Answer to Question #204584 in Physics for John Mark

Question

1. Density. A piece of copper whose density is 8.93 g/cm3 has a mass of 180 g in air and 162 g when submerged in a certain liquid. What is the mass density (in g/cm3) of the liquid?

2. Gauge Pressure. At what depth (in meters) in seawater is the gauge pressure 1.00 x 105 Pa? (density of seawater= 1025/kg/m3)

3. Pascal’s Principle. In a hydraulic press the small cylinder has a diameter of 8.00 cm, while the large piston has a diameter of 20.0 cm. If a force of 500 N is applied to the small piston, what is the force on the large piston, neglecting friction? 

Answer 1

(1) The apparent weight of the piece of copper when it placed in the liquid can be written as follows:

$$W_A=W-F_B.$$

Then, we can find the buoyant force acting on the piece of copper:

$$F_B=W-W_A=m_{air}g-m_{liquid}g=(m_{air}-m_{liquid})g,$$ $$F_B=(0.18\ kg-0.162\ kg)\cdot9.8\ \dfrac{m}{s^2}=0.1764\ N.$$

Let's find the volume of the piece of copper:

$$V=\dfrac{m}{\rho}=\dfrac{0.18\ kg}{8930\ \dfrac{kg}{m^3}}=2.01\cdot10^{-5}\ m^3.$$

By the definition of the buoyant force we have:

$$F_B=\rho_{liquid}V_{liquid}g=\rho_{liquid}Vg,$$ $$\rho_{liquid}=\dfrac{F_B}{Vg},$$ $$\rho_{liquid}=\dfrac{0.1764\ N}{2.01\cdot10^{-5}\ m^3\cdot9.8\ \dfrac{m}{s^2}}=895\ \dfrac{kg}{m^3}=0.895\ \dfrac{g}{cm^3}.$$

(2) By the definition of the gauge pressure we have:

$$P_{gauge}=P_{abs}-P_{atm}=\rho_{seawater}gh,$$ $$h=\dfrac{P_{gauge}}{\rho_{seawater}g}=\dfrac{1.0\cdot10^5\ Pa}{1025\ \dfrac{kg}{m^3}\cdot9.8\ \dfrac{m}{s^2}}=9.9\ m.$$

(3)

$$\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2},$$ $$F_2=F_1\dfrac{A_2}{A_1}=F_1\dfrac{\pi r_2^2}{\pi r_1^2},$$ $$F_2=500\ N\cdot\dfrac{\pi(0.1\ m)^2}{\pi(0.04\ m)^2}=3125\ N.$$