Challenge Questions

Welcome to October edition of Monthly Challenge Question from Specs & Techs by GlobalSpec:

You and your wife are out admiring the foliage in the late, cool days of fall in the Northeast U.S. Your wife says it feels cold; you tell her she must be getting sick because you are finding the temperature very comfortable. Who is correct?

The Answer will be posted right here on CR4 on November 4th. Can't wait that long? Well, check out these weekly challenges from CR4.

Mirror Mirror: CR4 Challenge (09/30/08)

Battery Powered Light Bulb: CR4 Challenge (09/23/08)

The Grazing Goat: CR4 Challenge (09/15/08)

This week's CR4 Challenge Question:

You are 4 ft. in front of a plane mirror, and you notice that you can see a full-size image of yourself in the mirror. What is the minimum length of the mirror?

Look for the answer right here on CR4, on October 7th!

This week's CR4 Challenge Question:

A light bulb is connected to the power line by a coiled wire. You insert an iron cylinder inside the coil of the coiled wire. How the light of the light bulb will be affected? Now you repeat this experiment, but this time you power the light bulb with a very powerful battery. What happens to the light of the light bulb?

Disclaimer: This is a theoretical challenge for discussion only – DO NOT ATTEMPT THIS EXPERIMENT! It is potentially dangerous.

And The Answer is....

In the first case, the light bulb will dim. Because the light bulb is connected to an AC source (the line voltage) the iron cylinder will produce a variable magnetic field which will produce a a voltage drop in the coil. Because, the power source is constant (110 V rms), by having a voltage drop in the coil, the voltage across the light bulb will diminish.

In the second case, the power source is DC voltage. The dc current flowing through the coil will not produce a variable magnetic field. Without a variable magnetic field in the coil no voltage drop will develop across it. Therefore, in this case the brightness of the light bulb remains unchanged.

This week's CR4 Challenge Question:

A goat is placed in a fenced-in circular field with radius r. It is tethered to a point on the perimeter of the field by a rope which allows it to graze as far as the radius R. What is the value of R to the nearest cm, if the goat can graze exactly ½ the circular field area, and r = 10m

Thanks to SlideRuler for this question!

And the Answer Is....

The grazing area is segment ABC + segment ABD

Therefore ½*r²*(2θ-sin(2θ))+½*R²*(2Φ-sin(2Φ))=½*π*r². ... eq(1)

also we have θ+Φ+Φ = π (isosceles Δ origin,C,A) ... eq(2)

and r*sin(θ) = R*sin(Φ) (both equal to ½*AB) ... eq(3)

Replacing 2Φ with (θ-π); and replacing R with r*sin(Φ)/sin(θ); and using the relationship sin²(Φ) = ½*(1-cos(2Φ)) = ½*(1-cos(π-θ)) = ½*(1+cos(θ)) in equation (1) gives

(π-θ-sin(θ))+(2θ-sin(2θ)-π)*(1+cos(θ)/(2*sin²(θ))=0 ... eq(4)

This has only one variable θ, and the root can be found using Newton's approximation. x_n+1 = x_n - y_n/y'_n

Using a starting value of θ=1 gives successively

θ =1 --> 1.2143 --> 1.2327 --> 1.2354 --> 1.2358 --> 1.2359

and the corresponding values of R by using eq(2) to get Φ and eq(3) to get R; (remembering r=10m) are

R= 10m --> 11.41m -->11.56m --> 11.58m --> 11.59m --> 11.59m

The answer to the given problem is therefore 11.59m

This week's CR4 Challenge Question:

How many people must be in a room in order for the probability to be greater than 1/2 that at least two of them have the same birthday? (By "same birthday", we mean the same day of the year; the year may differ.) Ignore leap years.

Thanks to Maths_Physics_Maniac for this question!

Answer:

Given n people, the probability, Pn, that there is not a common birthday among them is

The first factor is the probability that two given people do not have the same birthday. The second factor is the probability that a third person does not have a birthday in common with either of the first two. This continues until the last factor is the probability that the nth person does not have a birthday in common with any of the other n - 1 people.

We want Pn < 1/2. If we simply multiply out the above product with successive values of n, we find that P22 = 0.524, and P23 = 0.493

Therefore, there must be at least 23 people in a room in order for the odds to favor at least two of them having the same birthday.