Q.48. The height of liquid column raised in a capillary tube of certain radius when dipped in liquid ‘A’ vertically is 5 cm. If the tube is dipped in a similar manner in another liquid ‘B’ of surface tension and density double the values of liquid ‘A’, the height of liquid column raised in liquid ‘B’ would be (Assume angle of contact same)
A. 0.20 m
B. 0.5 m
C. 0.05 m
D 0.10 m
Answer :- C. 0.05 m
Explanation :-
Height of liquid column, \mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho g}
Given: \mathrm{h}_1=5 \mathrm{~cm}
Also \mathrm{T}_2=2 \mathrm{~T}_1 ; \rho_2=2 \rho_1
\mathrm{h}_1=\frac{2 \mathrm{~T}_1 \cos \theta}{\mathrm{r} \rho_1 \mathrm{~g}} \mathrm{h}_2=\frac{2 \mathrm{~T}_2 \cos \theta}{\mathrm{r} \rho_2 \mathrm{~g}} =\frac{2\left(2 \mathrm{~T}_1\right) \cos \theta}{\mathrm{r}\left(2 \rho_1\right) \mathrm{g}} =\frac{4 \mathrm{~T}_1 \cos \theta}{\mathrm{r}\left(2 \rho_1\right) \mathrm{g}} \frac{\mathrm{h}_1}{\mathrm{~h}_2}=\frac{5}{\mathrm{~h}_2}=\frac{2 \mathrm{~T}_1 \cos \theta}{\mathrm{r} \rho_1 \mathrm{~g}} \times \frac{\mathrm{r} 2 \rho_1 \mathrm{~g}}{4 \mathrm{~T}_1 \cos \theta} \frac{5}{\mathrm{~h}_2}=1 \mathrm{h}_2=5 \mathrm{~cm}=0.05 \mathrm{~m}