Pressure in Fluids and Atmospheric Pressure — Question 11

(a) As we know,

Pressure (P) = $\dfrac{\text {Thrust (F)}}{\text {Area (A)}}$

and by the principle of hydraulic machine

Pressure on neck = pressure on bottom of bottle

Hence,

$\dfrac{\text F_{1}}{\text A_{1}} = \dfrac{\text F_{2}}{\text A_{2}} \\[0.5em]$

Given,

Diameter of neck (d₁) = 2 cm

Hence, A₁ = 𝜋 $(\dfrac{2}{2})^{2}$ = 𝜋

Diameter of bottom of bottle (d₂) = 10 cm

Hence, A₂ = 𝜋 $(\dfrac{10}{2})^{2}$ = 25 𝜋

Force applied on the cork in the neck (F₁) = 1.2 kgf

Substituting the values in the formula above we get,

$\dfrac{1.2}{\text π} = \dfrac{\text F_{2}}{25\text π} \\[0.5em] \text F_{2} = 1.2 \times 25 \\[0.5em] \text F_{2} = 30 \text{ kgf} \\[0.5em]$

Hence, force exerted at the bottom of the neck = 30 kgf.

(b) The Pascal's law is applied to solve the part (a) which states that the pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.

Hence,

$\dfrac{\text F_{1}}{\text A_{1}} = \dfrac{\text F_{2}}{\text A_{2}} \\[0.5em]$