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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For disc `1 = MK^(2) = (MR^(2))/(2) rArr K = (R)/(sqrt(2)), l = R/2` `L = l + (K^(2))/(l) = (R)/(2) + (R^(2))/(2((R)/(2))) = R/2 + R = (3R)/(2) rArr T = 2pisqrt((L)/(g)) = 2pisqrt((3R)/(2g))`
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