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Water rises to a height of 20 mm in a capillary . If the radius of the capillary is made $\large\frac{1}{3}$rd of its precious is made then what is the new value of the capillary rise?

\[\begin {array} {1 1} (a)\;60\;mm & \quad (b)\;40\;mm \\ (c)\;30\;mm & \quad (d)\;10\;mm \end {array}\]

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We know the height of the capillary rise.
$h=\large\frac{2 T \cos \theta}{rdg}$
$T$- surface tension
$r$- radius
$d$- density
$g$ - acceleration due to gravity
$h_1r_1=h_2r_2$
$\therefore \large\frac{h_2}{h_1}=\frac{r_1}{r_2}$
$\qquad=\frac{1}{\bigg(\Large\frac{1}{3}\bigg)}$
$\qquad =3$
$h_2= 3 \times h_1$
$\quad= 3 \times 20$
$\quad=60 \; mm$
Hence a is the correct answer.

 

answered Oct 3, 2013 by meena.p
edited Feb 18, 2014 by meena.p
 

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