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Surface Energy

A piece of go...

Question

A piece of gold weighs 200gm in air and 150 gm in water. If there is a cavity inside the piece of gold then find its volume (cavity). Gold density = 20.

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Solution

Apparent weight = Weight in air buoyant force
$150 g = 299 g - F_{B} F_{B} = 50 g T o t a l v o l u m e o f t h e d i s p l a c e d w a t e r \times ρ_{W} \times g = 50 g (V_{c} + V_{g}) \times ρ_{W} \times g = 50 g V_{c} = 50 - \frac{200}{ρ_{g}} = 50 - \frac{200}{20} = 40 c m^{3} R e g a r d s$

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