AP Physics B & C - Multiple Choice Practice Questions on Direct Current Circuits

Equations to be remembered for answering questions on direct current circuits and RC transients were posted on this site on 31^st 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.

Therefore, R’ = R/(0.99)⁴ = R/[1 – (0.01)]⁴ = R [1 – (0.01)]^–4

This gives R’ = 1.04R, nearly.

(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.

Kinematics - Practice Questions (MCQ) for AP Physics B & C

“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).

Therefore, average speed = 2s/(s/v₁ + s/v₂) = 2v₁v₂/(v₁ + 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 v_average during during the time interval from 2 s to 2.1 s is given by

v_average = (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.

AP Physics B & C - Multiple Choice Practice Questions on Gravitation

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 the work 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)³ since 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

From the above relations (on dividing),

24/T = 2^3/2

Therefore, T = 24/2^3/2 = 24/(2√2) = 12/√2 = 6√2 hour.

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 = kR^3/2 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

W = mgR²(r₂–r₁)/ 2r₂r₁

AP Physics B & C - Multiple Choice Practice Questions on Gravitation

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 the work 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)³ since 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

From the above relations (on dividing),

24/T = 2^3/2

Therefore, T = 24/2^3/2 = 24/(2√2) = 12/√2 = 6√2 hour.

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 = kR^3/2 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 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 a 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

W = mgR²(r₂–r₁)/ 2r₂r₁