The balloon filled with hydrogen is rising because the weight of the balloon filled with hydrogen is less than the weight of air displaced by it i.e. upthrust of the atmosphere is acting on the balloon. As we go up in the atmosphere the density of air decreases that leads to decrease in the weight of displaced air i.e. decrease in the upthrust. As the weight of the hydrogen balloon is same to the weight of air displaced by it, it stops to rise.

Given,

Density of wood(d_{1})= 800kg/m^{3}

Density of water (d_{2})=100kg/m^{3}

Volume of wood(V)=1.6m^{3}

Mass of wood (m)= d_{1}×v

=800×1.6

=1280kg

Now,

Mass of displaced water=mass of wood=1280kg

\begin{align*} \text{Volume of wood, sinking on water}&= \text{volume of displaced water} \\&=\frac{\text{mass of displaced water} }{\text{density of water}}\\ &= \frac{1280kg}{1000kg/m^3}\\ &= 1.28 \: m^3\end{align*}

\begin{align*}\text{Sinking part of wood} &= \frac{\text{Volume of wood sinking on water}}{\text{Volume of wood}}\\ &=\frac{1.28m^3}{1.6m^3}\\&= 0.8 \end{align*}

Here,mass of a brick(m1) = 2 kg

Density of the brick (d_{1}) = 2.5 g/cm^{3} = \(\frac {2.5 × 10^{-3} kg}{10^{-6} m^3}\)

= 2500 kg/m^{3}

∴ Volume of the brick (V_{1}) = \(\frac{mass}{density}\)

= \(\frac{2}{2500}\)

∴ Mass of the displaced water is 0.8 kg.

= \(\frac{1}{1250}\)m^{3}

^{H}ere,mass of brick (m_{1}) = 2 kg

Density of brick(d_{1}) = 2.5g /cm^{3}

= 2500 kg/m^{3}

Density of water (d_{2}) = 1000 kg/m

Mass of water displaced = (m_{2}) = ?

For sinking bodies

V_{1}=V_{2}

or, \(\frac{m_1}{d_1}\)=\(\frac{m_2}{d_2}\) [∴d=\(\frac{m}{v}\) ]

∴ \(\frac{d_1}{d_2}\)=\(\frac{m_1}{m_2}\)

\(\frac{2500}{1000}\)

\(\frac{2}{m_2}\)

∴ m_{2} =\(\frac{1000×2}{2500}\)

= \(\frac{20}{25}\)

= \(\frac{4}{5}\)

= 0.8 kg

∴ Mass of the displaced water is 0.8 kg.

Pascal's law states that liquid exerts pressure equally in all directions.

Given,

Depth of water (h) = 6m

Density of water (d) =1000 kg /m^{3}

Acceleration due to gravity (g) = 9.8/sec^{2}

Pressure of water (p) =?

We know,

p=h× d× g

=6 × 1000×9.8

=58800Pa

Hence, the water pressure exerted at the bottom of the tank is 58800 Pascal.