You can access all posts involving magnetic fields and electromagnetic induction on this blog by trying a search using the search box at the top of this page or by clicking on the labels ‘magnetic field’ and ‘electromagnetic induction’ below this post. Today I give you a free response practice question on electromagnetism. This question is for AP Physics C aspirants:
A rectangular wire loop of length ℓ and breadth b havingnegligible resistance is arranged in the plane of an infinitely long straight vertical wire as shown, with the longer sides parallel to the wire. The wire loop contains a capacitor of capacitance C. A steady current ‘I’ flows upwards through the straight wire. Now answer the following questions:
(a) Calculate the magnetic flux through the wire loop.
(b) The loop is rotated through 180º about a central axis (fig) which is parallel to the straight wire. If the time taken for this rotation is ∆t, determine the magnitude of the emf induced in the loop.
(c) The loop is kept stationary and instead of the steady current I, a current ‘i’varying with time t as i = Im sinωt (where Im and ω are constants) is passed through the straight wire. Calculate the maximum value of the emf induced in the wire loop.
(d) Determine the maximum current induced in the wire loop under the conditions mentioned in part (c) above.
Try to answer this question. You can take 15 minutes for answering it and can score up to 15 points for the right answer. I’ll be back soon with a model answer for your benefit.
Equations to be remembered in the section ‘oscillations’ were discussed in the post dated 17thApril 2008.Some multiple choice practice questions in this section were discussed in the post dated 22nd April 2008, followed by a free response practice question in the post dated 2ndMay 2008. You can access all these posts by clicking on the label ‘oscillation’ below this post. Today we will discuss a few multiple choice practice questions involving simple pendulum:
(1) A simple pendulum arranged inside a train has period of oscillation T when the train is at rest. When the train moves along straight horizontal rails with uniform acceleration of x ms–2, the period of the pendulum is (assuming g = 10 ms–2)
(a) T ×g1/2/(g+x)1/2
(b) T ×g1/4/(g+x)1/4
(c) T ×g1/4/(g2+x2)1/4
(d) T ×g1/2/(g2+x2)1/2
(e)T ×g1/2/(g2+x2)1/4
When the train is at rest the period of oscillation T of the pendulum is given by
T = 2π (ℓ/g)1/2 ………….(i)
where ℓ is the length of the pendulum and g is the acceleration due to gravity.
When the train moves forward with acceleration x ms–2, an inertial backward force acts on the bob of the pendulum and supplies a backward acceleration of x ms–2 (fig.). The resultant acceleration of the bob is (g2+x2)1/2. The period T1of the pendulum is now given by
T1 =2π[ℓ/(g2+x2)1/2]1/2……..(ii)
Dividing eqn (ii) by eqn (i) we have
T1/T = g1/2/(g2+x2)1/4 from which T1 = T ×g1/2/(g2+x2)1/4 The correct option is (e)
[Suppose the train is moving with uniform velocity of x ms–1. What will be the period? No doubt, T itself since you cannot distinguish between state of rest and uniform motion].
(2) Two simple pendulums A and B have periods 2.1 s and 2 s respectively. They start oscillating at the same time in phase. They will be in phase instantly at the end of
(a) 42 s
(b) 40 s
(c) 22 s
(d) 21 s
(e) 20 s
Suppose the pendulums are in phase instantly at the end of n oscillations of pendulum A. Pendulum B should then execute n+1 oscillations. Therefore we have
n×2.1 = (n+1)×2
Therefore 0.1n = 2 so that n = 20
The time elapsed is therefore 20×2.1 = 42 s [Option (a)].
[The difference between the periods of the two pendulums is 0.1 s. So you require 20 oscillations of pendulum A to obtain a time difference equal to one period (2 s) of pendulum B so that the two pendulums will be instantly in phase].
(3) The bob of a simple pendulum is a hollow metal sphere filled with water. There is a small hole at the bottom of this hollow sphere and water drains out through the hole as the pendulum oscillates. After completely filling the bob with water, if this pendulum is made to oscillate for a long time, its period of oscillation will
(a) increase first and will reach a final constant value
(b) decrease first, reach a minimum value, then increase and will finally settle at the initial value
(c) increase first, reach a maximum value, then decrease and will finally settle at the initial value
(d) decrease first and will reach a final constant value
(e) remain unchanged
The correct option is (c). Initially the centre of gravity of the spherical bob is at its centre since it is completely filled with water. When the water flows out, the centre of gravity of the bob moves gradually downwards, reaches a minimum level and then moves up. When the water is fully drained out, the centre of gravity of the bob once again reaches the centre of the bob and remains there.
The length of the pendulum is the distance between the centre of gravity of the bob and the point of suspension. Therefore, the length of the pendulum gets increased initially, becomes a maximum, then gets decreased and finally settles at the initial value. Therefore, the period of the pendulum will increase first, reach a maximum value, then decrease and will finally settle at the initial value.
(4) A simple pendulum is taken to a location where the acceleration due to gravity is decreased by 0.1 %. If the period of oscillation is to be unaltered
(a) the mass of the bob is to be increased by 0.1 %
(b) the mass of the bob is to be decreased by 0.1 %
(c) the length of the pendulum is to be decreased by 0.316 %
(d) the length of the pendulum is to be decreased by 0.2 %
(e) the length of the pendulum is to be decreased by 0.1 %
Since the period of oscillation T is given by T = 2π√(ℓ/g), the ratio ℓ/g should be unaltered. Therefore, the length ℓ should be decreased by 0.1 %
The following questions are for AP Physics C aspirants only:
(5) The period of a simple pendulum is decreased by 0.02 s when the length of the pendulum is decreased by 1 cm. The original length of the pendulum is nearly (g = 10 ms–2)
-->
(a) 0.25 m
(b) 0.5 m
(c) 1 m
(d) 1.01 m
(e) 2 m
The period of oscillation T is given by T = 2π√(ℓ/g).
Therefore, ∆T/T = ½ ∆ℓ/ℓ – ½ ∆g/g
[Here ∆T,∆ℓ, and∆g represent the increments in the period, length and acceleration due to gravity respectively. We have written the above equation by taking the logarithm of the expression for period and then differentiating it]
The increment in T is – 0.02 s and the increment in ℓ is – 0.01 m (negative signs are because T and ℓ are decreased). There is no change in g.
Therefore we have – 0.02/T = – ½ × 0.01/ℓ
This gives T = 4ℓ.
But T = 2π√(ℓ/g) so that 2π√(ℓ/10) = 4ℓ
Or, π2/10 = 4ℓ
Since π2 is nearly equal to 10, we obtain ℓ = 0.25 m, nearly.
Equations to be remembered for answering questions on direct current circuits and RC transients were posted on this site on 31st March 2008. You can access them by clicking on the label ‘direct current circuit’ below this post or by trying a search using the search box at the top of this page. Today we will discuss some more multiple choice practice questions in this section:
(1) A uniform copper wire of radius r and length L has resistance R at temperature T.If this wire is carefully stretched so that its radius becomes kr where k is a constant, its resistance at the temperature T will be
(a) kR
(b) R/k2
(c) k2R
(d) k4R
(e)R/k4
If A is the area of cross section of the wire before the stretching, we have
R = ρL/A where ρ is the resistivity (specific resistance) of the material of the wire.
When the radius becomes kr, the area of cross section of the wire becomes k2A.Since the volume of the wire is unchanged, the length of the wire becomes L/k2. The resistance of the wire after stretching therefore becomes R’ given by
R’ = ρ(L/k2)/k2A = ρL/ k4A = R/k4
[Note that the resistance of the wire after stretching is greater than R since k < 1].
(2) In the above question suppose the wire was stretched so that its length becomes 1.01 L. The resistance of the wire after stretching then becomes nearly
(a) 1.01R
(b) R/1.02
(c) 1.02R
(d) 1.04R
(e)R/1.04
Note that if a wire of resistance R is stretched so that its length becomes n times its original length, its area of cross section becomes (1/n) times its original area of cross section. The resistance of the wire therefore becomes n2R.
Since n = 1.01 the resistance after stretching is (1.01)2R =1.02R, nearly [Option (c)].
Here is a variation of question no.1:
(3) If the radius of a uniform metal wire of resistance R is changed by 1% by stretching it carefully under isothermal conditions, its resistance becomes
(a) 1.04R
(b) R/1.04
(c) 1.02R
(d) R/1.02
(e)1.01R
We have R’ = R/k4 as shown in question no.1. Since the change (decrease) in the radius is 1%, k = 0.99.
(4) The resistance of an ammeter of range 2 A is R. The shunt resistance required to make its range 6 A is
(a) R
(b) R/5
(c) R/6
(d) R/2
(e)R/3
Let S be the shunt resistance required. When a current of 6 A is passed through the ammeter- shunt combination, 2 A should pass through the ammeter and the remaining 4 A should pass through the shunt (fig.). Therefore we have
2×R = 4×S, on equating the potential difference across the ammeter resistance R to the potential difference across the shunt resistance S.
This gives S = R/2
The following questions are for AP Physics C aspirants (even though AP Physics B aspirants also can try them):
(5) A power supply having internal resistance r drives a current through a load of resistance P. The heat developed in the load resistance is H. When the same power supply drives a current through another load of resistance Q for the same time t, the heat developed in the load resistance is H itself. The internal resistance of the power supply is for a time
(a) √(PQ)
(b) √[(P2 + Q2)/2]
(c) (P + Q)/2
(d) (P – Q)/2
(e) √[(P2 – Q2)/2]
Since H = I2Rt where I is the current through the load resistance R, we have
H = [V/(P+r)]2Pt = [V/(Q+r)]2Qt where V is the emf of the power supply.
[V/(P+r) is the current in the load resistance P and V/(Q+r) is the current in the load resistance Q].
From the above equation we have
P/(P+r)2 = Q/(Q+r)2
Or, P/(P2 +r2 +2Pr) = Q/(Q2+r2 + 2Qr)
Or, PQ2 + Pr2 + 2PQr = QP2 + Qr2 + 2PQr from which
r2(P – Q) = QP2 – PQ2
Or, r2(P – Q) = PQ(P – Q) so that r = √(PQ)
(6) A 12 V battery having internal resistance 0.2 Ω is being charged at a current of 2 A using the circuit shown in the figure. The generator which charges the battery has a terminal voltage of 16 V at this charging current. What is the value of the resistance R connected in the circuit?
(a) 0.2 Ω
(b) 0.8 Ω
(c) 1.8 Ω
(d) 2.6 Ω
(e) 3.4 Ω
The voltage between the terminals of the battery (terminal potential difference) while the battery is being charged (which is equal to the sum of the emf of the battery and the potential drop across its internal resistance) = 12 + 0.4 = 12.4 V.
[0.4 V is the voltage drop (2×0.2 volt) across the internal resistance of the battery].
Therefore, the voltage drop across the series resistance R is (16–12.4) V = 3.6 V.
Since the charging current is 2 A, we have 2 ×R = 3.6 from which R = 1.8 Ω.
You will find some useful multiple choice questions (with solution) in this section here.
“Clarity about the aims and problems of socialism is of greatest
significance in our age of transition.”
–Albert Einstein
You can access all posts on kinematics on this site by clicking on the label ‘kinematics’ below this post. Or, you may try searching for ‘kinematics’ using the search box at the top of this page.
Today I give you three more multiple choice practice questions with solution:
(1) Tom runs from his school to his home with uniform speed v1 and returns to his school with uniform speed v2. The average speed of his round trip is
(a) (v1 + v2)/2
(b) [(v12 + v22)/2]1/2
(c) (v1v2)1/2
(d) v1v2/(v1 + v2)
(e)2v1v2/(v1 + v2)
The average speed is the ratio of the total distance traveled to the total time taken.
If the distance between Tom’s school and his home is s, the total distance for the round trip is 2s and the total time for the round trip is (s/v1 + s/v2).
(2) The displacement ‘y’ of a particle thrown vertically down is given by the equation, y= 2t + 5t2, where y is in metre and t is in second. The average velocity during the time interval from 2 s to 2.1 s is
(a) 12.5 ms–1
(b) 16 ms–1
(c) 20.5 ms–1
(d) 22.5 ms–1
(e) 32 ms–1
The velocity ‘v’ of the particle at the instant t is given by
v = dy/dt = 2 + 10t
Therefore, velocities at instants 2 s and 2.1 s are respectively 2 + 10×2 = 22 ms–1 and 2 + 10×2.1 = 23 ms–1.
The average velocity vaverage during during the time interval from 2 s to 2.1 s is given by
vaverage = (22 + 23)/2 = 22.5 ms–1
[The above question can be answered without using calculus (as in the case of some of the AP Physics B aspirants) like this:
The displacement y is in the form for uniformly accelerated motion in one dimension: y = v0t + ½ at2 where v0 is the initial velocity and ‘a’ is the acceleration. Therefore, v0 = 2 ms–1 and a = 10 ms–2.
The velocity ‘v’ at the instant t is given by v = v0 + at = 2 + 10t].
The following question is meant specifically for AP Physics C aspirants:
(3) At the instant t,the position ‘x’ of a particle moving along the x-axis is given by x = 12t2 – 2t3 where x is in metre and t is in second. What will be the position of this particle when it moves with the maximum speed along the positive x direction?
(a) 32 m
(b) 36 m
(c) 40 m
(d) 48 m
(e) 52 m
The speed v of the particle is given by
v = dx/dt = 24t – 6t2.
When the speed is maximum we have dv/dt = 0
Therefore, 24– 12t = 0 from which t = 2 s.
The position of the particle at 2 seconds will be 12×(2)2 – 2×(2)3 = 48 – 16 = 32 m (as obtained from x = 12t2 – 2t3).
Useful posts in this section can be seen at physicsplus.
The percentage goals for AP Physics B and AP Physics C (Mechanics) are 6% and 18% respectively in the case of the topics under oscillations and gravitation. These topics were discussed earlier on this site. You can access those posts making use of the ‘search blog’ box.
Today we will discuss a few more multiple choice practice questions on gravitation. The following questions are useful for AP Physics B as well as AP Physics C aspirants:
(1) The velocity of escape from the earth’s surface is √(2GM/R) and thework done by earth’s gravitational force to bring unit mass from infinity to a height h from the earth’s surface is GM/(R+h) where ‘G’ is the gravitational constant and ‘M’ is the mass of the earth. If a body of mass m is projected up with the escape velocity from the earth’s surface, how high will it rise? Neglect air resistance.
(a) R/2
(b) R
(c) 2R
(d) 8R
(e) ∞
The velocity of escape is the velocity to be imparted to the body to make it escape from the gravitational pull (of the earth in the present case). So the height attained will be infinite [Option (e)].
(2) If the body in the above problem is projected up with 20 %of the escape velocity, how high will it rise?
(a) R/24
(b) R/20
(c) R/16
(d) R/8
(e) R/4
The kinetic energy supplied to the body is ½ m[0.2×√(2GM/R)]2 = 0.04 GMm/R.
The above energy is used up in increasing the gravitational potential energy of the body. If the maximum height reached is h, the increase in the gravitational potential energy is – GMm/(R+h) – (– GMm/R) = GMm/R – GMm/(R+h)
[Note that the gravitational potential energy is negative]
Therefore, we have
0.04 GMm/R = GMm/R – GMm/(R+h)
Or, 0.04/R = 1/R – 1/(R+h) = h/[R(R+h)]
This gives h = 0.04(R+h) from which h = 0.04 R/0.96 = R/24.
(3) The height of geostationary (synchronous) satellites above the surface of the earth is approximately 6R where R is the radius of the earth. The orbital period of a research satellite at an altitude of 2.5R above the surface of the earth will be approximately
(a) 16 hour
(b) 12√2 hour
(c) 8√2 hour
(d) 6√2 hour
(e) 6 hour
According to Kepler’s third law, the square of the orbital period of a satellite is directly proportional to the cube of the mean radius of the orbit of the satellite. Therefore, in the case of of geostationary (synchronous) satellites we have
242 α (7R)3since the orbital radius is 6R+R = 7R and the orbital period is 24 hour.
In the case of the research satellite the period T is related to its orbital radius by
T2 α (3.5R)3 since its orbital radius is 2.5R+R = 3.5R
The following questions are meant solely for AP Physics C aspirants:
(4) Two satellites have orbital radii R and 1.01 R. Their orbital periods differ by
(a) 1%
(b) 1.5%
(c) 2%
(d) 2.5%
(e) 3%
We have from Kepler’s third law, T2 α R3 where T and R are the orbital period and orbital radius respectively.
Therefore, T = kR3/2 where k is the constant of proportionality.
Taking logarithms, ln T = ln k + (3/2) ln R
Differentiating, dT/T = 0 + (3/2) dR/R. This says that the fractional change in orbital period is 3/2 times the fractional change in orbital radius. In other words, the percentage change in orbital period is 3/2 times the percentage change in orbital radius.
In the present problem the orbital radii differ by 1%. Therefore, the orbital periods will differ by (3/2)×1% = 1.5% [Option (b)].
(5) If the orbital radius of a satellite moving around the earth is increased by 2%, the orbital speed will
(a) remain unchanged
(b) decrease by 2%
(c) increase by 2%
(d) increase by 1%
(e) decrease by 1%
The orbital speed v of a satellite moving around the earth in an orbit of radius r is given by
v = √(GM/r) where G is the gravitational constant and M is the mass of the earth.
[You can easily obtain this by equating the centripetal force to the gravitational pull: mv2/r = GMm/r2 where m is the mass of the satellite].
This equation shows that the orbital speed will decrease when the orbital radius is increased.
Since G and M are constants, we have
dv/v = – ½ dr/r
[You will get this by taking logarithms and by differentiating, as we did in question No.4].
This means that percentage increase in orbital speed = – (½)×percentage increase in orbital radius. The negative sign indicates the decrease in the orbital speed because of the increase in orbital radius.
Since the orbital radius of the satellite is increased by 2%, the orbital speed is decreased by 1% [Option (e)].
(6) A communication satellite of mass m is placed initially in a temporary orbit of radius r1 around the earth. How much work is to be done to shift it from this orbit to a permanent orbit of greater radius r2?(Acceleration due to gravity at the sea level = g. Radius of the earth = R)
(a) mgR2(r2–r1)/ r2r1
(b) mgR (r2–r1)/ 2r2
(c) mgR2(r2–r1)/ 2r2r1
(d) mgR2(r2–r1)/ 2r2r1
(e) 2mgR2(r2–r1)/ r2r1
Total energy (K.E. + P.E.) of a satellite of mass m in an orbit of radius r is – GMm/2r where M is the mass of the earth.
Work (W) done in shifting the satellite from one orbit to the other is the difference between the energies in the orbits.
Therefore, W = – GMm/2r2 – (– GMm/2r1)
[Note that the satellite has greater energy in the orbit of greater radius since the energy is negative by a smaller amount. When the radius is infinite, the energy is maximum and is equal to zero].
Thus W = (GMm/2)(1/r1 – 1/r2) = (GMm/2)[(r2–r1)/ r2r1]
Since g = GM/R2 (as you can obtain from mg = GMm/R2), we have
