An underwater fish is moving horizontally in sea and is below 10
An underwater fish is moving horizontally in sea and is below 10 m from the surface of water. A pitot tube just place in front of it and along its axis is connected to the two limbs of U-Tube containing mercury. The difference of mercury level is found to be 100 mm. The speed of diver is
Take density of mercury = 13600 kg/m3 and density of sea water = 1020 kg/m3;
A. 2.91 m/s
B. 3.91 m/s
C. 4.91 m/s
D. 1.91 m/s
Please scroll down to see the correct answer and solution guide.
Right Answer is: C
SOLUTION
Concept:
The velocity is determined by measuring the rise of the liquid in the pitot tube.
\({V_{1}} = {C_v}\sqrt {2gh}\)
\(h = x\left[ {\frac{{{s_g}}}{{{s_0}}} - 1} \right] \)
x = differential manometric reading
sg = specific gravity of the manometric fluid
s0 = specific gravity of sea water
Calculation:
Given:
sg = 13.6, s0 = 1.02, x = 0.1 m
Cv = 1 (not given, so assumed as 1)
\(h = x\left[ {\frac{{{s_g}}}{{{s_0}}} - 1} \right] = 0.1\left( {13.33 - 1} \right) = 1.23 m\)
\(\therefore Velocity\ of\ flow = \sqrt {2 \times 9.8 \times 1.23} = 4.91 \ {m}/{s}\)