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(a) The diameter of neck and bottom of a bottle are 2 cm and 10 cm respectively. The bottle is completely filled with oil. If the cork in the neck is pressed in with a force of 1.2 kgf, what force is exerted on the bottom of the bottle ?

(b) Name the law/principle you have used to find the force in part (a)

Fluids Pressure

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Answer

(a) As we know,

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

and by the principle of hydraulic machine

Pressure on neck = pressure on bottom of bottle

Hence,

F1A1=F2A2\dfrac{F{1}}{A{1}} = \dfrac{F{2}}{A{2}} \\[0.5em]

Given,

Diameter of neck (d1) = 2 cm

Hence, A1 = 𝜋 (22)2(\dfrac{2}{2})^{2} = 𝜋

Diameter of bottom of bottle (d2) = 10 cm

Hence, A2 = 𝜋 (102)2(\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,

1.2π=F225πF2=1.2×25F2=30 kgf\dfrac{1.2}{\pi} = \dfrac{F{2}}{25\pi} \\[0.5em] F{2} = 1.2 \times 25 \\[0.5em] 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,

F1A1=F2A2\dfrac{F{1}}{A{1}} = \dfrac{F{2}}{A{2}} \\[0.5em]

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