$rho$ is equal to $""mass"/"volume"$ so in our case the mass is $15g$ while the volume should be: $30-20=10ml=10cm^3$ The density of the material of the bead should then...
A more interesting question is the value of the nugget (more interesting to me anyway). As a heavy metal, gold is incredibly dense, and has $rho=19.3*g*mL^-1$; this is almost twice...
Use the formula to determine the volume of the piece of metal. $"density"="mass"/"volume"$ Rearrange the equation to isolate volume. $"volume" ="mass"/"density"$ $"volume"=(147 cancel"g")/(7.00cancel"g"/"mL")="21.0 mL"$ The final volume in the cylinder...
By definition, $"density"$, $rho$ $-=$ $"Mass"/"Volume"$, and we know that this $227*g$ displaces an $100*mL$ volume of water.... And thus.......... $rho=(227*g)/((150-50)*mL)=2.27*g*mL^-1$
Do you agree? The copper displaces the given volume of water. Now $rho_"Cu"=8.90*g*cm^3$ OR $rho_"Cu"=8.90*g*mL^-1$, i.e. $1*mL-=1*cm^3$ But by definition, $rho_"density"="mass"/"volume"$ And thus $"mass"=rhoxx"volume"=8.90*g*cancel(mL^-1)xx23.4*cancel(mL)$ $=208.3*g$. Do you follow?