Question bank/JEE Main/Physics

A long capillary tube of mass$'\pi^{'}\ gm$, radius 2 mm and
negligible thickness, is partially immersed in a liquid of surface
tension 0.1 N/m. Take angle of contact zero and neglect buoyant force of
liquid. The force required to hold the tube vertically, will be-
$\left( g = 10m/s^{2} \right)$
eksVkbZ dh,d yEch
i`"B ruko okys æo:i ls Mqcksrs
gSaA s.k ysrs gq, rFkk æo ds mRIykou cy dks ux.;
ekurs cy

A$10.4\ \pi mN$B$10.8\ \pi mN$C$0.8\ \pi mN$D$4.8\ \pi mN$

Show answer and solution▾

The free body diagram of the capillary tube is as shown in the
figure. Net force F required to hold tube is
F = force due to surface tension at cross-section
$\left( S_{1} + S_{2} \right) +$weight of tube.
dsf'kdk uyh dk cy js
fy, dqy cy F gS
F = vuqizLFk dkV ij i`"B ruko j.k cy
$\left( S_{1} + S_{2} \right) +$V~ Hkkj
$= 4\pi \times 2 \times 10^{- 3} \times 0.1 + \pi \times 10^{- 3} \times 10 = 10.8\pi mN$

Source: Grass-G / Ph-26 Physics paper 1 15 Jan.docx