A submarine moves horizontally in sea & has its axis 15m below the surface of water. A pitot tube properly placed infront of submarine & along its axis is connected to the two limbs of a U tube containing mercury. The difference of mercury level in two limbs is 170mm. If sea water is 1.26 times as dense as compared to normal water & take specific gravity of mercury 13.6. The speed of submarine in (m/sec) is

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Solution

The correct option is A 6
h = 170 mm
Density of sea water = 1.26 times density of normal water
$= 1.26 \times 1000 = 1260$

$ρ_{m} = manometric fluid density = 13600 (\frac{k g}{m^{3}})$

$V = \sqrt{\frac{2 Δ P}{ρ}} = \sqrt{\frac{2 \times ρ_{m} g h}{ρ}} = \sqrt{\frac{2 \times 13600 \times 9.81 \times 170 \times 10^{- 3}}{1260}}$

$V = 6 (m / s e c)$

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The correct option is A 6 h = 170 mm Density of sea water = 1.26 times density of normal water =1.26×1000=1260 ρm=manometric fluid density=13600(kgm3) V=√2ΔPρ=√2×ρmghρ=√2×13600×9.81×170×10−31260 V=6(m/sec)