Question

A cubical block of wood weighing 200 g has a lead piece fastened underneath. Find the mass of the lead piece which will just allow the block to float in water. Specific gravity of wood is 0.8 and that of lead is 11.3.

Answer 1

$\rho$ w = 0.8 gm/cc

= density of wood

${\rho }_{p}b$ = 11.3 gm/cc

= density of lead

$m_w$ = 200g

= mass of wood

Here, $(m_w+m_{p}b)g=(V_w+V_{p}b)\rho \times g$

[where, $\rho$ = density of water]

$\Rightarrow (m_w+m_{p}b)=(\frac{m_w}{{\rho }_w}+\frac{m_{p}b}{\rho pb})\rho$

$\Rightarrow (200+m_{p}b)=(\frac{200}{0.8}+\frac{m_{p}b}{11.3})\times 1$

$\Rightarrow m_{p}b-\frac{m_{p}b}{11.3}$= 250-200

$\Rightarrow \frac{{10.3}_{p}b}{11.3}$ = 50

$\Rightarrow m_{p}b=\frac{50\times 11.3}{10.3}$ = 54.8 gm