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Question

A sphere of solid material of relative density 9 has a concentric spherical cavity and just floats in water. If the radius of the sphere be R, then the radius of the cavity (r) will be related to R as:


A
r3=89R3
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B
r3=23R3
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C
r3=83R3
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D
r3=23R3
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Solution

The correct option is A $$r^3 = \displaystyle \frac {8}{9} R^3$$
According to law of floatation, a body floats if weight of the body is equal to the weight of water displaced
$$\Rightarrow W_{ s }=W_{ w }\\ \Rightarrow V_{ s }\times \rho _{ s }\times g=V_{ w }\times \rho _{ w }\times g\\ \Rightarrow \frac { 4 }{ 3 } \pi \left( R^{ 3 }-r^{ 3 } \right) \times 9\rho _{ w }=\frac { 4 }{ 3 } \pi R^{ 3 }\times \rho _{ w }\\ \Rightarrow 9\left( R^{ 3 }-r^{ 3 } \right) =R^{ 3 }\\ \Rightarrow r^{ 3 }=\dfrac { 8 }{ 9 } R^{ 3 }$$

Physics

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