shown by blowing a bubble five or six inches in diameter and then remove the pipe from the mouth, when the bubble will rapidly contract driving all the air it contains out through the pipe stem. As compared with other shapes a sphere has the least surface for the same volume of substances, and so, as the film is trying to make its area as small as possible it puts its contents in the form of a sphere. This is well shown in falling drops of rain, which are round because of the surface tension of water, and when this surface tension bears a certain ratio to the mass of water within, one drop will rebound on striking another and it cannot be easily broken up. So that, under ordinary conditions, we are neither deluged by great masses of water coming down from the clouds altogether, nor does the rain fall as fine mist. Even the shot-maker takes advantage of this principle. If his melted lead be separated into small globules, they will be spherical while freely falling and if they fall far enough will harden before they strike the water at the bottom.

The surface tension of water is much greater than that of the soap solution, but the soap solution has another property called surface viscosity by virtue of which it can be enormously stretched as in in blowing a large bubble. Water has this property only in a slight degree.

Let a drop of petroleum fall up

on a basin of water and instantly it will spread over the whole surface. Now, the surface of water acts very much as if a thin elastic membrane were tightly stretched over the water and fastened all around to the sides of the basin and so when the oil broke this, it dragged the oil out with it in all directions.

The old experiment of floating a needle on water is a clear demonstration of this strong surface tension or water-skin.

In the soap solution it is the surface tension which keeps the film stretched on both sides from point to point of its support. This can be nicely shown by preparing a wire as shown in Fig. 2 and ty

violet 760X1012, and as light travels only 185,000 miles in one second, each wave is very short and when white light falls upon the film a part is reflected directly toward the eye, while another part passes through and is reflected from the posterior side of the film, so that whenever the film is of such a thickness that the wave which passes twice through lags one-half a wave length behind that reflected from the front those waves will be destroyed, and we receive no longer the sensation of white light but only that of the other components which have not been destroyed. These bright bands of light appear first at the top where the film is thinnest and gradually spread over all; but in a short time new bands will cease to appear at the top, a light uniform tint will take their place, soon to be succeeded by a gray tint, when the film always breaks. This gray tint indicates a degree of tenuity where the film is one molecule in thickness and to stretch it any further would be to convert the liquid into a gas. There is evidence that the film in the region of the gray was about I-500000000 of an inch thick, which, from other evidence, is about the thickness of a molecule.

ARITHMETIC.

By Ed. M. Mills.

The following problems will not be found too difficult for very ordinary eighth grade pupils, and they will serve herewith to illustrate the

Take A B equal to the greater number, and A E equal to the less; E B will then be equal to 53. Complete the square A B C D, and take B M equal to E B. Then draw E N parallel to B C, and R M parallel to A B.

The square R Q N D is the square of the smaller number; and the two equal rectangles, Q M C N and A E QR, together with the square E B M Q, is the difference between the squares of greater and smaller numbers.

The area of the square E B M Q=53' 2809. Then 10759-2809 =7950, the area of the two equal

4800, for it is the difference of the squares of the two numbers. Removing part K from the figure, there would remain two equal squares, SQ GE and C B MO, and the two equal rectangles, O MD N and A COR, whose combined area is equal to 14400-4800, or 9600. It will now be evident that one of these small squares and one of the rectangles, as C B MO and O M D N, will form a rectangle whose area will be equal to of 9600, or 4800, and whose side BD is 120. Then, C B=4800÷ 120, or 40, the smaller number; and A C=120-40=80, the larger number.

40.

The required numbers are 80 and

Let A C represent the greater of the two numbers, and C B, in the same straight line, represent the smaller. Complete the square A BD E, and its area will be equal to 120 14400. The square RON E is the square of the greater of the two numbers in question, and square C B M O is the square of the smaller number.

On the square R ON E, lay off the square S Q G E, equal to the square C B M O; then will the part marked (K) be equal, in area, to

Solution.

Let A B be the length of the rectangle and A D its breadth, and complete the figure. Divide A B into five equal spaces, and A D into three spaces each equal to one of the spaces of line, A B. At these

points of division draw lines parallel to the sides and ends of the rectangle, thus dividing it into fifteen small squares, as Q. Hence, 15 Q=135 square rods.

Q=1-15 of 135=9 sq. rods, and Aa=V9=3 rods; then

A B=5X3=15 rods, and

A D=3X3=9 rods.

15 rods length of rectangle, and 9 rods breadth.

ANSWERS TO QUESTIONS.

[The following solutions to problems published at the request of a subscriber in November, are the first to reach us. We are also in receipt of solutions by B. F. Finkel, C. L. Martzolff and Anson McKinney, which we are compelled to omit for want of space. - It is not our purpose to go into the "Notes and Queries Business" to any great extent, because we believe that there is something. better for teachers than poring over the conundrums and their answers, which are so often found under the head of "Notes and Queries" in some of the school journals of the day. If we can, however, at any time, help a subscriber who is honestly seeking light on some difficult point which comes up in his work as a teacher, we shall be glad to give a small amount of space for that purpose. -Ed.]

cu. ft.

1225.3+ cu. ft.

Width street = √1225.3+ =35+

ft.