CR4 - The Engineer's Place for News and Discussion ®
Login | Register for Engineering Community (CR4)

Challenge Questions

Stop in and exercise your brain. Talk about this month's Challenge from Specs & Techs or similar puzzles.

So do you have a Challenge Question that could stump the community? Then submit the question with the "correct" answer and we'll post it. If it's really good, we may even roll it up to Specs & Techs. You'll be famous!

Answers to Challenge Questions appear by the last Tuesday of the month.

Permanent Magnet: Newsletter Challenge (May 2016)

Posted May 01, 2016 12:00 AM
Pathfinder Tags: challenge questions Magnets

This month's Challenge Question: Specs & Techs from IHS Engineering360:

You have two steel bars. One is a permanent magnet and the other is not magnetized. Without using any tool, meter or instrument of any type, how can you find out which one is the permanent magnet?

And the answer is:

See the figure below. If you arrange the two bars in a T shape by putting the permanent magnet as the top of the T (second figure) and the non-magnetized bar exactly at the middle there is no way the magnet will magnetize the bar, because there is no direct unique pole in contact with the bar. So, there is no attraction between the bars.

On the other hand, if the non-magnetized bar is the top of the T structure, then a clear pole (in the first figure and in this case, it is the south pole) will be in contact with the bar and it will magnetize the bar by creating a North pole in the bottom of the bar and a South pole in the top. Therefore, in this condition the two bars will attract.

46 comments; last comment on 05/24/2016
View/add comments

Gliese 581g: Newsletter Challenge (April 2016)

Posted April 01, 2016 12:00 AM
Pathfinder Tags: challenge questions

This month's Challenge Question: Specs & Techs from IHS Engineering360:

Earth has sent a team of scientists to explore Gliese 581g, a super-Earth exoplanet orbiting a Red Dwarf and located 20 light-years from Earth. Upon approach they marvel that Gliese 581g is almost a perfect sphere considering earlier data. They decide to land in the equatorial region to save fuel. Why?

And the answer is:

The scientists were surprised to learn that Gliese 581g was spherical because of previous data which had told them that it was spinning incredibly rapidly. So rapidly in fact that the centrifugal force at the equator is strong enough to measurably weaken the effect of gravity at its equator, reducing the fuel needed for landing and lift off.

A similar yet much smaller effect occurs on Earth, where the rotation at the equator generates a centrifugal force that reduces effective gravity by one third of one percent.

17 comments; last comment on 04/19/2016
View/add comments

Liquid in Motion: Newsletter Challenge (March 2016)

Posted February 29, 2016 4:59 PM
Pathfinder Tags: challenge questions

This month's Challenge Question: Specs & Techs from IHS Engineering360:

Entering a laboratory, you notice a motionless clear cylinder resting on a table that is filled with a liquid circulating around inside. After hours of observation, you are puzzled to find that the liquid's motion has not changed. Knowing that no outside force or energy has entered the cylinder since you began observing, how is this possible?

And the answer is:

The liquid in the cylinder is superfluid helium-4.

Helium becomes a liquid when cooled to a temperature below its boiling point of 4.22 K (-268.93° C). Taken down a couple more degrees past its "lambda point" of 2.17 K (-270.98° C), liquid helium starts to exhibit unusual properties. It is at this point that a fraction of the helium has transitioned into a "superfluid." In this curious state of matter, the helium behaves like a fluid with zero viscosity. With no friction to slow its motion, the superfluid helium flows past the surface of the cylinder, continuing to circulate endlessly (or until it warms up and transitions back into its more regular liquid or gaseous states).

Unlike most liquids, helium doesn't turn into a solid when cooled down. Its atoms are light and weakly attracted to each other, so even when cooled to temperatures at which regular heat vibrations are absent, helium doesn't settle into a solid. At low temperatures its atoms wiggle with zero-point motion, a slight momentum bestowed by the quantum uncertainty principle. Instead of settling in a solid state, liquid helium undergoes a transformation known as Bose-Einstein condensation. Its atoms start acting in harmony, behaving like one big particle, no longer colliding together. It is these quantum effects that grant superfluids their remarkable properties.

For more on superfluid helium, including a detailed description of how superfluid helium is actually a mixture, with normal and superfluid components, see The physics of superfluid helium[pdf].

40 comments; last comment on 04/10/2016
View/add comments

Baseball Spin: Newsletter Challenge (February 2016)

Posted February 01, 2016 12:00 AM
Pathfinder Tags: baseball challenge questions

This month's Challenge Question: Specs & Techs from IHS Engineering360:

Suppose you are a professional baseball batter (maybe Alex Rodriguez?). You are looking at the pitcher (Sandy Koufax?), and you claim that you can see the spin on the ball throughout its trajectory from its release from the pitcher's hand to when you strike the ball with the bat. Is this true or it is simply a "ball-park" lie?

And the answer is:

Let's say only that you are not lying, but you are exaggerating. It is almost impossible to track the trajectory of the ball at the speed at which a professional pitcher pitches it (not to mention the spinning). The fastest pitched baseball on record is around 105 miles per hour or almost 47 meters per second.

Let's assume, although it's lower than typical speeds in professional baseball, a typical speed of 60 miles per hour (or 27 meters per second). For you to follow the trajectory of this ball your head has to turn at an angular speed of around 500 degrees per second, a feat that is physically impossible for a human.

35 comments; last comment on 02/23/2016
View/add comments

Handel's Tune: Newsletter Challenge (January 2016)

Posted January 01, 2016 12:00 AM
Pathfinder Tags: challenge questions

This month's Challenge Question: Specs & Techs from IHS Engineering360:

A time traveler went back to 1742 and kidnapped George Frideric Handel so he could bring him to the Kennedy Center in NYC in December of 2015 to enjoy his masterpiece Messiah, played by the National Symphony Orchestra. As soon as the first notes are played Handel cringes and shakes his head. What's wrong?

And the answer is:

Handel doesn't like the pitch, which to his trained ear sounds sharp. Modern orchestras use a tuning standard of 440 Hz for A above middle C. However, this has not always been the case. This standard has varied by as much as 50 Hz over the years. A tuning fork from 1740 associated with Handel has been found to have a frequency of 422.5 Hz. Handel's Messiah was composed in 1741 and likely followed this tuning convention. Thus the modern orchestra would sound sharp to Handel. Still, once Handel adjusted to the higher pitch, he no doubt would have appreciated the National Symphony Orchestra's playing of his work. Music has much less to do with pitch than it has to do with the intervals (differences between pitches).

44 comments; last comment on 02/23/2016
View/add comments

Three Spheres: Newsletter Challenge (December 2015)

Posted November 30, 2015 12:00 AM
Pathfinder Tags: challenge questions

This month's Challenge Question: Specs & Techs from IHS Engineering360:

Consider three identical metal spheres, with diameters of 20 cm. Arrange them in a straight line, as shown. Each two consecutive spheres are connected by an extremely small diameter conducting wire.

If the potential at the center of each sphere is the same, determine the charge in each sphere. The sum of the three charges is Q.

And the answer is:

The potential at a charged sphere is given by

where R is the radius of the sphere.

Because of the symmetric arrangement spheres A and C must have the same charge, if both have the same potential. Let q be this charge; so qA = qC = q

Let the charge at sphere B be qB. Then the potential at its center is given by

The potential at each spheres A and C is given by

We also know that


So equating the first two equations yields



but from the total charge equation, we have

Substitute this into the previous equation and solve for q to get



16 comments; last comment on 01/27/2016
View/add comments

Previous in Blog: Tropical Cyclone No Fly Zone: Newsletter Challenge (November 2015)  
Show all Blog Entries in this Blog


New Privacy Policy

We have adopted new policies. Please read each one carefully.