Questions on Gravitation

You should definitely remember the following relations to ensure good score in gravitation:
(1) Acceleration due to gravity at a height ‘h’ is given by

g’ = GM/(R+h)², with usual notations.
Surface value of acceleration due to gravity, g = GM/R²
If ‘h’ is small compared to the radius ‘R’ of the earth, g' = g(1-2h/R)
(2) Acceleration due to gravity at a depth ‘d’ is given by g'' = g (1-d/R)
Note that this is true for all values of ‘d’.
(3) Gravitational potential energy of a mass ‘m’ at a height ‘h’ is given by U= -GMm/(R+h)
This can be written as U = -GMm/r where ‘r’ is the distance from the centre of the earth.

(4) Escape velocity from the surface of earth (or any planet or star),

v_e = √(2GM/R) = √(2gR)
Escape velocity from a height ‘h’ = √[2GM/(R+h)] = √[2g'(R+h]
(5) Kinetic energy and total energy of a satellite are equal in magnitude. But K.E. is positive where as total energy is negative. The potential energy of a satellite is negative and is equal to twice the total energy.( Note that this is true in all central field motion under inverse square law force, as for example, the energy of the electron in the hydrogen atom.)
In the case of a satellite of mass ‘m’ in an orbit of radius ‘r’:
Potential energy = -GMm/r
Kinetic energy = +GMm/2r
Total energy = -GMm/2r

(6) As per Kepler’s law, T² α r³
(7) Orbital speed ‘v’ of a satellite in an orbit of radius ‘r’ is independent of its mass and is given by v = √(GM/r) = √(g'r) where g' is the acceleration due to gravity at the orbit and M is the mass of the earth (or planet).
Let us now discuss the following question which appeared in the Kerala Medical Entrance Test paper of 2002:
The escape velocity of a body on an imaginary planet which has thrice the radius of the earth and twice the mass of the earth is (where v_e is the escape velocity on the earth)
(a) √(2/3).v_e (b) √(3/2).v_e (c) √(2).v_e/3 (d) 2v_e/√3 (e) 2v_e/3
We have v_e = √(2GM/R). Replacing R with 3R and M with 2M, we obtain the answer as√(2/3).ve [option (a)].
Consider now the following question which may confuse some of you:
The orbital velocity of an artificial satellite near the surface of the moon is increased by 41.4%. The satellite will
(a) move in an orbit of radius greater by 41.4% (b) move in an orbit of radius twice the original value (c) move in an elliptical orbit (d) fall down (e) escape into outer space
The correct option is (e). The orbital speed of any satellite moving round any heavenly body is √(GM/r) where as the escape velocity is √(2GM/r). This means that the escape velocity is √2 times the orbital speed or 1.414 times the orbital speed. Therefore, when the orbital speed is increased by 41.4% the satellite will escape into outer space.
Consider now the question which appeared in the Kerala Engineering Entrance Test paper of 2001:
The orbital speed of an artificial satellite very close to the surface of the earth is V0. Then the orbital speed of another artificial satellite at a height equal to 3 times the radius of the earth is
(a) 4V0 (b) 2V0 (c) V0 (d) 0.5V0 (e) 2V0/3
We have V0 =√(GM/R). At a height equal to three times the radius of the earth, the orbital velocity is obtained by replacing R with R+3R = 4R. The answer is 0.5R [option (d)].
The following simple question appeared in the IIT 2001 test paper:
A simple pendulum has a time period T1 when on earth’s surface, and T2 when taken to a height R above the earth's surface, where R is the radius of the earth. The value of T2/T1 is
(a) 1 (b) √2 (c) 4 (d) 2
The required ratio is [2π√(L/g’)] / [2π√(L/g)] = √(g/g’). But g = GM/R^2 and g’ = GM/(2R)^2 so that g/g' = 4. The answer therefore is 2 [option(d)].

Consider now the following question which appeared in H.P.P.M.T.2005:
If a body of mass ‘m’ is raised from the surface of the earth to a height ‘h’ which is comparable to the radius of the earth R, the work done is
(a) mgh (b) mgh[1-(h/R)] (c) mgh[1+(h/R)] (d) mgh/[1+(h/R)]
Note that ‘g’ is the acceleration due to gravity on the surface of the earth. The work done for raising the body is the difference between the gravitational potential energies at the height ‘h’ and at the surface. Therefore, work done, W = -GMm/(R+h) – (-GMm/R) where M is the mass of the earth. Therefore, W = GMm/R - GMm/(R+h) = mgR – mgR/[1+(h/R)], on substituting g=GM/R².
Thus, W = mgR[1- 1/1+(h/R)] = mgh/[1+(h/R)]

Questions from Optics- Cartesian Sign Convention

The sign convention adopted widely in Optics is the Cartesian convention. The ray incident on the curved surface is to be considered as proceeding in the positive X-direction and you have to measure all distances from the pole, which is supposed to be the origin.
Occasionally, you may be given a ray diagram in which the incident ray may be proceeding from right to left. To avoid confusion, imagine that the direction of the incident ray is still in the positive X-direction. You can even redraw the diagram to make the incident ray proceed from left to right if you want. The signs given to the distances are as in the Cartesian coordinate system: pole to right positive and pole to left negative. Distances measured (from pole) upwards are positive and those measured downwards are negative but you will mostly encounter problems with leftward and rightward measurements.
While solving problems, you should apply the signs to all known quantities. The unknown quantities are left as they are in the formulae. You will be able to arrive at conclusions by interpreting the sign of the unknown quantity you finally arrive at as the answer. For instance, if the distance of an image is obtained as negative, you will immediately understand that the image is on the same side of the curved surface as the object is.
Let us consider the following M.C.Q.:
A convex lens made of crown glass of refractive index 1.5 has focal length 20 cm. What will be its focal length within water (refractive index 4/3)?
(a) 10 cm (b) 20 cm (c) 40 cm (d) 60 cm (e) 80 cm
Writing lens maker’s equation 1/f = [(μ₂/μ₁) -1] [(1/R₁) - (1/R₂)] for the two cases, we have,
1/20 = [1.5-1] [(1/R₁) - (1/R₂)] for the first case, where f = +20cm (for convex lens), μ₂ = 1.5 and μ₁ = 1 (for air) and
1/f = [1.125 – 1] [(1/R₁)-(1/R₂)] for the second case where μ₂= 1.5 and μ₁ = 4/3 (for water).
Note that we have ignored the signs of the unknown quantities f, R₁ and R₂. On dividing the first equation by the second, we obtain f = 80 cm[option (e)].
The result shows that within water, the lens is still a converging lens. If the liquid has refractive index more than that of the lens, the lens will become diverging within the liquid. Let us consider such a case:
A converging lens of focal length 20cm is made of glass of refractive index 1.5. How will this lens behave if it is immersed in a liquid of refractive index 1.6?
(a) As a converging lens of focal length 80cm (b) As a diverging lens of focal length 80cm (c) As a diverging lens of focal length 120cm (d) As a diverging lens of focal length 160cm (e) As a converging lens of focal length 120cm
The correct option to this question is (d). You can check this by writing the lens maker’s equation for the two cases as in the previous question. The first equation is unchanged: 1/20 = [1.5-1] [(1/R) - (1/R₂)]. The second equation is
1/f = [0.9375 – 1] [(1/R₁) - (1/R₂)] since μ₂/μ₁) = 1.5/1.6 = 0.9375.
Dividing the first equation by the second, we obtain f = -160cm. As we get a negative value for the focal length, the lens behaves as a diverging lens within the given liquid [option (d)].
Let us discuss one more question:
The plane face of a planoconvex lens is silvered. If the refractive index of the material of the lens is μ, and the radius of curvature of its curved face is R, then the system behaves like a concave mirror of radius of curvature
(a) Rμ (b) R/μ (c) R/ (μ-1) (d) R/2(μ-1) (e) R(μ-1)
When one surface of a lens is silvered, the rays of light entering through the un-silvered surface are refracted first, reflected by the silvered surface next and finally are once more refracted. The effective focal length (F) of the system is given by
1/F = 1/f + 1/fm + 1/f where ‘f’ is the focal length of the un-silvered lens and fm is the focal length of the silvered surface which makes a mirror. In the present problem since the silvered surface is plane, fm is infinity so that the above equation becomes
1/F = 1/f + 0 + 1/f = 2/f. Therefore, F = f/2.
But we have for the original un-silvered lens, 1/f = (μ – 1) [(1/R₁) - (1/R₂)] where R₁ = R and R₂ = infinity. Therefore, f = R/(μ – 1)
Since F = f/2, the system has a focal length of R/2(μ – 1).The radius of curvature of a concave mirror is twice its focal length. Therefore, the system behaves like a concave mirror of radius of curvature R/(μ – 1). [Option (c)]. You will find multiple choice questions (with solution) on refraction at plane surfaces here as well as here

Questions on Rotational motion

In most of the Medical and Engineering Entrance test papers you will encounter at least a couple of questions on rotational motion. Let us consider the following questions:

(1) A solid cylinder initially at rest rolls down an inclined plane of angle θ and height ‘h’ without slipping. The linear velocity with which it will reach the bottom of the plane is

(a) √(3gh/4) (b) √(4gh/3) (c) √(4ghsinθ/3) (d) √(5gh/4) (e) √(2ghsinθ/3)

If ‘v’ and ‘ω’ are the linear and angular velocities respectively, ½Mv² + ½Iω² = Mgh where I is the moment of inertia and M is the mass of the cylinder. Substituting I = ½ MR² and ω = v/R , we get v = √(4gh/3). Note that the angle θ mentioned in the problem is just a distraction.

(2) A simple pendulum bob of mass ‘m’ is drawn to one side so that the string is horizontal. It is then let free. When it crosses the mean position, the tension in the string is

(a) mg (b) 0.5mg (c) 1.5mg (d ) 3mg (e) 5mg

In problems of this type, you should note that the centripetal force (which is equal to mv²/r) acting on the body executing circular motion is the net force on the body. Therefore, centripetal force, mv²/r = T – mg, where r is the length of the pendulum and T is the tension in the string.Therefore, T= mv²/r + mg. But, ½ mv² = mgr, on equating the initial gravitational potential energy of the bob to its kinetic energy in the mean position. From this, mv²/r = 2mg and hence T = 2mg + mg = 3mg.

Let us consider the following two questions involving the rotation of the earth:
(1) If the radius of the earth is changed to 1/√3 times the present value, the duration of the day (in hours) will be
(a) 72 (b) 41.6 (c) 24 (d) 12 (e) 8
This question is set to check your understanding of the law of conservation of angular momentum: I₁ω₁= I₂ω₂ where I₁and I₂ are the moments of inertia and ω₁ and ω₂ are the angular velocities of the earth before and after the contraction respectively. Substituting for I₁ (= 2MR²/5) and I₂ [= 2M (R²/3)/5] we obtain ω₂ = 3ω₁. Since the angular velocity changes to 3 times the initial value, the spin period of the earth (T= 2π/ω) changes two one-third of the initial value. So, the duration of the day will become 24/3 = 8 hours.
Questions of this type are often found in Medical and Engineering Entrance Test papers. Generally, if the radius of the earth becomes ‘n’ times the present value, the duration of the day becomes 24n² hours. Remember this equation and write the answer in no time!
(2) If your weight while standing on the earth’s surface at the equator is to become zero, the earth should spin at nearly ------ times the present speed.
(a) 12 (b) 14 (c) 17 (d) 24 (e) 37
This is a simple question. If you are to become weightless due to the spin of the earth, the gravitational pull on you is to be balanced by the centrifugal force so that, mg = mRω². From this ω=√(g/R) and the spin period T=2π/ω=2π√(R/g). On substituting for R = 6400 km and g = 9.8ms^-2, the period works out to be 84.6 minutes. This is one-seventeenth the present period of 24 hours. So, the earth should spin at 17 times the present speed. [Option (c)].

Two Questions from Electrodynamics

(1) A straight conductor carries a current of 10A. An electron traveling with a speed of 2.5×10^6m/s parallel to the conductor at a distance of 1cm from the conductor experiences a force of
(a) 8×10^-17N (b) 8×10^-18N (c) 8×10^-19N (d) 4×10^-20N (e) 4×10^-18N
Force on the electron, F=qvB, where B= μoI/2πr with usual notations.
Note that the direction of the magnetic field produced by the conductor is perpendicular to the direction of motion of the electron since the magnetic field lines are in the form of concentric circles.
Therefore, F= (1.6×10^-19×2.5×10^6×4π×10^-7×10) / 2π×0.01 = 8×10^-17N.
(2) A galvanometer having resistance 3Ω is converted into an ammeter by connecting a 3Ω shunt resistance across it. In order to double the range of this ammeter, the additional shunt resistance to be connected across it is
(a) 3Ω (b) 2.5Ω (c) 2Ω (d) 1.5Ω (e) 6Ω
With usual notations, Ig G = (I-Ig)S. Since G = S =3Ω, Ig = I/2. If S′ is the additional shunt required, the effective shunt for doubling the range is SS′/(S+S′) . Therefore, when the range is doubled, IgG = (2I-Ig) SS′/(S+S′) . Substituting G=S=3 and Ig= I/2 , 3S'/(3+S') =1. Therefore,S'=1.5Ω.
Instead of following the above mathematical steps, you can easily arrive at the answer as follows:
The ammeter with the original shunt has an effective resistance of 1.5Ω. If you want to pass twice the maximum allowed current into the ammeter, half of this doubled current has to be diverted through another path. So you have to provide a shunt path of resistance 1.5Ω.

An Interesting Post on Friction

Here is a useful link which will take you to an interesting post on friction:
physicsplus: Friction Moves the Car and Friction Stops the Car

Questions on Simple Pendulum

The period of oscillation of a simple pendulum is given by the simple equation,
T = 2π√(L/g).
Questions based on this equation can be seen in test papers. The following question appeared in the I.I.T. screening test of 2005:

The point of suspension of a simple pendulum with normal time period T₁ is moving upward according to the equation, y=kt² where k=1 m/s². If the new time period is T, the ratio T₁²/ T² will be

(a) 2/3        (b) 5/6        (c) 6/5        (d) 3/2

             This is a simple question. But you should note that the acceleration due to gravity ‘g’ is to be replaced by the net acceleration (g+a) since the pendulum as a whole is moving up with an acceleration ‘a’ which is obtained by differentiating the equation y = kt² twice. Therefore, a = 2k = 2 since k=1. The new period is given by,

            T = 2π√[L/(g+a)] = 2π√[L/(10+2)] = 2π√(L/12).

The normal period of the pendulum is

            T1 = 2π√(L/10).

Therefore, T₁²/ T² =12/10 = 6/5  [Option (c)]

Let us consider another question in which en electric force modifies the effective weight of the bob of the pendulum, thereby changing the period of oscillation:
A simple pendulum of length ‘L’ has a small spherical bob of mass ‘m’ that carries a positive charge ‘q’. The pendulum is located in a uniform electric field ‘E’ directed vertically upwards. If the electric force is less than the gravitational force, the period of oscillation of this pendulum is
(a) 2π√(L/g) (b) 2π√[L/(g-E)] (c) 2π√[L/(g+Eq/m)] (d) 2π√[L/(g-Eq/m)]
(e) 2π√[L/(g+E)]
Here also you have to replace ‘g’ (in the expression for the period) by the net acceleration, as in the previous question. But the net acceleration in the present case is (g-a) where a = Eq/m, which is the acceleration produced by the electric force Eq. Therefore ‘g’ is to be replaced by (g-Eq/m). The correct option therefore is (d).
Note that the real weight of the bob is mg. The apparent weight of the bob is (mg-Eq) since the electric force is upwards. The net downward acceleration therefore is (g-Eq/m).
A simple pendulum will not work on an artificial satellite orbiting round the earth since the pendulum bob becomes weightless and hence there is no restoring force mgsinθ. But you can use a spring loaded with a mass as an oscillator even on an artificial satellite or for that matter, even in a region of space where there is no gravitational force. The period of oscillation of such a spring-mass system, as you might be remembering is
T = 2π√(m/k), where m is the mass and k is the spring constant.
This equation is devoid of g and so the system works even in weightless situations.
Now suppose that the bob of a simple pendulum of length ‘L’is immersed in a non-viscous liquid of density equal to one-tenth the density of the material of the bob. The apparent weight of the bob is now reduced to nine-tenth of the real weight(because of the upthrust of the liquid). The period of oscillation of the pendulum therefore increases as
T = 2π√(10L/9g)
[The above equation is easily obtained if you remember that the mass of the bob is vρ and the apparent weight of the bob is v(ρ-σ)g so that the net value of the downward acceleration of the bob is v(ρ-σ)g /vρ = (1- σ/ρ)g = (1- 1/10)g = (9/10)g.]
In the case of a spring-mass system, there is no change in the period if the oscillating mass is immersed in a non-viscous liquid, since the period is independent of ‘g’.

Simple Pendulum of Infinite Length
The period of oscillation of a simple pendulum, as you know, is given by

T=2π√(l/g), with usual notations. If the length of the pendulum is not negligible compared to the radius(R) of the earth, the period is given by
T = 2π√[Rl/(R+l)g]. This equation shows that in the case of a simple pendulum of infinite length(or, to be more realistic, in the case where the length is large compared to the radius of the earth), the period is
T = 2π√(R/g)
On substituting for R = 6400 km (=64×10^5m) and g = 9.8m/s^2, the period works out to be 5078seconds or, 84.6 minutes.
Remember the above equation. The period of oscillation of a stone dropped into an imaginary hole drilled along the diameter of the earth and the orbital period of a satellite moving close to the earth’s surface also are given by this equation.

Elastic Potential Energy

When you elongate or contract a rod or wire by exerting a force, you do work. This work is stored as elastic potential energy in the rod or wire. The elastic potential energy per unit volume is easy to remember and is equal to ½ stress×strain. Since the Young’s modulus Y = stress/strain, we can modify the above expression as ½ Y×strain². Another form of the same expression is ½ (stress)²/Y.

Consider the following M.C.Q. which appeared in the Kerala Engineering Entrance test paper of 2006:

A work of 2×10^-2 J is done on a wire of length 50cm and area of cross section 0.5mm². If the Young’s modulus of the material of the wire is 2×10¹⁰ N/m², then the wire must be

(a) elongated to 50.1414cm (b) contracted by 2mm (c) stretched by 0.707mm (d) of length changed to 49.293cm (e) of length changed to 50.2cm

Since the total work done is involved in this prblem, we write,

½ (Y×strain²) ×volume = 2×10^-2

Here, strain = δ/0.5 where ‘δ’ is the elongation (or contraction) and volume = AL = (0.5×10^-6) ×0.5

Substituting proper values, δ works out to 1.414×10^-3 m = 1.414mm = 0.1414cm. The wire therefore elongates to 50.1414cm [option(a)].

An expression you should remember for obtaining the total work done in stretching or contracting a rod or wire through ‘l’ by exerting a force ‘F’ is, W = ½ F × l.

Breaking Strength and Breaking Stress

The breaking strength of a given wire depends on its area of cross section where as the breaking stress is a constant for a given material. Consider the following M.C.Q.:
A cable of length 10m and diameter 2cm can support a maximum load of 800 kg. If the length and the diameter of the cable are reduced to 5m and 1cm respectively, it will be able to support a maximum load of
(a) 200kg (b) 400kg (c) 800kg (d) 1600kg (e)100kg
The breaking strength is independent of the length but is directly proportional to the area of cross section. This follows from the expression for Young’s modulus (Y):
Y = Stress/strain. It is the strain which determines the breaking point of a cable.The strain at the breaking point is a constant for a given material. Since the Young’s modulus is a constant for a given material, the breaking stress also is a constant for a given material. Since the breaking stress is the ratio of the breaking strength to the area of cross section, it follows that the breaking strength is directly proportional to the area of cress section.
In the above problem, the breaking strength of the cable of diameter 2cm is 800 kgwt or(800×g) newton. Since the breaking stress is constant for the material of the cable, we can write, (800×g)/ (π×0.02^2) = mg/(π×0.01^2) where ‘m’ is the load the cable of half the diameter can support. Therefore, m = 200kg [option (a)]. It will be more convenient to write, 800/2^2 = m/1^2 to find ‘m’.
Suppose the breaking stress of a steel cable is ‘s’. What is the breaking stress of a steel cable of double the length and three times the diameter?
The answer, as you know, is ‘s’ since the breaking stress is a constant for a given substance.
Here is a very simple question which may mislead you if you are overconfident:
The Young’s modulus of a piano wire is Y. The Young’s modulus of another piano wire of half the thickness and twice the length is
(a) 2Y (b) 4Y (c) 8Y (d) Y/8 (e) Y
The correct option is (e) since the Young's modulus of a given substance is a constant.

You may visit physicsplus.blogspot.com for more multiple choice questions with solution.

Two Questions from Modern Physics

(1) The energy of a photon is 20eV. Its momentum in kg m/s is
(a) 2.56×10^-27 (b) 5.33×10^-27 (c) 1.066×10^-26 (d) 2.13×10^-26 (e) 3.18×10^-26
Since E=mc^2, momentum of the photon, p = E/c = (20×1.6×10^-19)/(3×10^8). Note that we have converted the energy in eV into joule. The answer is 1.066×10^-26 kg m/s given in option (c).
(2) The wave length associated with an electron having kinetic energy 6eV is
(a) 9A.U. (b) 5A.U. (c) 2.5A.U. (d) 1.5A.U. (e) 0.5A.U.
In the case of electrons, de Broglie wave length, λ = √ [ 150/V] A.U. where V is the accelerating voltage for the electron (= 6 volt since the energy is 6 eV).
Therefore, λ = √25 = 5 A.U.

Two Questions from Properties of Matter:
The following question appeared in the Kerala Engineering Entrance test paper of 2006:
The pressure inside two soap bubbles is 1.01 and 1.02 atmosphere respectively. The ratio of their respective volumes is
(a) 2 (b) 4 (c) 6 (d) 8 (e) 10
To the surprise of this author, a comparatively bright student omitted this question, which is quite simple. You know that the excess of pressure inside a bubble is 2T/r, where T is the surface tension and r is the radius. The ratio of the excess pressures inside the bubbles is P1/P2 = r2/r1. But, P1/P2 =0.01/0.02. [Since the actual pressures are 1.01 and 1.02 atmosphere]
So, we have r2/r1 = 1/2. The ratio of volumes, V1/V2 = (r1/r2)^3 = 2^3 = 8.
Consider now the following question which appeared in Kerala Medical Entrance test paper of 2006:
To what depth below the surface of sea should a rubber ball be taken so as to decrease its volume by 0.1%? [Take: Density of sea water = 1000kg/m^3, bulk modulus of rubber = 9×10^8 N/m^2, acceleration due to gravity = 10m/s^2]
(a) 9m (b) 18m (c) 180m (d) 90m (e) 900m
We have bulk modulus B = P/(dv/v) so that P = B(dv/v) = (9×10^8) ×0.001 = 9×10^5 pascal.
Therefore, hdg = 9×10^5 from which, h = (9×10^5)/(1000×10) = 90m. [Option (d)]