A crown made of gold and copper weights $210 g$ in air and $198 g$ in water. The weight of gold in crown is-[given : density of gold $= 19.3 g m / c m^{3}$ and density of copper $= 8.5 g m / c m^{3}$ .]

93 g

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100 g

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150 g

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193 g

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Solution

The correct option is D $193 g$

Let volume of gold be $V_{g}$ and volume of copper be $V_{c}$
$19.3 V_{g} + 8.5 V_{c} = 210 . . . . . . . . . . . . . . . . . . . . . (1)$
Buoyancy force in water results in a loss of $12$ gm.
Buoyancy force $(B) = 12 g m \times g (9.8 m / s^{2})$
$B = (V_{g} + V_{c}) g = 12 g$
So, $(V_{g} + V_{c}) = 12 . . . . . . . . . . . . . . . . . (2)$
On, solving the two equations,
$V_{g} = 10; V_{c} = 2$
Weight of gold $= 10 \times 19.3 = 193 g m$

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19.3 g c m^{- 3}

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= 19.3 {g/cm}^{3}]

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