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Question 2.11: A block of steel (sp. gr. 7.85) floats at a mercury water in...

A block of steel (sp. gr. 7.85) floats at a mercury water interface as in Fig. 2.34. What is the ratio of a and b for this condition? (sp. gr. of mercury is 13.57).

2.11
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Let the block have a uniform cross-sectional area A.

Under the condition of floating equilibrium as shown in Fig. 2.34,

Weight of the body = Total Buoyancy force acting on it

A \times(a+b)(7850) \times g=(b \times 13.57+a) \times A \times g \times 10^{3}

Hence 7.85(a+b)=13.57 b+a

or, \frac{a}{b}=\frac{5.72}{6.85}=0.835

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