By Philip J. Carter
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Additional resources for Challenging IQ Tests
The key to the solution is the fact that the pressure in a liquid is determined by the depth below a free surface where the liquid and the atmosphere are in direct contact. The upper part of the water column inside the tube T is in direct contact with the atmosphere, which gives a constant pressure there. When water ﬂows out from the opening O, the pressure of the air trapped in the upper part of the bottle decreases because it is conﬁned to a volume that becomes larger. Then the upper water level inside the tube is pressed down below that of the water in the bottle.
Rather than immediately trying to solve the problem with seven steps, it is a good idea ﬁrst to consider ladders with only one or two steps. 75R where R is the resistance of a single resistor. The ladder with three steps is only slightly more complicated. 733R For each step that is added, the total resistance gets smaller, but at a rapidly decreasing rate. It seems clear that if we only want a result with an error less than 1 %, we can stop here. In practice, the resistance of a ladder with three or more steps is the same as that of an inﬁnite ladder.
It seems clear that if we only want a result with an error less than 1 %, we can stop here. In practice, the resistance of a ladder with three or more steps is the same as that of an inﬁnite ladder. To make that point even more clear, we derive a recursion relation that relates the resistance RN+1 of a ladder with N + 1 steps to that of a ladder with N steps. The ladder with N + 1 steps is constructed by adding one “step unit” to the front end of a ladder with N steps (rather than adding it to the far end of the ladder; see ﬁg.