Properties of Matter 4 Question 17
19. When the gas bubble is at a height $y$ from the bottom, its temperature is
(a) $T _0\left(\frac{p _0+\rho _l g H}{p _0+\rho _l g y}\right)^{2 / 5}$
(b) $T _0\left(\frac{p _0+\rho _l g(H-y)}{p _0+\rho _l g H}\right)^{2 / 5}$
(c) $T _0\left(\frac{p _0+\rho _l g H}{p _0+\rho _l g y}\right)^{3 / 5}$
(d) $T _0\left(\frac{p _0+\rho _l g(H-y)}{p _0+\rho _l g H}\right)^{3 / 5}$
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Answer:
Correct Answer: 19. $\frac{1}{720} N-s / m^{2}$
Solution:
- When the tube is not there,
$$ \begin{aligned} & p+p _0+\frac{1}{2} \rho v _1^{2}+\rho g H=\frac{1}{2} \rho v _2^{2}+p _0 \\ & \therefore \quad p+\rho g H=\frac{1}{2} \rho\left(v _2^{2}-v _1^{2}\right) \\ & A _1 v _1=A _2 v _2 \\ & v _1=\frac{A _2 v _2}{A _1} \\ & \therefore \quad p+\rho g H=\frac{1}{2} \times \rho\left[v _2^{2}-\left(\frac{A _2}{A _1} v _2\right)^{2}\right] \\ &=\frac{1}{2} \times \rho \times v _2^{2}\left[1-\left(\frac{\pi(0.3)^{2}}{\pi(0.9)^{2}}\right)^{2}\right] \\ &=\frac{1}{2} \times \rho \times(10)^{2}\left[1-\frac{1}{81}\right] \\ &=\frac{4 \times 10^{3} \rho}{81} \\ &=\frac{4 \times 10^{3} \times 900}{81} \\ &=\frac{4}{9} \times 10^{5} N / m^{2} \end{aligned} $$
This is also the excess pressure $\Delta p$.
By Poiseuille’s equation, the rate of flow of liquid in the capillary tube
$$ Q=\frac{\pi(\Delta p) a^{4}}{8 \eta l} $$
$$ \begin{array}{rlrl} \therefore & 8 \times 10^{-6} & =\frac{\left(\pi a^{2}\right)(\Delta p)}{8 \eta}\left(\frac{a^{2}}{l}\right) \\ \therefore \quad \eta & =\frac{\left(\pi a^{2}\right)(\Delta p)\left(\frac{a^{2}}{l}\right)}{8 \times 8 \times 10^{-6}} \end{array} $$
Substituting the values, we have
$$ \begin{aligned} \eta & =\frac{\left(10^{-6}\right)\left(\frac{4}{9} \times 10^{5}\right)\left(2 \times 10^{-6}\right)}{8 \times 8 \times 10^{-6}} \\ & =\frac{1}{720} N-s / m^{2} \end{aligned} $$
