A rod with rectangular cross section oscillates about a horizontal axis passing through one of its ends and it behaves like a seconds pendulum, its length will be
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Because ocillating rod behaves as a second's pendulum so its time period will be `2` second. `T = 2pisqrt((l + (K^(2))/(l))/(g)) = 2s rArr l + (K^(2))/(l) = 1"….."(i)[:' pi^(2) = g]` Assume length of rod is L, because axis passes through one end So `l = (L)/(2)` and `K^(2) = (L^(2))/(12)`
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