Saurabh Kumar
Last Activity: 10 Years ago
We know, centre of mass of the rod of length be ‘L’ would be at L/2,
Placing each rod on preceeding one would shift cenrre of mass by L/2+ X,
If right end of the rod shift by, more than L/2, the whole system would fall, so we can count the no. of rods that can be placed safely..
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