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Four identical bricks, each of length l, are put on the top of one another in such a way that part of each extends beyond the one beneath. For the largest equilibrium extensions,

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The total over hanging length from the edge of the bottom brick is $\frac{11}{12} l$

Second brick from top over hanging the one below l/4

Third brick from top over hanging by bottom one by l/6

Top brick over hanging the one below by l/2

answer is A, B, C, D.

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Detailed Solution

For nth block $X_{n} = \frac{1}{2 n}$

If n = 1, $X_{1} = \frac{l}{2}; n = 2, X_{2} = \frac{l}{4}; n = 3, X_{3} = \frac{1}{6}$

Total over hanging length $= \frac{l}{2} + \frac{l}{4} + \frac{l}{6} = \frac{11 l}{12}$

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