Question

Barometers are made of two capillary tubes of radii $1.44\times {10}^{-3}m$ and $7.2\times {10}^{-4}$ respectively. If the height of a liquid in narrow tube is $0.2m$ more than the height of it in broad tube, then what will be the exact pressure difference? (Density of liquid $={10}^{3}kg{m}^{-3}$, surface tension $=72\times {10}^{-3}N/m$ and $g=8m{s}^{-2}$).

• A. $1750N{m}^{-2}$
• B. $1860N{m}^{-2}$
• C. $1570N{m}^{-2}$
• D. $1680N{m}^{-2}$

Answer 1

The correct option is B $1860N{m}^{-2}$

Let the pressures in wide and narrow limbs of $P_1$ and $P_2$, respectively. If $R_1$ and $R_2$ be the radii of the meniscus in wide and narrow limb pressure just below the meniscus of wide tub $=P_1-\frac{2T}{R_1}$

and pressure just below the meniscus of narrow limb $=P_2-\frac{2T}{R_2}$

Therefore, the difference in these pressures

$(P_1-\frac{2T}{R_1})-(P_2-\frac{2T}{R_2})=\rho gh$

$P_1-P_2=\rho gh-2T(\frac{1}{R_2}-\frac{1}{R_1})$

For the water and glass surface, taking in the angle of contact $\theta$ to be zero, we have $R_1=\frac{r_1}{\cos \theta }\approx r_1$ and $R_2=\frac{r_2}{\cos \theta }\approx r_2$

where r_1 and r_2 are radii of wide and narrow limbs, respectively.

$P_1-P_2=\rho gh-2T(\frac{1}{r_2}-\frac{1}{r_1})$

$P_1-P_2={10}^3\ast 9.8\ast 0.2-2\times 72\times {10}^{-3}(\frac{1}{7.2\times {10}^{-4}}-\frac{1}{1.44\times {10}^{-3}})$

$P_1-P_2=1.96\times {10}^3-1.44\times {10}^{-1}\left(\frac{2-1}{1.44\times {10}^{-3}}\right)$

$P_1-P_2=1.96\times {10}^3-{10}^2$

$P_1-P_2=1960-100=1860N{m}^{-2}$

Hence option 'B' is correct.