Question 16: Properties of Matter

Question

Water flows through a horizontal tube as shown in the figure. The difference in height between the water columns in vertical tubes is 5 cm and the area of cross-sections at $A$ and $B$ are $6 \mathrm{~cm}^2$ and $3 \mathrm{~cm}^2$ respectively. The rate of flow will be $\_\_\_\_$ $\mathrm{cm}^3 / \mathrm{s}$. (take $g=10 \mathrm{~m} / \mathrm{s}^2$ )

Accepted Answer

Applying the equation of continuity
The rate of flow ( Q ) is constant throughout the tube.
$$ \mathrm{Q}=\mathrm{A}_1 \mathrm{v}_1=\mathrm{A}_2 \mathrm{v}_2 $$
Where :
$\mathrm{A}_1=$ cross-section area at $\mathrm{A}, \mathrm{A}_1=6 \mathrm{~cm}^2$
$\mathrm{A}_2=$ cross-section area at $\mathrm{B}, \mathrm{A}_2=3 \mathrm{~cm}^2$
$\mathrm{V}_1, \mathrm{~V}_2$ are velocities at A and B .
$$ 6 v_1=3 v_2 \Rightarrow v_2=2 v_1 $$
Applying Bernoulli's principle between two points,
$$ P_1+\frac{1}{2} ho v_1^2+ho g h_1=P_2+\frac{1}{2} ho v_2^2 $$
For a horizontal tube, the height is the same, $\mathrm{h}_1=\mathrm{h}_2$
$$ P_1-P_2=\frac{1}{2} ho\left(v_2^2-v_1^2ight) $$
The pressure difference ( $P_1-P_2$ ) is measured by the difference in the heights of the water columns ( $h$ ) :
$$ \mathrm{P}_1-\mathrm{P}_2=ho \mathrm{gh} $$
From relations (ii) and (iii),
$$ ho g h=\frac{1}{2} ho\left(v_2^2-v_1^2ight) $$
$$\Rightarrow $$ $$ 2 g h=v_2^2-v_1^2 $$
Substituting $\mathrm{v}_2=2 \mathrm{v}_1$ from relation (i) :
$$ 2 g h=\left(2 v_1ight)^2-v_1^2 $$
$$\Rightarrow $$ $2 g h=4 v_1^2-v_1^2=3 v_1^2$
$$ \Rightarrow v_1=\sqrt{2 g h / 3} $$
The given values are,
$\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2=1000 \mathrm{~cm} / \mathrm{s}^2$
$\mathrm{h}=5 \mathrm{~cm}$
Putting the values,
$$ v_1=\sqrt{\frac{2 	imes 1000 	imes 5}{3}}=\sqrt{\frac{10000}{3}}=\frac{100}{\sqrt{3}} \mathrm{~cm} / \mathrm{s} $$
So, the rate of flow (Q),
$$ Q=A_1 \cdot v_1 $$
$$\Rightarrow $$ $$ Q=6 \cdot \frac{100}{\sqrt{3}}=2 	imes \sqrt{3} 	imes \sqrt{3} 	imes \frac{100}{\sqrt{3}}=200 \sqrt{3} \mathrm{~cm}^3 / \mathrm{s} $$
Therefore, the rate of flow of water through the horizontal tube is $200 \sqrt{3} \mathrm{~cm}^3 / \mathrm{s}$.
Hence, the correct option is (D).

Answer

$100 \sqrt{3}$

Answer

$\frac{200}{\sqrt{3}}$

Answer

$200 \sqrt{6}$

Answer

$200 \sqrt{3}$

Properties of Matter

Choose the correct answer: