Question

A pitot's tube is attached at one of the wing's of an aeroplane in order to determine its velocity with respect to air. If the difference of two liquid levels in manometer is $10$ cm and density of liquid is $0.8gm/c{m}^3$, then the velocity of plane with respect to air will be (given density of air = $1.293\times {10}^{-3}gm/c{m}^3$).

• A. $34.82cm/s.$
• B. $3.48cm/s.$
• C. $348.2cm/s.$
• D. $3482cm/s.$

Answer 1

The correct option is D $3482cm/s.$

Velocity of liquid/gas as measured from pitot's tube is given by
$v=\sqrt{\frac{2h{\rho }_{l}g}{\rho }}$, where
$h=$ difference in the liquid levels in manometer attached to pitot's tube
${\rho }_l=$ density of liquid is pitot's tube
$\rho =$ density of moving fluid (liquid/gas )
$\Rightarrow v=\sqrt{\frac{2\times 10\times 0.8\times 980}{1.293\times {10}^{-3}}}=3482cm/s$