Question

An ornament weighing 50 g in air weighs only 46 g is water assuming that some copper is mixed with gold to prepare the ornament.Find the amount of copper in it.Specific Gravity of gold is 20 and that of copper is 10.​

Answer 1

Solution

mass of copper [m(cu)]=m gm

mass of gold [m(au)]=(50-m)gm

now....

volume of copper[v(cu]=(m/10) cm³

volume of gold [v(au)]=[(50-m)/20] cm³

now.....

loss in weight=upthrust

=>(50-46)×g=[v(cu)+v(au)]×1×g

[as, specific density of water =1gm/cm³]

$= > 4 = \frac{m}{10} + \frac{50 - m}{20} \\ = > 4 = \frac{2m + 50 - m}{20} \\ = > m + 50 = 80 \\ = > m = 80 - 50 \\ = > m = 30 \: \: gm$

therefore amount of copper in it is 30 gm

Answer 2

$\huge{\underline{\sf{\red{Answer\leadsto 30g}}}}$

_____________________________

GiveN:

$Ornament\:Weight _{initial}=46g \\ Ornament\: Weight _{final} = 50g$

$\rho_{sp(gold) }= 20 \rho_{sp(cu)}=10$

_____________________________

SoluTion:

Let M be the mass of copper in ornament then mass of gold in it is (50 - m)g

Volume of copper $(V_1) =\dfrac{m}{10} \: \: \bigg(volume = \dfrac{mass}{density}\bigg)$

Volume of gold$(V_2) = \dfrac{50-m}{20}$

Now this when immersed in water

$\implies$ $\rho _w = 1g/cm^3)$

Decrease in weight = upthrust

$\therefore (50-46)g=(V_1+V_2)\rho_w g \\ 4 = \dfrac{m}{10} + \dfrac{50-m}{20} \\ 80=2m+50-m \\ \\ \boxed{m= 30g}$