Question

The lower end of a capillary tube of dimeter 2 mm is dipped 8 cm below the surface of water in a beaker. The pressure required in the tube to blow a bubble at its end in water is :
$[T_{water}=7.3\times {10}^{-3}N{m}^{-1},{\rho }_{water}=1000kg{m}^{-3}and1atm=1.01\times {10}^{5}Paandg=9.8m{s}^{-2}]$

• A. $2.19\times {10}^{5}Pa$
• B. $1.02\times {10}^{5}Pa$
• C. $4.9\times {10}^{5}Pa$
• D. $7\times {10}^{5}Pa$

Answer 1

The correct option is B $1.02\times {10}^{5}Pa$

The excess pressure in a bubble of gas inn a liquid is given by $\frac{25}{r}$
we know that

$P_0=P_a+hdg$

$P_0=1.01\times {10}^5+0.08\times 1000\times 9.80$

$P_0=1.01784\times {10}^{5}pa$

Now $P_i=P_0+\frac{25}{r}$

$P_i=P_0+\frac{25}{r}{P}_i=1.01784\times {10}^5+\frac{2\times 7.9\times {10}^{-2}}{{10}^{-3}}{p}_i=1.02\times 0^{5}Pa$

Hence (B) option is correct