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One end of a long iron chain of linear m...

One end of a long iron chain of linear mass density `lambda` is fixed to a sphere of mass `m` and specific density `1//3` while the other end is free. The sphere slong with the chain is immersed in a deep lake. If specific density or iron is `7`, the height `h` above the bed of the lake at which the lake at which the sphere will float in equillibrium is (Assume that the part of the chain lying on the bottom of the lake exerts negligible force on the upper part of the chain.).

Text Solution

Verified by Experts

The correct Answer is:
`(7m)/(3lambda)`
Weight of sphere `+` chain `= (m + lambdah)g`
Buoyant force `= (3m + (lambdah)/(7))g`
for equilibrium, weight `=` Buoyant force or, `m + lambdah = 3m + (lambdah)/(7)` or `h = (7m)/(3lambda)`
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