Properties of Matter 3 Question 6
12. After the drop detaches, its surface energy is
(a) $1.4 \times 10^{-6} J$
(b) $2.7 \times 10^{-6} J$
(c) $5.4 \times 10^{-6} J$
(d) $8.1 \times 10^{-9} J$
Integer Answer Type Questions
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Answer:
Correct Answer: 12. $29025 J / m^{3},-29400 J / m^{3}$
Solution:
- Given, $A _1=4 \times 10^{-3} m^{2}, A _2=8 \times 10^{-3} m^{2}$,
$$ \begin{aligned} & h _1=2 m, h _2=5 m, \\ & v _1=1 m / s \text { and } \rho=10^{3} kg / m^{3} \end{aligned} $$
From continuity equation, we have
$$ \begin{aligned} & A _1 v _1=A _2 v _2 \quad \text { or } \quad v _2=\left(\frac{A _1}{A _2}\right) v _1 \\ \text { or } & v _2=\left(\frac{4 \times 10^{-3}}{8 \times 10^{-3}}\right)(1 m / s) \\ \Rightarrow \quad & v _2=\frac{1}{2} m / s \end{aligned} $$
Applying Bernoulli’s equation at sections 1 and 2
$$ \begin{aligned} & p _1+\frac{1}{2} \rho v _1^{2}+\rho g h _1=p _2+\frac{1}{2} \rho v _2^{2}+\rho g h _2 \\ & \text { or } \quad p _1-p _2=\rho g\left(h _2-h _1\right)+\frac{1}{2} \rho\left(v _2^{2}-v _1^{2}\right) \end{aligned} $$
(a) Work done per unit volume by the pressure as the fluid flows from $P$ to $Q$
$W _1=p _1-p _2$
$$ \begin{aligned} & =\rho g\left(h _2-h _1\right)+\frac{1}{2} \rho\left(v _2^{2}-v _1^{2}\right) \\ & ={\left(10^{3}\right)(9.8)(5-2)+\frac{1}{2}\left(10^{3}\right)\left(\frac{1}{4}-1\right) } \\ & =[29400-375]=29025 J / m^{3} \end{aligned} $$
(b) Work done per unit volume by the gravity as the fluid flows from $P$ to $Q$.
$$ \begin{aligned} W _2 & =\rho g\left(h _1-h _2\right) \\ & ={\left(10^{3}\right)(9.8)(2-5) } \\ \text { or } \quad W _2 & =-29400 J / m^{3} \end{aligned} $$
