One end of a long iron chain of linear mass density $λ$ is fixed to a sphere of mass $m$ and specific density 1/3 while the other end is free as shown in the figure. The sphere along with the chain is immersed in a deep lake. If specific density of iron is 7, the height $h$ above the bed of the lake at which the sphere will float in equilibrium is $h = \frac{A \times m}{3 λ}$ . Then the value of $A$ 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):

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answer is 7.

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Detailed Solution

Weight of sphere + chain $= (m + λ h) g$
Buoyant force $= (ρ_{l} \frac{m}{ρ_{s}} g + ρ_{l} \frac{λ h}{ρ_{c}} g) = (3 m + \frac{λ h}{7}) g$
For equilibrium,
Weight = Buoyant force
Or, $m + λ h = 3 m + \frac{λ h}{7}$
Or, $h = \frac{7 m}{3 λ}$