Question

The lower end of a glass capillary tube is dipped in water. Water rises to a height of 8 cm. The angle of contact is $0^{\circ }$. the tube is then broken at a height of 6 cm. The height of water column and angle of contact will be.

• A. $6m,{sin}^{-1}\left(\frac{3}{4}\right)$
• B. $6m,{sin}^{-1}\left(\frac{\sqrt{7}}{4}\right)$
• C. $4m,{sin}^{-1}\left(\frac{\sqrt{7}}{8}\right)$
• D. $6m,0^{\circ }$

Answer 1

The correct option is B $6m,{sin}^{-1}\left(\frac{\sqrt{7}}{4}\right)$

we know that $h=\left(2Tcos\theta \right)/\left(\rho rg\right)$

As radius and content of tube is same.

$T,r,\rho ,g$ = constant

$\therefore h/\left(cos\theta \right)$= constant

$\therefore \left(h1/h2\right)=\left(cos{\theta }_1\right)/\left(cos{\theta }_2\right)$

Initial height = $h1=8cm$

As capillary is broken at height 6 cm, meaning that water will rise to height 6 cm hence $h2=6cm$

$\left(h1/h2\right)=\left(cos0\right)/\left(cos{\theta }_2\right)$ --------- initial angle = 0°

$\therefore \left(8/6\right)=1/\left(cos{\theta }_2\right)$

$cos{\theta }_2=3/4$

${\theta }_2=co{s}^{–1}\left(3/4\right)$

∴ height = $6cm$ & angle = $co{s}^{–1}\left(3/4\right)$

or $si{n}^{–1}\left(\sqrt{7}/4\right)$