Question

The difference of two liquid levels in a manometer is $10cm$ and its density is $0.8gm/c{m}^3$. If the density of air is $1.3\times {10}^{3}gm/c{m}^3$ then the velocity of air will be (in $cm/s$)

• A. $347$
• B. $34.7$
• C. $3470$
• D. $0.347$

Answer 1

Answer: (C)

$3470$

The manometer here gives the stagnation pressure i.e. the velocity of liquid is zero at point B.Hence the pressure drop is given by $\Delta P=\rho gh=P_B-P_A\rho =0.8gm/{cm}^3=800kg/m^3,h=0.1m,{\rho }_{air}=1.3\times {10}^{3}gm/{cm}^3=1.3kg/m^3\Delta P=800\times 9.81\times 0.1\Delta P=784.8N/m^2$

From Bernoulli's equation for stagnation pressure

$P_A+\frac{{\rho }_{air}{V}^2}{2}=P_{B}V=\sqrt{\frac{2\Delta P}{{\rho }_{air}}}V=\sqrt{\frac{2\times 784.8}{1.3}}V=34.74m/sV\cong 3470cm/s$