Question

A liquid rises to a height of $1.8 \,cm$ in a glass capillary ‘$A$’. Another glass capillary ‘$B$’ having diameter $90\%$ of capillary ‘$A$’ is immersed in the same liquid. The rise of liquid in capillary ‘$B$’ is

• A. $1.4\, cm$
• B. $1.8\, cm$
• C. $2.0\, cm$
• D. $2.2\, cm$

Answer 1

Answer: (C)

Height of liquid in a capillary tube, $h=\frac{2 S \cos \theta}{rp g}$
where, $S=$ surface tension
$\rho=$ density of the liquid
$r=$ radius of the capillary tube
For capillary $A, \,\,\,h_{A}=\frac{2 S \cos \theta}{r_{A} \rho g} \,\,\,\,\,..(i)$
For capillary $B, \,\,\, h_{B}=\frac{2 S \cos \theta}{r_{B} \rho g}\,\,\,\,...(ii)$
From Eqs. (i) and (ii), we get
$h_{A} r_{A}=h_{B} r_{B}$
$\Rightarrow \,\,(1.8\, cm )\left(r_{A}\right)=\left(h_{B}\right)\left(\frac{90}{100} \times r_{A}\right)$
or $\,\,\,(1.8 \,cm )=h_{B} \times \frac{9}{10}$
or $h_{B}=\frac{10}{9} \times 1.8 \,cm =2 \,cm$