91Ó°ÊÓ

i) The mass of the wood block, $m=3.67kg$

ii) The density of wood block,${\mathrm{ÒÏ}}_{wood}=600\frac{\mathrm{kg}}{{\mathrm{m}}^3}$

iii) The density of lead,${\mathrm{ÒÏ}}_{lead}=1.14×{10}^4\frac{\mathrm{kg}}{{\mathrm{m}}^3}$

iv) The submerged volume of a block, $V_{submerged}=0.900m^3$

We can find the lead mass if the lead is fitted to the top of the block by using Archimedes’ principle in which the volume of lead will not contribute to the buoyant force. Also, we can find the lead mass if the lead is fitted to the bottom of the block in which the volume of lead will contribute to the buoyant force.

Formulae:

Force applied on body (or weight),$F_b=m_{f}g$(i)

Density of a substance,$\mathrm{ÒÏ}=\frac{m}{V}$ (ii)

When the lead mass is fitted at the top, its volume does not contribute to the buoyant force.

Using equations (i) & (ii) and the given values, we can get

$$\begin{gathered} F_b={\mathrm{ÒÏ}}_w{V}_{i}g \\ =0.900{\mathrm{ÒÏ}}_w{V}_{wood}g \\ =0.900{\mathrm{ÒÏ}}_{w}g\frac{m_{wood}}{{\mathrm{ÒÏ}}_{wood}}...................................\left(iii\right) \end{gathered}$$

The block and lead is floating on water.Hence, the buoyant force equals to the weight of wood and lead that implies:

$F_b=\left(m_{wood}+m_{lead}\right)g.............................\left(iv\right)$

From equation (iii) and (iv), we get

$$\begin{gathered} 0.900{\mathrm{ÒÏ}}_{w}g\frac{m_{wood}}{{\mathrm{ÒÏ}}_{wood}}=\left(m_{wood}+m_{lead}\right)g \\ 0.900{\mathrm{ÒÏ}}_w\frac{m_{wood}}{{\mathrm{ÒÏ}}_{wood}}=\left(m_{wood}+m_{lead}\right) \\ {m}_{lead}=0.900{\mathrm{ÒÏ}}_w\frac{m_{wood}}{{\mathrm{ÒÏ}}_{wood}}-m_{wood} \\ =0.900×1000\frac{3.67kg}{600{\displaystyle \frac{kg}{m^3}}}-3.67kg \\ =1.835kg \end{gathered}$$

Hence, the mass of the lead is$1.835kg$

When lead is fitted to the block’s bottom, the volume of lead contributes to the buoyant force.

$$\begin{gathered} F_b=0.900{\mathrm{ÒÏ}}_{w}g{V}_{wood}+{\mathrm{ÒÏ}}_{w}g{V}_{lead} \\ =0.900{\mathrm{ÒÏ}}_{w}g\frac{m_{wood}}{{\mathrm{ÒÏ}}_{wood}}+{\mathrm{ÒÏ}}_{w}g\frac{m_{lead}}{{\mathrm{ÒÏ}}_{lead}}..................................\left(v\right) \end{gathered}$$

The block and lead are floating on water.Hence, buoyant force equals to the weight of wood and lead.

$F_b=\left(m_{wood}+m_{lead}\right)g.............................\left(vi\right)$

From equations (v) and (vi), we get

$$\begin{gathered} 0.900{\mathrm{ÒÏ}}_{w}g\frac{m_{wood}}{{\mathrm{ÒÏ}}_{wood}}+{\mathrm{ÒÏ}}_{w}g\frac{m_{lead}}{{\mathrm{ÒÏ}}_{lead}}=\left(m_{wood}+m_{lead}\right)g \\ 0.900{\mathrm{ÒÏ}}_w\frac{m_{wood}}{{\mathrm{ÒÏ}}_{wood}}+{\mathrm{ÒÏ}}_w\frac{m_{lead}}{{\mathrm{ÒÏ}}_{lead}}=m_{wood}+m_{lead} \\ 0.900{\mathrm{ÒÏ}}_w\frac{m_{wood}}{{\mathrm{ÒÏ}}_{wood}}-m_{wood}=m_{lead}-{\mathrm{ÒÏ}}_w\frac{m_{lead}}{{\mathrm{ÒÏ}}_{lead}} \\ 0.900{\mathrm{ÒÏ}}_w\frac{m_{wood}}{{\mathrm{ÒÏ}}_{wood}}-m_{wood}=m_{lead}\left(1-\frac{{\mathrm{ÒÏ}}_w}{{\mathrm{ÒÏ}}_{lead}}\right) \\ {m}_{lead}=\frac{0.900×1000{\displaystyle \frac{kg}{m^3}}×{\displaystyle \frac{3.67kg}{600{\displaystyle \frac{kg}{m^3}}}}-3.67kg}{\left(1-{\displaystyle \frac{1000{\displaystyle \frac{kg}{m^3}}}{1.14×{10}^4{\displaystyle \frac{kg}{m}}}}\right)} \\ =2.01kg \end{gathered}$$

Hence, the value of lead mass is$2.01kg$