Question

Consider a small surface area of 1 mm2 at the top of a mercury drop of radius 4.0 mm. Find the force exerted on this area (a) by the air above it (b) by the mercury below it and (c) by the mercury surface in contact with it. Atmospheric pressure = 1.0 × 105 Pa and surface tension of mercury = 0.465 N m−1. Neglect the effect of gravity. Assume all numbers to be exact.

Answer 1

(a). The force exerted on this area is 0.10 N.

(b). The force on the area by mercury is 0.10023 N.

(c). The force by the mercury surface in contact with it is 0.00023 N.

Explanation:

Given that,

Area of small surface = 1 mm²

Radius of drop = 4.0 mm

(a). We need to calculate the force exerted on this area

Using formula of force

$P = \dfrac{F}{A}$

$F=P\times A$

Where, P = pressure

A = area

Put the value into the formula

$F=1.0\times10^{5}\times1\times10^{-6}$

$F=0.10\ N$

(b). We need to calculate the excess pressure inside the drop

Using formula of excess pressure

$\Delta P=\dfrac{2S}{r}$

Where, S = surface tension

Put the value into the formula

$\Delta P=\dfrac{2\times0.465}{4.0\times10^{-3}}$

$\Delta P=232.5\N/m^2$

We need to calculate the total pressure inside the drop

$P=\Delta P+P_{atm}$

$P=232.5+1.0\times10^{5}$

We need to calculate the force on the area by mercury

Using formula of force

$F=P\times A$

Put the value into the formula

$F=(232.5+1.0\times10^{5})\times1\times10^{-6}$

$F=0.10023\ N$

(c).  the mercury surface in contact with the area has a net pressure

We need to calculate the net pressure

Using formula of net pressure

$P=P_{m}-P_{atm}$

$P=232.5+1.0\times10^{5}-1.0\times10^{5}$

$P=232.5\ N/m^2$

We need to calculate the force by the mercury surface in contact with it

Using formula of force

$F=P_{net}\times A$

Put the value into the formula

$F=232.5\times1\times10^{-6}$

$F=0.00023\ N$

Hence, (a). The force exerted on this area is 0.10 N.

(b). The force on the area by mercury is 0.10023 N.

(c). The force by the mercury surface in contact with it is 0.00023 N.

Learn more :

Topic : Force

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Answer 2

Answer:

The area of the given surface, A = 1 mm² =1x10⁻⁶ m²

(a) The force exerted by the air above it = A x atmospheric pressure  

= 1x10⁻⁶ m²*1.0x10⁵ N/m²  = 0.10 N  

(b) The surface tension of the mercury, S =0.465 N/m  

The excess pressure inside the drop, Δp = 2S/r  

Δp = 2*0.465/(4/1000) = 465/2 = 232.5 N/m²  

Total pressure inside the drop =Δp + Atmospheric pressure  

= 232.5 +1.0x10⁵ N/m²  

Force on the area by the mercury =A x mercury pressure  

= 1x10⁻⁶(232.5 +1.0x10⁵) N/m²

= 2.325x10⁻⁴+0.10 N/m²

= 0.00023 + 0.10 N/m²

= 0.10023 N  

(c) The mercury surface in contact with the area has a net pressure equal to the pressure in mercury minus the atmospheric pressure outside.

i.e. excess pressure inside the drop = 2S/r  

= 2*0.465/(4/1000) N/m²

= 465/2 N/m²  =232.5 N/m²

Hence the force = Area x Pressure  

= 1x10⁻⁶*232.5 N  

= 0.00023 N