Two narrow bores of diameter 5.0 , mm and 8.0 | vedclass

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(B) The pressure at points $A$ and $B$ at the same horizontal level must be equal,so $P_A = P_B$.The pressure just below the meniscus in the two limbs

Vedclass · Accepted Answer

$2.19$

Vedclass · Answer

(B) The pressure at points $A$ and $B$ at the same horizontal level must be equal,so $P_A = P_B$.The pressure just below the meniscus in the two limbs is given by $P_{atm} - \frac{2T}{r_1}$ and $P_{atm} - \frac{2T}{r_2}$ respectively.Equating the pressures at the same horizontal level:$P_{atm} - \frac{2T}{r_1} + ho g(x + \Delta h) = P_{atm} - \frac{2T}{r_2} + ho g x$$ho g \Delta h = 2T \left( \frac{1}{r_1} - \frac{1}{r_2} ight)$Given $r_1 = 2.5 \, mm = 2.5 	imes 10^{-3} \, m$ and $r_2 = 4.0 \, mm = 4.0 	imes 10^{-3} \, m$.$\Delta h = \frac{2T}{ho g} \left( \frac{1}{r_1} - \frac{1}{r_2} ight)$$\Delta h = \frac{2 	imes 7.3 	imes 10^{-2}}{1.0 	imes 10^3 	imes 10} \left( \frac{1}{2.5 	imes 10^{-3}} - \frac{1}{4.0 	imes 10^{-3}} ight)$$\Delta h = \frac{14.6 	imes 10^{-2}}{10^4} 	imes 10^3 \left( \frac{1}{2.5} - \frac{1}{4.0} ight)$$\Delta h = 14.6 	imes 10^{-3} 	imes (0.4 - 0.25) = 14.6 	imes 10^{-3} 	imes 0.15 = 2.19 	imes 10^{-3} \, m = 2.19 \, mm$.

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