Question

A ring cut with an inner radius $4.85\, cm$ and outer radius $4.95\, cm$ is supported horizontally from one of the pans of a balance so that it comes in contact with the water in a vessel. If surface tension of water is $70 \times 10^{-3}\, Nm^{-1}$, then the extra mass in the other pan required to pull the ring away from water is

• A. 2 g
• B. 3 g
• C. 4.4 g
• D. 15 g
• E. 10 g

Answer 1

Answer: (C)

Given, $r_{1}=4.85 \,cm , \, r_{2}=4.95 \,cm $ and
$T=70 \times 10^{-3} \,Nm ^{-1}$
Here, $mg=$ force due to surface tension
$m g=2 \pi\left(r_{1}+r_{2}\right) \times T $
$ m=\frac{2 \pi\left(r_{1}+r_{2}\right) \times T}{g} $
$m=\frac{2 \times 3.14 \times(4.85 \times 4.95) \times\left(10^{-2}\right) \times 70 \times 10^{-3}}{9.8} $
$ m=\frac{2 \times 3.14 \times 9.8 \times 10^{-2} \times 70 \times 10^{-3}}{9.8} $
$m=6.28 \times 70 \times 10^{-5}$
$=439.6 \times 10^{-5} \cong 4.4 g $