Gears and Gear Trains: Lesson Test

Answer: A

The contact between the external tooth surfaces reverses angular direction when both shafts are viewed from the same side. This response follows the specified gear geometry and the assumptions in the question.

• A (correct): The contact between the external tooth surfaces reverses angular direction when both shafts are viewed from the same side.
• B: A normal gear pair rotates continuously in one direction unless the input itself is reversed.
• C: Equal direction would be characteristic of some other connections, such as a conventional open belt or internal mesh.
• D: A properly engaged pair continues transmitting rotation while a driving input is supplied.

Answer: D

Correct answer: They run more quietly and smoothly. Helical teeth are cut at an angle, so each tooth comes into contact gradually instead of all at once. More teeth share the load, which makes the drive quieter and smoother than a spur pair, at the cost of axial thrust on the bearings.

• A: The angled teeth actually push each gear along its shaft, creating axial thrust the bearings must resist.
• B: Self-locking is a feature of many worm sets, not of helical gears.
• C: Joining intersecting shafts is the job of bevel gears; helical gears connect parallel shafts.
• D (correct): Angled teeth engage gradually and share the load among more teeth, so helical gears are quieter and smoother.

Answer: A

Driver identifies the gear receiving the stated input power or motion; the designation depends on operating direction. This response follows the specified gear geometry and the assumptions in the question.

• A (correct): Driver identifies the gear receiving the stated input power or motion; the designation depends on operating direction.
• B: The pitch circle is a geometric reference, not a separate physical component receiving input.
• C: An idler is an intermediate gear and need not be the original input member.
• D: The driven gear is the receiver at the contact, not necessarily the motor-side input.

Answer: D

Correct answer: Worm and wheel. A worm meshing with a worm wheel links shafts that cross at 90 degrees without intersecting. Because one worm turn advances the wheel by only one or two teeth, the set gives a large reduction in little space.

• A: Bevel gears connect shafts whose axes intersect, so they do not suit shafts that pass by each other without meeting.
• B: A rack and pinion turns rotation into straight-line motion rather than linking two rotating shafts.
• C: Spur gears connect parallel shafts, not shafts that cross at right angles.
• D (correct): A worm and wheel joins non-intersecting shafts at right angles and gives a large reduction because each worm turn moves the wheel only a tooth or two.

Answer: D

Rigidly fixed gears co-rotate with the shared shaft even when their pitch diameters and tooth counts differ. This response follows the specified gear geometry and the assumptions in the question.

• A: At equal angular speed, tangential speed increases with radius, so differently sized gears need not match.
• B: Torque can be introduced or extracted at different positions; shaft balance cannot be inferred from gear size alone.
• C: Gears of different diameters and tooth counts can be fixed on the same shaft.
• D (correct): Rigidly fixed gears co-rotate with the shared shaft even when their pitch diameters and tooth counts differ.

Answer: A

Correct answer: It turns rotary motion into straight-line motion. The straight toothed bar is a rack and the round gear is a pinion. As the pinion turns, the rack moves in a straight line, so the pair converts rotation into linear motion.

• A (correct): The straight bar is a rack, and a pinion turning against it drives the rack along a straight path, which slides the gate.
• B: Linking parallel shafts is the job of spur or helical gear pairs, not a straight toothed bar.
• C: Turning the drive around a corner between intersecting shafts is what bevel gears do.
• D: Self-locking is a feature of many worm sets, not the main purpose of a rack and pinion.

Answer: C

Correct answer: It makes the output turn the same direction as the driver. An idler adds one more external mesh, so the output now turns the same way as the input. Its tooth count cancels in the ratio, so it changes direction but not the overall speed in the ideal model.

• A: The idler's tooth count cancels out when the two mesh ratios are multiplied, so it does not set the speed ratio.
• B: Torque follows the overall speed ratio, which depends only on the first and last gears, so the idler adds no torque gain.
• C (correct): The idler adds a second external mesh, and two reversals bring the output back to the driver's direction while the speed ratio stays the same.
• D: The output speed is set by the driver and output tooth counts alone; adding an idler does not speed it up.

Answer: A

Correct answer: Interlocking teeth keep an exact speed ratio without slipping. Gear teeth interlock, giving positive transmission: the speed relationship is fixed by the tooth counts and cannot drift through slipping. That exact timing is why clocks and machine tools use gears.

• A (correct): Each driver tooth pushes one tooth space of the driven gear, so the ratio is fixed by the tooth counts and does not drift with load.
• B: Gears require precise mounting and lubrication, so this is not a reason to choose them.
• C: Interlocking teeth cannot slip, which is why overload can break gear teeth; slipping protection belongs to friction drives.
• D: Gears, especially spur gears, can be noisy at speed, so quietness is not their main advantage here.

Answer: B

Correct answer: Clockwise. Count the external meshes from the first gear to the third: there are two. An even number of meshes keeps the original direction, so the third gear turns clockwise.

• A: This is the direction of the second gear, after only one mesh.
• B (correct): Each external mesh reverses direction; two meshes reverse it twice, so the third gear matches the first.
• C: Gears in a single chain on their own shafts turn freely; nothing in the train is held still.
• D: A gear driven at a steady direction keeps that direction; meshes fix the sense of rotation, not a back-and-forth motion.

Answer: D

Correct answer: 2 mm. Module is pitch diameter per tooth.

• A: This is half the pitch diameter, the pitch radius, not the module.
• B: This divides the tooth count by the diameter, which is the inverse of the module formula.
• C: This is the circular pitch for a 2 mm module, found by multiplying the module by pi.
• D (correct): Module equals pitch diameter divided by tooth count: 72 mm divided by 36 teeth gives 2 mm.

Answer: C

Use n_out =

the larger output gear rotates more slowly. This response follows the specified gear geometry and the assumptions in the question.

• A: Multiplying by 45/15 reverses the driver-to-output tooth ratio.
• B: Equal rpm would require equal tooth counts, absent another constraint.
• C (correct): Use n_out = 600(15/45) = 200 rpm; the larger output gear rotates more slowly.
• D: A tooth count is not itself a rotational speed and cannot replace the speed formula.

Answer: D

The 48-tooth driver moves three output circumferences per turn, so

This response follows the specified gear geometry and the assumptions in the question.

• A: This is the reciprocal reduction calculation, which would suit a small driver and large output.
• B: Unequal tooth counts on a direct mesh do not produce one-to-one speeds.
• C: 144 rpm does not follow from the tooth ratio 48/16 = 3; the output must be 90 × 3 = 270 rpm.
• D (correct): The 48-tooth driver moves three output circumferences per turn, so 90(48/16) = 270 rpm.

Answer: C

Correct answer: 500 rpm clockwise. Skip the idler for speed:

There are two external meshes, an even number, so the output turns clockwise like the driver.

• A: The speed is right, but two external meshes return the output to the driver's direction instead of reversing it.
• B: This inverts the ratio, multiplying by 30/20 even though the larger output must turn slower than the driver.
• C (correct): The idler cancels, so speed is 750 times 20/30 = 500 rpm, and two external meshes keep the clockwise direction.
• D: This uses the idler as if it were the output, 750 times 20/50, rather than the last gear in the train.

Answer: B

Correct answer: 40 rpm. For a worm set, the ratio is wheel teeth divided by starts:

The wheel speed is

• A: This assumes the wheel matches the worm speed, ignoring the large reduction a worm set gives.
• B (correct): A single-start worm gives a ratio equal to the wheel's tooth count, 45:1, so 1,800 divided by 45 is 40 rpm.
• C: This doubles the reduction to 90:1, as if each worm turn moved the wheel only half a tooth.
• D: This treats the worm as two-start, dividing 45 by 2 to get a smaller ratio.

Answer: B

The chosen reduction convention gives

meaning one output turn per three input turns. This response follows the specified gear geometry and the assumptions in the question.

• A: That reciprocal describes output/input speed, not the defined reduction ratio i.
• B (correct): The chosen reduction convention gives i = 60/20 = 3, meaning one output turn per three input turns.
• C: The tooth count of the output gear alone omits division by input teeth.
• D: The tooth count of the input gear alone does not establish the pair ratio.

Answer: C

Ten times 20/50 = four turns of the larger driven wheel. This response follows the specified gear geometry and the assumptions in the question.

• A: Using 50/20 incorrectly treats the output as the smaller faster wheel.
• B: One-to-one revolutions require equal tooth counts in this simple pair.
• C (correct): Ten times 20/50 = four turns of the larger driven wheel.
• D: Halving the input turns ignores the actual 20-to-50 ratio.

Answer: A

Correct answer: 45 teeth. Use

Check:

• A (correct): Teeth times speed must match: 30 times 600 is 18,000, and 18,000 divided by 400 is 45 teeth.
• B: This is a guess based on the 200 rpm drop rather than the tooth-speed product rule.
• C: A 60-tooth gear would halve the speed to 300 rpm, which is slower than required.
• D: This sets up the ratio the wrong way; a slower driven gear must be larger than the driver, not smaller.

Answer: B

Correct answer: 180 rpm clockwise. Speed still follows tooth counts:

Because the pinion meshes on the inside of the ring, both gears turn the same way, so the ring turns clockwise.

• A: This multiplies by the tooth ratio the wrong way; the larger ring must turn slower than the pinion.
• B (correct): The tooth ratio gives 900 times 15/75, or 180 rpm, and an internal mesh keeps the same direction.
• C: The speed is right, but an internal mesh does not reverse direction the way an external mesh does.
• D: This flips the tooth ratio and also applies the external-mesh reversal to an internal mesh.

Answer: D

Correct answer: 314.2 mm. Rack travel per turn is the pinion's pitch circumference, $\pi d$.

• A: This uses 2 pi times the diameter, which counts the circumference twice per turn.
• B: This multiplies the diameter by the number of turns and leaves out pi.
• C: This uses the 20 mm radius in place of the diameter in the circumference formula.
• D (correct): Each turn moves the rack one pitch circumference, pi times 40 mm, and 2.5 turns gives about 314.2 mm.

Answer: A

Correct answer: 36 N·m. Multiply input torque by the ratio, then by the efficiency.

N·m.

• A (correct): Ideal torque is 4 times 12 = 48 N·m, and 75 percent of that is 36 N·m.
• B: This divides the input torque by the ratio, which is what a speed-increasing drive would do.
• C: This is the ideal torque; it ignores the 75 percent efficiency.
• D: This divides the ideal torque by the efficiency, making the output larger than the ideal value.

Answer: C

Correct answer: 60 mm. The pitch diameters are

• A: This uses the difference of the tooth counts, which applies to no external pair.
• B: This halves the total tooth count but never multiplies by the 2 mm module.
• C (correct): Center distance is the module times the sum of the teeth, halved: 2 times 60 divided by 2 = 60 mm.
• D: This adds the two pitch diameters (36 mm and 84 mm) but forgets to halve the sum.

Answer: C

Correct answer: 200 rpm. Stage 1:

on the middle shaft. The 16-tooth gear shares that shaft, so it also turns at 500 rpm. Stage 2:

• A: This adds the stage ratios (3 + 2.5 = 5.5) instead of multiplying them.
• B: This treats the 36-tooth and 16-tooth gears as an idler step and uses only 12/40, but those gears share a shaft rather than mesh.
• C (correct): The middle shaft turns at 1,500 times 12/36 = 500 rpm, and the output turns at 500 times 16/40 = 200 rpm.
• D: This is the speed of the intermediate shaft; the second mesh still reduces the speed.

Answer: A

Correct answer: 1,005 W. Convert speed:

rad/s. Then

• A (correct): Angular velocity is 2 pi times 1,200 / 60, about 125.7 rad/s, and 8 times 125.7 is about 1,005 W.
• B: This multiplies torque by rpm directly without converting rpm to radians per second.
• C: This includes 2 pi but forgets to divide by 60 seconds per minute.
• D: This converts minutes to seconds but leaves out the 2 pi radians in each revolution.

Answer: B

Correct answer: A chain running on two sprockets. A chain drive meshes with sprocket teeth, so it cannot slip and its ratio follows the tooth counts. Both sprockets turn in the same direction, and the chain can span a long distance between shafts.

• A: Bevel gears connect intersecting shafts, not parallel shafts far apart.
• B (correct): A chain gives positive transmission with no slip, both sprockets turn the same way, and it suits long shaft spacing.
• C: An open belt keeps the direction and spans the distance, but friction belts can slip, so the ratio is not exact.
• D: An external gear pair reverses direction and would need very large gears to span 1.5 m.

Answer: D

Conical bevel gear geometry is suitable for intersecting shafts and changing shaft direction. This response follows the specified gear geometry and the assumptions in the question.

• A: A coupling joins shaft ends and does not itself provide a meshing angle change.
• B: A rack translates and does not constitute a pair of intersecting rotating shafts.
• C: Standard spur pairs generally use parallel rather than intersecting axes.
• D (correct): Conical bevel gear geometry is suitable for intersecting shafts and changing shaft direction.

Answer: B

Correct answer: 300 rpm. With the ring fixed, sun in and carrier out:

The carrier turns at

• A: This uses only the ring-to-sun ratio of 3 and leaves out the added 1 in the planetary formula.
• B (correct): With the ring fixed, the reduction is 1 + 72/24 = 4, so the carrier turns at 1,200 / 4 = 300 rpm.
• C: This multiplies by the ring-to-sun ratio, treating a reduction as a speed increase.
• D: This multiplies the input by the reduction of 4 instead of dividing by it.

Answer: B

Equal tooth counts give a one-to-one speed magnitude; external contact reverses direction. This response follows the specified gear geometry and the assumptions in the question.

• A: A doubled speed would require unequal tooth counts or another transmission stage.
• B (correct): Equal tooth counts give a one-to-one speed magnitude; external contact reverses direction.
• C: Neither halving of speed nor same-direction rotation follows the equal external pair.
• D: Equal counts do not make the motion vanish; they simply produce equal speed magnitudes.

Answer: D

Correct answer: 80 N·m at 300 rpm. For an ideal reducer,

N·m and

Torque rises by the same factor that speed falls, so power is unchanged.

• A: The speed is right, but a reduction raises torque; it does not lower it.
• B: Raising both torque and speed would create power from nothing; a reducer lowers the speed.
• C: This describes a 1:4 speed increaser, the reverse of the gearbox described.
• D (correct): A 4:1 reduction divides speed by 4 and, in the ideal case, multiplies torque by 4.

Answer: A

Both gear wheels are locked to the same shaft and must turn together. This response follows the specified gear geometry and the assumptions in the question.

• A (correct): Both gear wheels are locked to the same shaft and must turn together.
• B: Rigid attachment transmits rotation rather than halting it.
• C: Reversal occurs across suitable external tooth contacts, not through one rigid axle.
• D: Tooth-count ratio applies at a mesh, not between rigidly attached wheels.

Answer: C

Pitch diameters and tooth counts are proportional for compatible gears, so the doubled output diameter turns at half the input speed. This response follows the specified gear geometry and the assumptions in the question.

• A: That ratio would apply if the input diameter were twice the output diameter instead.
• B: Equal rotational speeds require equal pitch diameters in a direct mesh.
• C (correct): Pitch diameters and tooth counts are proportional for compatible gears, so the doubled output diameter turns at half the input speed.
• D: Squaring the diameter ratio has no basis in a simple gear-pair kinematic relation.

Answer: A

Correct answer: 800 rpm clockwise. The 35-tooth and 50-tooth gears are idlers, so only the first and last gears set the speed:

There are three external meshes, an odd number, so the output turns clockwise, opposite to the input.

• A (correct): The two middle gears act as idlers and cancel, so speed is 1,200 times 20/30 = 800 rpm, and three meshes reverse the direction.
• B: This stops at the 50-tooth gear, 1,200 times 20/50, instead of using the last gear.
• C: This inverts the first-to-last ratio, multiplying by 30/20 instead of 20/30.
• D: The speed is right, but three external meshes is an odd count, so the output reverses.

Answer: C

Correct answer: 54 N·m. Stage ratios are

N·m.

• A: This is the ideal torque from the 12:1 ratio; it skips the 90 percent efficiency.
• B: This adds the stage ratios (3 + 4 = 7) instead of multiplying them, then applies the efficiency.
• C (correct): The stage ratios multiply, 3 times 4 = 12, so ideal torque is 60 N·m, and 90 percent of that is 54 N·m.
• D: This adds the stage ratios to get 7 and also ignores the efficiency.

Answer: C

Correct answer: 17:1. Solve

for the ratio.

The ratio must be at least this, so round up to 17:1, which gives

N·m.

• A: This ignores the 80 percent efficiency; 14:1 would give only 0.80 times 14 times 3, or 33.6 N·m.
• B: This multiplies the target by the efficiency instead of dividing by it, giving far too small a ratio.
• C (correct): The ratio must be at least 40 divided by (0.80 times 3), or about 16.7, so the next whole number, 17, is needed.
• D: This rounds 16.7 down; 16:1 delivers 0.80 times 16 times 3, or 38.4 N·m, which falls short.

Answer: D

The first stage reduces by three, but the second stage increases speed by two;

This response follows the specified gear geometry and the assumptions in the question.

• A: This wrongly treats the second stage as a 2:1 reduction instead of a speed increase.
• B: This includes only the second stage and omits the first-stage reduction.
• C: This counts only the first mesh and ignores the 30-to-15 tooth second stage.
• D (correct): The first stage reduces by three, but the second stage increases speed by two; i = 3 × 0.5 = 1.5.

Answer: B

Correct answer: 204 mm. For a 3:1 reduction the driven gear needs

with the same 3 mm module. The outside diameter of a standard gear is

• A: This adds only one module to the pitch diameter instead of one module on each side.
• B (correct): A 3:1 reduction needs 3 times 22 = 66 teeth, and the outside diameter is 3 times (66 + 2) = 204 mm.
• C: This is the outside diameter of the 22-tooth pinion, 3 times 24, not of the mating gear.
• D: This is the pitch diameter of the 66-tooth gear, 3 times 66; the tips stick out beyond the pitch circle.

Answer: A

Correct answer: 377.0 mm. Teeth passing the mesh:

so the 40-tooth gear and its pinion turn

times.

• A (correct): The 40-tooth shaft turns 5 times 16/40 = 2 times, and each turn moves the rack pi times 60 mm, about 377.0 mm in all.
• B: This gets the 2 turns right but uses the 30 mm radius in place of the diameter.
• C: This inverts the gear ratio, making the 40-tooth shaft turn 12.5 times instead of 2.
• D: This lets the rack pinion turn 5 times, ignoring the reduction from the 16-tooth to the 40-tooth gear.

Answer: C

Correct answer: 150 rpm clockwise. First stage:

counterclockwise. The 18-tooth gear shares that shaft, so it also turns at 450 rpm counterclockwise. Second stage:

reversed again to clockwise. The overall reduction is

• A: This adds the stage ratios (4 plus 3 equals 7) instead of multiplying them.
• B: This treats the two gears on the shared shaft as if they meshed, adding a false stage and a false reversal.
• C (correct): The stages give 1,800 times 12/48 times 18/54, or 150 rpm, and two external meshes return the original direction.
• D: The speed is right, but there are two external meshes, an even number, so the output turns the same way as the input.

Answer: B

Correct answer: 250 rpm and 55.2 N·m. Output speed is set only by the tooth ratio:

Efficiency lowers the torque, not the speed:

N·m.

• A: This applies the efficiency to the speed, but teeth cannot slip, so speed stays exactly as the ratio sets it.
• B (correct): Speed follows the tooth ratio exactly, 1,000 / 4 = 250 rpm, and torque is 0.92 times 4 times 15 = 55.2 N·m.
• C: These are the ideal values; the 92 percent efficiency must reduce the torque.
• D: This reduces both speed and torque, counting the friction loss twice.

Answer: D

Correct answer: 60 rpm and 30 N·m. For a worm,

Speed:

Torque:

N·m. Efficiency lowers torque but not the speed ratio.

• A: This divides the ideal torque by the efficiency; losses lower output torque, so multiply by 0.60.
• B: The speed is right, but 50 N·m is the ideal torque; the 60 percent efficiency must reduce it.
• C: This treats the worm as single-start, giving a 50:1 ratio instead of 25:1.
• D (correct): The ratio is 50/2, or 25, so the wheel turns at 1,500/25 = 60 rpm, and torque is 0.60 times 25 times 2, or 30 N·m.

Answer: B

Correct answer: 40 teeth turning at 600 rpm. From

Speed:

Check: pitch diameters 60 mm and 120 mm, and half their sum is 90 mm.

• A: The tooth count is right, but a larger driven gear turns slower than the pinion, not faster.
• B (correct): The tooth total is 2 times 90 divided by 3, or 60, leaving 40 teeth; 1,200 times 20/40 is 600 rpm.
• C: 60 is the combined tooth count of both gears; the pinion's 20 teeth must be subtracted.
• D: This divides the center distance by the module without doubling it, so the tooth total comes out as 30 instead of 60.

Answer: C

Correct answer: Excess backlash between mating teeth. Backlash is the small gap between the back of one tooth and the face of its mating tooth. When the input reverses, the teeth must cross that gap before they push again, so the output lags and may clunk. Some backlash is normal, but wear makes it grow.

• A: Thrust from helical teeth pushes a gear along its shaft; it does not cause a lag on reversal.
• B: Slip belongs to smooth belt drives; meshing teeth cannot slip past each other.
• C (correct): The gap between teeth lets the input turn a little before it engages the output after a reversal.
• D: Self-locking stops a wheel from driving a worm backward; it does not produce a short delay and clunk.

Answer: A

Correct answer: 140 rpm counterclockwise. Work stage by stage, carrying speed across each shared shaft. Stage 1:

(counterclockwise). Stage 2:

(clockwise). Stage 3:

There are three external meshes, an odd number, so the output turns opposite to the input: 140 rpm counterclockwise. The overall ratio is

• A (correct): Multiplying the stage ratios gives 15:1, so 2,100 ÷ 15 = 140 rpm, and three external meshes reverse the direction.
• B: The speed is right, but three external meshes is an odd count, so the output turns opposite to the clockwise input.
• C: This adds the stage ratios (3 + 2.5 + 2 = 7.5) instead of multiplying them.
• D: This is the speed and direction of the middle shaft; the motion still passes through one more mesh.

Answer: D

Correct answer: 60 N·m. The ideal output torque is

N·m. Because the input and output shafts turn in opposite directions, their torques on the housing add rather than cancel.

The mounting must resist

N·m.

• A: Subtracting applies when the shafts turn the same way; here they turn opposite ways, so the torques add.
• B: This is only the output torque; the input torque also acts on the housing.
• C: Input and output torques are not equal in a reducer, so they cannot balance; the housing must be bolted down.
• D (correct): With opposite shaft rotation, input and output torques add: 10 + 50 = 60 N·m.

Answer: C

Correct answer: About 9.4 m/s. Convert to revolutions per second:

rev/s. Surface speed is

A mating gear shares this same speed at the contact point.

• A: This uses 900 rpm directly without converting minutes to seconds.
• B: This uses the radius, 0.10 m, in place of the diameter in v = π d n.
• C (correct): With speed in revolutions per second, π × 0.20 × 15 ≈ 9.4 m/s.
• D: This leaves out π; each turn carries a pitch-circle point a full circumference, π d.

Answer: C

Correct answer: Two external spur gears meshing directly. When two gears mesh on the outside of each other, the teeth at the contact point move together, which spins the two gears in opposite directions. An open belt, a chain on sprockets and a pinion inside a ring gear all keep the same direction.

• A: With an open belt, both pulleys turn the same way.
• B: A chain drive, like an open belt, turns both sprockets in the same direction.
• C (correct): An external mesh always reverses direction, whatever the gear sizes.
• D: An internal mesh makes the pinion and the ring turn in the same direction.

Answer: B

Correct answer: About 43 N·m. Output power =

Output speed =

rad/s.

Torque = power ÷ angular velocity = 1,350 ÷ 31.4 ≈ 43 N·m.

• A: This uses the full 1,500 W; a 90 percent efficient gearbox delivers only 1,350 W.
• B (correct): Output power 1,350 W divided by about 31.4 rad/s gives about 43 N·m.
• C: This is the motor's torque reduced for losses; it leaves out the 6:1 torque gain.
• D: This divides power by rpm; power must be divided by angular velocity in rad/s.

Answer: B

Correct answer: No gear can turn, and the teeth are heavily stressed. In an ordinary fixed-axis train, every gear meshed with a locked gear is also locked, because the teeth cannot slip past each other. The input and output cannot turn, and the stalled motor loads the teeth heavily, which can break them.

• A: The output receives motion only through the middle gear; if it is jammed, nothing reaches the output.
• B (correct): Interlocked teeth cannot slip, so a jammed gear locks every gear meshed with it.
• C: The input meshes directly with the jammed gear, so it cannot turn either.
• D: Teeth cannot slip to give a partial speed; a jammed gear stops the whole chain.

Answer: D

Correct answer: At 240 rpm, on a shaft at a right angle to the crank shaft. Bevel gears connect shafts whose axes intersect, usually at a right angle. With equal tooth counts the ratio is

so the speed is unchanged. The pair only turns the drive around the corner.

• A: The speed is right, but bevel gears join shafts whose axes meet, typically at 90 degrees.
• B: Equal tooth counts do not change speed; halving would need a gear with twice as many teeth.
• C: Doubling the speed would need a driven gear with half as many teeth; these counts are equal.
• D (correct): Equal tooth counts give a 1:1 ratio, and bevel gears connect intersecting shafts.

Answer: A

Correct answer: About 7.85 mm. The pitch circle's circumference πd is shared equally by N teeth, so p = πd/N. Since d/N is the module, p = πm.

• A (correct): Circular pitch equals π times the module: π × 2.5 ≈ 7.85 mm.
• B: This is the module itself, the pitch diameter per tooth, not the spacing along the circle.
• C: This multiplies by 2π; the formula is p = πm, not 2πm.
• D: This divides the module by π instead of multiplying.

Answer: A

Correct answer: 165 rpm. A chain gives positive transmission, so the same number of teeth pass each sprocket:

The smaller sprocket turns faster.

• A (correct): Matching teeth passed: 60 × 44 ÷ 16 = 165 rpm.
• B: This flips the ratio; the smaller rear sprocket must turn faster than the pedals, not slower.
• C: Sprockets with different tooth counts turn at different speeds; only the chain speed is shared.
• D: This multiplies by 44 but never divides by the 16 teeth on the rear sprocket.