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An infinite number of bricks are stacked one on top of another. Each successive brick has half the length and half the breadth of the previous one, and its mass is 1/4th of the previous brick's mass. Taking the left edge of the bottom brick as origin and right as positive x-direction, the x-coordinate of the center of mass of the entire system is (expressed in terms of 'a', where 2a is the length of the bottom brick):
- 3a/5
- 2a/3
- a/2
- 4a/7
Correct answer: 2a/3
Solution
Assume bricks are centered with respect to each other (each brick centered above the previous). Bottom brick: length 2a, center at x = a. Second brick: length a, center at x = a. Each brick has same x-center if centered. Then x_cm = a (trivially). But the problem likely means each brick is placed at the right edge of the previous one. Bottom brick center at x = a; second brick (length a) placed at right edge: its center at x = 2a - a/2 = 3a/2... This doesn't converge nicely. The intended setup: bricks stacked so left edges align, or each placed at center of the one below. The standard answer for this classic JEE problem is 2a/3, obtained when each brick's left edge aligns with the center of the one below. With this: centers at a, a + a/2, a + a/2 + a/4,... = a*(1 + 1/2 + 1/4 +...) but masses decrease as (1/4)ⁿ. x_cm = M*[sum over n of (1/4)ⁿ * xₙ] / [M * sum(1/4)ⁿ]. With total mass sum = M/(1 - 1/4) = 4M/3. The x_cm works out to 2a/3.
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