Water flows through the tube shown in figure. The areas of cross section of the wide and the narrow portions of the tube are 5 cm² and 2 cm² respectively. The rate of flow of water through the tube is 500 cm³ s^-1. Find the difference of mercury levels in the U-tube. Assume this tube lying horizontally on a table.

1.93 cm

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1.93 m

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2.58 cm

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2.58 m

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Solution

The correct option is A
1.93 cm

As clear from the sitution,
$P_{1} - P_{2} =, ρ g h$
h= difference in height of the mercury column.
Using Bernoulli's equation,
$P_{1} + ρ g H + \frac{1}{2} V^{2} = c o n s t a n t$
As both the points are at the same horizontal level.
$P_{1} + \frac{1}{2} V_{1}^{2} = P_{2} + \frac{1}{2} V_{2}^{2} ρ$
$P_{1} - P_{2} = \frac{1}{2} [V_{2}^{2} - V_{2}^{2}] ρ$
$\Rightarrow \frac{1}{2} ρ (V_{2}^{2} - V_{1}^{2}) =, ρ g h . . . . . (1)$

From equation of continuity,
$A_{1} V_{1} = A_{2} V_{2}$
$5 V_{1} = 2 V_{2}$
$V_{2} = \frac{5}{2} V_{1} . . . . . . (2)$
$a l s o A_{1} V_{1} = A_{2} V_{2} = 500 \frac{c m^{3}}{s}$
$\Rightarrow V_{1} = 100 \frac{c m}{s} = 1 \frac{m}{s}$ ......(3)
from (3) &(1)
$\frac{1}{2} ρ {\frac{21}{4}} =, ρ g h$
$\frac{1}{2} \times 1000 {\frac{21}{4}} = 13600 \times 10 \times h$
$h = 0.0193 m$
$h = 1.93 c m$

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