A drop of liquid pressed between two glass plates spreads to a circle of diameter $10 c m .$ Thickness of the liquid film is $0.5 m m$ and surface tension is $70 \times 10^{- 3} N m^{- 1}$ . The force required to pull them apart is:

0.044 N

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0.011 N

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0.022 N

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0.036 N

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Solution

The correct option is B $0.044 N$
Force due to surface tension is given by:
$F = T l$ where $l$ is the length of contact
Here, $l = 2 π r$
Hence, force on one plate.
$F = 2 π \frac{5}{100} \times \frac{70}{1000} N$
$F = 0.022 N$
Total force required is $F^{'} = 2 F = 0.044 N$

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A drop of liquid pressed between two glass plates spreads to a circle of diameter 10 cm. Thickness of the liquid film is 0.5 mm and surface tension is 70×10−3 Nm−1. The force required to pull them apart is:

The correct option is B 0.044NForce due to surface tension is given by:F=Tl where l is the length of contactHere, l=2πrHence, force on one plate.F=2π5100×701000NF=0.022NTotal force required is F′=2F=0.044N

A drop of liquid pressed between two glass plates spreads to a circle of diameter $10 c m .$ Thickness of the liquid film is $0.5 m m$ and surface tension is $70 \times 10^{- 3} N m^{- 1}$ . The force required to pull them apart is:

The correct option is B $0.044 N$
Force due to surface tension is given by:
$F = T l$ where $l$ is the length of contact
Here, $l = 2 π r$
Hence, force on one plate.
$F = 2 π \frac{5}{100} \times \frac{70}{1000} N$
$F = 0.022 N$
Total force required is $F^{'} = 2 F = 0.044 N$