Question

A ring cut with an inner radius $4.85cm$ and outer radius $4.95cm$ 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 $70\times 10^{-3}N{m}^{-1}$, 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, $r_1=4.85cm,r_2=4.95cm$ and
$T=70\times 10^{-3}N{m}^{-1}$
Here, $mg=$ force due to surface tension
$mg=2\pi \left(r_1+r_2\right)\times T$
$m=\frac{2\pi \left(r_1+r_2\right)\times T}{g}$
$m=\frac{2\times 3.14\times (4.85\times 4.95)\times \left(10^{-2}\right)\times 70\times 10^{-3}}{9.8}$
$m=\frac{2\times 3.14\times 9.8\times 10^{-2}\times 70\times 10^{-3}}{9.8}$
$m=6.28\times 70\times 10^{-5}$
$=439.6\times 10^{-5}\cong 4.4g$