Question

A crown made of gold and copper weights $210g$ in air and $198g$in water, the weight of gold in crown is [given: density of gold $= 19.3gm/c{m^3}$and density of copper $= 8.5gm/c{m^3}]$
A. $93g$
B. $100g$
C. $150$
D. $193g$

Answer 1

The difference in weights of crown in air and water is due to buoyancy.
From Archimedes' principle and law of floatation, one can find the apparent weight in water.

b>Complete step by step answer:**
Let the volume of gold be ${V_g}$and volume of copper be ${V_c}$ is crown
${\rho _g}$be density of gold $= 19.3g/c{m^3}$
${\rho _c}$be density of copper $= 8.5g/c{m^3}$
${\rho _w}$be density of water$= 1g/c{m^3}$
So, total actual downward weight $=$ weight of gold $+$weight of copper
Now, we know that weight $= mass \times gravity$
And, density $= \dfrac{{mass}}{{volume}}$
So, weight $= density \times mass \times gravity$
So, weight of gold $+$weight of crown $=$total actual weight
$({\rho _g}{V_g} + {\rho _c}{V_c})g = 210g$
$\Rightarrow {\rho _g}{V_g} + {\rho _c}{V_c} = 210$ …..(i)
Now, in water, the weight reduces to $198g$
This is due to upthrust acting on crown and this weight will be apparent weight which is given by
$Apparent\,\,weight = actual\,\,weight - upthrust$
$198g = ({\rho _g}{V_g} - {\rho _c}{V_c})g - ({V_g} + {V_c}){\rho _w}$
$\Rightarrow ({\rho _g} - {\rho _w}){V_g} + ({\rho _C} - {\rho _w}){V_c} = 198$ …..(ii)
Putting the values of density in equation (i) and (ii)
Equation (i) becomes
${\rho _g}{V_g} + {\rho _c}{V_c} = 210$
$19.3{V_g} + 8.5{V_c} = 210$ …..(iii)
Equation (ii) becomes
$({\rho _g} - {\rho _w}){V_g} + ({\rho _c} - {\rho _w}){V_c} = 198$
$(19.3 - 1){V_g} + (8.5 - 1){V_c} = 198$ …..(iv)
From (iii) ${V_c} = \dfrac{{210 - 19.3{V_g}}}{{8.5}}$
Put in (iv) , $18.3 - 7.5 \times \left( {\dfrac{{210 - 19.3{V_g}}}{{8.5}}} \right) = 198$
$\Rightarrow 155.55{V_g} - 1575 + 144.75{V_g} = 1683$
$\Rightarrow 155.55{V_g} + 144.75{V_g} = 1683 + 1575$
$\Rightarrow 300.3{V_g} = 3258$
${V_g} = \dfrac{{3258}}{{300.3}}$
${V_g} = 10.849$
${V_g} \simeq 10.85c{m^3} \simeq 10c{m^3}$
Weight of gold $= 10 \times 19.3$
$= 193g$

So, the correct answer is “Option D”.

Note:
Weight $= mass \times gravity$ and upthrust always reduces the actual weight. So, when the crown is in water, the concept of upthrust is used.Then after getting equations and solving them, the weight of gold in the crown can be calculated.