Question

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

• A. $0.97\text{cm}$
• B. $1.96\text{cm}$
• C. $0.67\text{cm}$
• D. $4.67\text{cm}$

Answer 1

The correct option is B $1.96\text{cm}$

Given $Q=500{\text{cm}}^3/\text{sec}$
Using continuity equation, we have
$Q=A_1{V}_1=A_2{V}_2$
$500=5V_1=2V_2$
Then, we get
$V_1=100\text{cm/sec}=1\text{m/sec}$
and $V_2=250\text{cm/sec}=2.5\text{m/sec}$
Applying bernoulli's theorem between (1) and (2),
$P_1-P_2=\frac{1}{2}\rho (V_2^2-V_1^2)$
$=\frac{1}{2}\times 1000\left[\right(2.5{)}^2-\left(1{)}^2\right]$
$=\frac{1}{2}\times 1000[6.25-1]$
$=\frac{1}{2}\times 1000\times 5.25$
$=2625{\text{N/m}}^2$
We know that for U tube manometer,
$P_1-P_2=2625=\rho gh$
$2625=13.6\times 1000\times 9.8\times h$
$\frac{2625}{13.6\times 1000\times 9.8}=h$
$h=1.96\text{cm}$