Question

A glass capillary tube of inner diameter $0.28mm$ is lowered vertically into water in a vessel.The pressure to be applied on the water in the capillary tube so that water level in the tube is same as the vessel in $N/m^2$ is(surface tension of water $=0.07N/m$ atmospheric pressure $={10}^{5}N/m^2$ )

• A. ${10}^3$
• B. $99\times {10}^3$
• C. $100\times {10}^3$
• D. $101\times {10}^3$

Answer 1

Answer: (A)

${10}^3$

Pressure due to h height column of water in tube $\rho gh$

and By formula of rise in capillary tube of $h=\frac{2T\cos \theta }{\rho gr}$

Taking cos$\theta$ = 1

we get Pressure = $\rho gh=\frac{2T}{r}$

$=\frac{2\left(0.07\right)}{\frac{0.28}{2}\times {10}^{-3}}$

= $1\times {10}^{3}N/m^2$

so, pressure due to h height is $1\times {10}^{3}N/m^2$ and

so, to maintain the same level in a vessel as well as in capillary tube applied pressure should be ${10}^{3}N/m^2$