In the experimental arrangement shown in figure, the area of cross-section of the wide and narrow portions of the tube are $5 {cm}^{2}$ and $2 {cm}^{2}$ respectively. The rate of flow of water through the tube is $500 {cm}^{3} s^{- 1}$ . The difference of mercury levels in the U-tube is:

0.97 cm

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1.96 cm

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0.67 cm

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4.67 cm

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Solution

The correct option is B $1.96 cm$
Given $Q = 500 {cm}^{3} / sec$
Using continuity equation, we have
$Q = A_{1} V_{1} = A_{2} V_{2}$
$500 = 5 V_{1} = 2 V_{2}$
Then, we get
$V_{1} = 100 cm/sec = 1 m/sec$
and $V_{2} = 250 cm/sec = 2.5 m/sec$
Applying bernoulli's theorem between (1) and (2),
$P_{1} - P_{2} = \frac{1}{2} ρ (V_{2}^{2} - V_{1}^{2})$
$= \frac{1}{2} \times 1000 [(2.5)^{2} - (1)^{2}]$
$= \frac{1}{2} \times 1000 [6.25 - 1]$
$= \frac{1}{2} \times 1000 \times 5.25$
$= 2625 {N/m}^{2}$
We know that for U tube manometer,
$P_{1} - P_{2} = 2625 = ρ g h$
$2625 = 13.6 \times 1000 \times 9.8 \times h$
$\frac{2625}{13.6 \times 1000 \times 9.8} = h$
$h = 1.96 cm$

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In the experimental arrangement shown in figure, the area of cross-section of the wide and 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. The difference of mercury levels in the U-tube is:

In the experimental arrangement shown in figure, the area of cross-section of the wide and narrow portions of the tube are $5 {cm}^{2}$ and $2 {cm}^{2}$ respectively. The rate of flow of water through the tube is $500 {cm}^{3} s^{- 1}$ . The difference of mercury levels in the U-tube is: