What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20 °C) is 4.65 × 10^{–1} N m^{–1}. The atmospheric pressure is 1.01 × 10^{5} Pa. Also give the excess pressure inside the drop.

Asked by Pragya Singh | 1 year ago | 64

Given:

Surface tension of mercury, S = 4.65 × 10^{-1} N m^{-1}

Radius of the mercury drop, r = 3.00 mm = 3 × 10^{-3} m

Atmospheric pressure, P_{0} = 1.01 × 10^{5} Pa

We know:

Total pressure inside the mercury drop

= Excess pressure inside mercury + Atmospheric pressure

=\( \dfrac{2S}{r}\) + P_{0 }= \( \left (\dfrac{ 2 \times 4.65 \times 10^{ -1 } }{ 3 \times 10^{ -3 } } \right ) + 1.01 \times 10^{ 5 }\)

\( = 1.0131 \times 105 Pa\)

= 1.01 x 10^{5} Pa

Excess pressure = \( \dfrac{ 2S }{ r }\)

=\( \left (\dfrac{ 2 \times 4.65 \times 10^{ -1 } }{ 3 \times 10^{ -3 } } \right )= 310 Pa\)

Answered by Abhisek | 1 year ago**Answer the following questions:-**

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**(b)** A zeppelin of volume 1500 m^{3} is filled with helium and it is lifting a mass of 400 kg. Assuming that the radius of the zeppelin remains constant as it ascends. How high will it rise?[ y_{0} = 8000 m and ρH_{e} =0.18 kg m^{-3}].

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