles of Physics with enthusiasm. Finding answers to questions in Physics will then become a pleasing task for you. Be confident. Don’t be under the impression that Physics is a tough subject. It is a very interesting subject. You can definitely score high marks in Physics if you show genuine interest in it.
Many among you might be interested more in Physics than in Medicine or Engineering, but you may be forced to take up these professional courses by the pressure of circumstances. Don’t be discouraged. You will definitely find scope for applying the principles of Physics in almost all situations where you might be destined to work.

Your comments on the posts will be most welcome. Please do not hesitate to post crtical comments and suggestions