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QJEEAdv20Phf56d: JEE Advanced 2020 Physics - Class 11 JEEAdv20Phf56d Solution: When water is filled carefully in a glass, one can fill it to a …

When water is filled carefully in a glass, one can fill it to a height h above the rim of the glass due to the surface tension of water. To calculate h just before water starts flowing, model the shape of the water above the rim as a disc of thickness h having semicircular edges, as shown schematically in the figure. When the pressure of water at the bottom of this disc exceeds what can be withstood due to the surface tension, the water surface breaks near the rim and water starts flowing from there. If the density of water, its surface tension and the acceleration
due to gravity are 103 kg m−3 , 0.07 Nm−1 and 10 ms−2 , respectively, the value of h (in mm) is _________. JEE Advanced 2020 Paper 1 Offline Physics - Properties of Matter Question 32 English

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JEE Advanced 2020 Paper 1 Offline Physics - Properties of Matter Question 32 English Explanation

Pressure at the bottom of disc = Pressure due to surface tension

ρgh=T(1R1+1R2)\rho gh = T\left( {{1 \over {{R_1}}} + {1 \over {{R_2}}}} \right)

{R_1} > > > {R_2}

So, {1 \over {{R_1}}} < < < {1 \over {{R_2}}} and R2=h/2{R_2} = h/2

∴\therefore ρgh=T(1R1+1R2)=T(0+1h/2)\rho gh = T\left( {{1 \over {{R_1}}} + {1 \over {{R_2}}}} \right) = T\left( {0 + {1 \over {h/2}}} \right)

h2=2Tρg⇒h=2Tρg{h^2} = {{2T} \over {\rho g}} \Rightarrow h = \sqrt {{{2T} \over {\rho g}}}

⇒ =2×0.07103×10=14×100104×100\Rightarrow \, = \sqrt {{{2 \times 0.07} \over {{{10}^3} \times 10}}} = \sqrt {{{14 \times 100} \over {{{10}^4} \times 100}}}

h=14h = \sqrt {14} mm = 3.741

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