Question

A meat baster consists of a squeeze bulb attached to a plastic tube. When the bulb is squeezed and released, with the open end of the tube under the surface of the basting sauce, the sauce rises in the tube to a distance h, as the drawing shows. Using 1.013 × 105 Pa for the atmospheric pressure and 1470 kg/m3 for the density of the sauce, find the absolute pressure PB in the bulb when the distance h is (a) 0.20 m and (b) 0.12 m.



(a) PB =
(b) PB =

Answer 1

(a)

In the first trail \(h=0.20 \mathrm{~m}\) Upward Pressure due to sauce, \(P=h \rho g\)

$$ \begin{array}{l} =(0.20 \mathrm{~m})\left(1470 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right) \\ =2881.2 \mathrm{~Pa} \end{array} $$

So total pressure in the bulb is \(=P_{0}-P\)

$$ \begin{array}{l} =\left(1.013 \times 10^{5} \mathrm{~Pa}\right)-\left(0.028812 \times 10^{5}\right) \mathrm{Pa} \\ =0.9842 \times 10^{5} \mathrm{~Pa} \\ =9.842 \times 10^{4} \mathrm{~Pa} \end{array} $$

(b)

In the second trail \(\quad h=0.12 \mathrm{~m}\) Upward Pressure due to sauce, \(P=h \rho g\)

$$ \begin{array}{l} =(0.12 \mathrm{~m})\left(1470 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right) \\ =1728.72 \mathrm{~Pa} \\ =0.017287 \times 10^{5} \mathrm{~Pa} \end{array} $$

So total pressure on the bulb is \(=P_{0}-P\)

$$ \begin{array}{l} =\left(1.013 \times 10^{5}\right)-\left(0.017287 \times 10^{5}\right) \mathrm{Pa} \\ =0.9957 \times 10^{5} \mathrm{~Pa} \end{array} $$