Why does the shadow change and almost disappear?
Published: Thu 10 Mar 2016, 11:00 PM
Last updated: Fri 18 Mar 2016, 10:02 AM
E4
If you're one of those licensed astrophysicists - you know, the type we call when some physics or the other in our house is not working right, and the plum-ber's sick or the carpenter's on leave - and you then have to repair the hyperspace in a hurry yourself, if you're of THAT calling, don't even try the problem being presented here this week because you'll solve it in a few seconds and keep crowing into the sunset for days. But if you're a person who's just interested in the subject for leisure, and maybe at the most discovers a passing law of thermodynamics or two from time to time, you're welcome.
Meaning, hold on to your hats, then hold a pencil three inches above a white sheet of paper under a fluorescent tube light so that the pencil is parallel to the tubelight and observe its shadow. Now rotate the pencil through 90 degrees. Why does the shadow change and almost disappear?
DEAR MS
(The older problem concerned how steam engine locomotives with crankshafts that could come to rest in any position of the wheels they were attached to, could generate the required torque while starting again. - MS)
Wheels-Within-Wheels Dept:
To overcome the problem, the wheels and crankshaft were so designed that there would always be a difference of 90 degrees between the crankshafts of the left and right wheels. So even if one set of wheels came to a halt with shafts at exactly 3 o'clock and 9 o'clock, the position of the crankshafts of the other set of wheels would be exactly at 12 o'clock or 6 o'clock providing ample torque to start.
Steam locomotives usually have two cylinders, with cranks set at right angles; so one piston is always off the dead centre, so no starting assistance is needed.
(The other problem was about seven diamond thieves trying to divide an undisclosed loot equally among ?themselves without too much success until. - MS)
The-Forever Dept:
Let x be the number of diamonds, where x is a multiple of 7, but leaves a remainder 1 when 2, 3, 4, 5 and 6 are divided. That is, x-1 is divisible by 2, 3, 4, 5 and 6. Or, x-1 is a multiple of the LCM of 2, 3, 4, 5 and 6. LCM = 60. By trial, x-1 = 300 and as 301 is a multiple of 7, x = 301.
Look for a multiple of 7 that is an odd number (to get a remainder when divided by 2) that ends in 1 (to get a remainder of 1 when divided by 5). Divide the multiple successively by 3, 4 and 6 and check to see if the remainder is 1. An iterative trial of 21, 91, 161. ends when 301 fulfils all the requirements.
(Among the first five who also got it correct are: Uma Jaishankar, uma.jaishankar@gmail.com; Elle Rocky, wwefan1068@gmail.com; Harmeet Kohli, harmeet09@gmail.com; Akash Jain, kajain99@gmail.com; Hozefa K Berberawala, hozefakb@gmail.com)
(The third problem was: "527 F in Celsius is 275; in other words, the same number with the first digit moved to the last position. What is the next lowest temperature with this property?" - MS)
Degrees-Of-Freedom Dept:
The solution I got for this problem is a mind boggling figure. (But you nailed it! - MS.) I wonder if it's correct. I am sticking my neck out and sending my answer. It's a 19-digit number 5294117647-058823527 F, which gives 2941176470588235275 C.
ENDGAME(S)
- We have an angle of 1.5°. How big will it look through a magnifying glass that magnifies 3.5 times?
- There is exactly one astronomer on every planet of a certain system. He/she watches the closest planet. ?The number of the planets is odd and all the ?distances are different. Prove there is one planet not being watched.
(To get in touch with Mukul, mail him at mukul.mindsport@gmail.com)
iPhone | iPad
Android
Huawei
