Schaum’s Outline of College Physics

Eugene Hecht

Chapter 13

Fluids at Rest - all with Video Answers

Educators

Chapter Questions

Problem 1

An $80-\mathrm{kg}$ metal cylinder, $2.0 \mathrm{~m}$ long and with each end of area $25 \mathrm{~cm}^{2}$, stands vertically on one end. What pressure does the cylinder exert on the floor?
$$
P=\frac{\text { Normal force }}{\text { Area }}=\frac{(80 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}{25 \times 10^{-4} \mathrm{~m}^{2}}=3.1 \times 10^{5} \mathrm{~Pa}
$$

Dominique Jan Tan

Dominique Jan Tan

Numerade Educator

Problem 2

Atmospheric pressure is about $1.0 \times 10^{5} \mathrm{~Pa}$. How large a force does the still air in a room exert on the inside of a window pane that is $40 \mathrm{~cm} \times 80 \mathrm{~cm}$ ?
The atmosphere exerts a force normal to any surface placed in it. Consequently, the force on the window pane is perpendicular to the pane and is given by
$$F=P A=\left(1.0 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(0.40 \times 0.80 \mathrm{~m}^{2}\right)=3.2 \times 10^{4} \mathrm{~N}$$
Of course, a nearly equal force due to the atmosphere on the outside keeps the window from breaking.

Dominique Jan Tan

Dominique Jan Tan

Numerade Educator

Problem 3

Find the pressure due to the fluid at a depth of $76 \mathrm{~cm}$ in still $(a)$ water $\left(\rho_{w}=1.00 \mathrm{~g} / \mathrm{cm}^{3}\right)$ and
(b) mercury ( $\left.\rho=13.6 \mathrm{~g} / \mathrm{cm}^{3}\right)$.
(a) $P=\rho_{w} g h=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.76 \mathrm{~m})=7450 \mathrm{~N} / \mathrm{m}^{2}=7.5 \mathrm{kPa}$
(b) $P=\rho g h=\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.76 \mathrm{~m})=1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2} \approx 1.0 \mathrm{~atm}$

Arpit Gupta

Numerade Educator

Problem 4

When a submarine dives to a depth of $120 \mathrm{~m}$, to how large a total pressure is its exterior surface subjected? The density of seawater is about $1.03 \mathrm{~g} / \mathrm{cm}^{3}$.
$$
\begin{aligned}
P &=\text { Atmospheric pressure }+\text { Pressure of water } \\
&=1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}+\rho g h=1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}+\left(1030 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(120 \mathrm{~m}) \\
&=1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}+12.1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}=13.1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.31 \mathrm{MPa}
\end{aligned}
$$

Arpit Gupta

Numerade Educator

Problem 5

How high would water rise in the essentially open pipes of a building if the water pressure gauge shows the pressure at the ground floor to be $270 \mathrm{kPa}$ (about $40 \mathrm{lb} / \mathrm{in.}^{2}$ )?
Water pressure gauges read the excess pressure just due to the water, that is, the difference between the absolute pressure in the water and the pressure of the atmosphere. The water pressure at the bottom of the highest column that can be supported is $270 \mathrm{kPa}$. Therefore, $P=\rho_{w} g h$ gives
$$
h=\frac{P}{\rho_{w} g}=\frac{2.70 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}}{\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}=27.5 \mathrm{~m}
$$

Dominique Jan Tan

Dominique Jan Tan

Numerade Educator

Problem 6

A reservoir dam holds an $8.00-\mathrm{km}^{2}$ lake behind it. Just behind the dam, the lake is $12.0 \mathrm{~m}$ deep. What is the water pressure $(a)$ at the base of the dam and $(b)$ at a point $3.0 \mathrm{~m}$ down from the lake's surface?
The area of the lake behind the dam has no effect on the pressure against the dam. At any point, $P=\rho_{w} g h$.
(a) $P=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(12.0 \mathrm{~m})=118 \mathrm{kPa}$
(b) $P=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(3.0 \mathrm{~m})=29 \mathrm{kPa}$

Dominique Jan Tan

Dominique Jan Tan

Numerade Educator

Problem 7

A mass (or load) acting downward on a piston confines a fluid of density $\rho$ in a closed container, as shown in Fig. 13-2. The combined weight of the piston and load on the right is $200 \mathrm{~N}$, and the cross-sectional area of the piston is $A=8.0 \mathrm{~cm}^{2}$. Find the total pressure at point- $B$ if the fluid is mercury and $h=25 \mathrm{~cm}\left(\rho_{\mathrm{Hg}}=13600 \mathrm{~kg} / \mathrm{m}^{3}\right) .$ What would an ordinary pressure gauge read at $B ?$
Recall what Pascal's principle tells us about the pressure applied to the fluid by the piston and atmosphere:
This added pressure is applied at all points within the fluid. Therefore, the total pressure at $B$ is composed of three parts:
Pressure of the atmosphere $=1.0 \times 10^{5} \mathrm{~Pa}$ Pressure due to the piston and load $=\frac{F_{W}}{A}=\frac{200 \mathrm{~N}}{8.0 \times 10^{-4} \mathrm{~m}^{2}}=2.5 \times 10^{5} \mathrm{~Pa}$
Pressure due to the height $h$ of fluid $=h p g=0.33 \times 10^{5} \mathrm{~Pa}$
In this case, the pressure of the fluid itself is relatively small. We have Total pressure at $B=3.8 \times 10^{5} \mathrm{~Pa}$ The gauge pressure does not include atmospheric pressure. Therefore, Gauge pressure at $B=2.8 \times 10^{5} \mathrm{~Pa}$

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Numerade Educator

Problem 8

In a hydraulic press such as the one shown in Fig. $13-3$, the large piston has cross-sectional area $A_{1}=200 \mathrm{~cm}^{2}$ and the small piston has cross-sectional area $A_{2}=5.0 \mathrm{~cm}^{2} .$ If a force of $250 \mathrm{~N}$ is applied to the small piston, find the force $F_{1}$ on the large piston.
By Pascal's principle,
Pressure under large piston $=$ Pressure under small piston $\quad$ or $\quad \frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{2}}$
so that
$$F_{1}=\frac{A_{1}}{A_{2}} F_{2}=\frac{200}{5.0} 250 \mathrm{~N}=10 \mathrm{kN}$$
Note that atmospheric pressure acting on both pistons cancels out of the calculation.

Arpit Gupta

Numerade Educator

Problem 9

For the system shown in Fig. $13-4$, the cylinder on the left, at $L$, has a mass of $600 \mathrm{~kg}$ and a crosssectional area of $800 \mathrm{~cm}^{2}$. The piston on the right, at $S$, has a cross-sectional area of $25 \mathrm{~cm}^{2}$ and a negligible weight. If the apparatus is filled with oil $\left(\rho=0.78 \mathrm{~g} / \mathrm{cm}^{3}\right)$, find the force $F$ required to hold the system in equilibrium as shown.
The pressures at points $H_{1}$ and $H_{2}$ are equal because they are at the same level in a single connected fluid. Therefore,
Pressure at $H_{1}=$ Pressure at $H_{2}$
$$\begin{array}{l}
\left(\begin{array}{l}
\text { Pressure due to } \\
\text { left piston }
\end{array}\right)=\left(\begin{array}{c}
\text { Pressure due to } F \\
\text { and right piston }
\end{array}\right)+(\text { Pressure due to } 8.0 \mathrm{~m} \text { of oil }) \\
\frac{(600)(9.81) \mathrm{N}}{0.0800 \mathrm{~m}^{2}}=\frac{F}{25 \times 10^{-4} \mathrm{~m}^{2}}+(8.0 \mathrm{~m})\left(780 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)
\end{array}$$
from which $F=31 \mathrm{~N}$

Arpit Gupta

Numerade Educator

Problem 10

A barrel will rupture when the gauge pressure within it reaches $350 \mathrm{kPa}$. It is attached to the lower end of a vertical pipe, with the pipe and barrel filled with oil $\left(\rho=890 \mathrm{~kg} / \mathrm{m}^{3}\right)$. How long can the pipe be if the barrel is not to rupture?
From $P=\rho g h$ we have
$$
h=\frac{P}{\rho g}=\frac{350 \times 10^{3} \mathrm{~N} / \mathrm{m}^{2}}{\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\left(890 \mathrm{~kg} / \mathrm{m}^{3}\right)}=40.1 \mathrm{~m}
$$

Dominique Jan Tan

Dominique Jan Tan

Numerade Educator

Problem 11

A vertical test tube has $2.0 \mathrm{~cm}$ of oil $\left(\rho=0.80 \mathrm{~g} / \mathrm{cm}^{3}\right)$ floating on $8.0 \mathrm{~cm}$ of water. What is the pressure at the bottom of the tube due to the liquid in it?
$\begin{aligned} P &=\rho_{1} g h_{1}+\rho_{2} g h_{2}=\left(800 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.020 \mathrm{~m})+\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.080 \mathrm{~m}) \\ &=0.94 \mathrm{kPa} \end{aligned}$

Arpit Gupta

Numerade Educator

Problem 12

As shown in Fig. $13-5$, a column of water $40 \mathrm{~cm}$ high supports a $31-\mathrm{cm}$ column of an unknown liquid. What is the density of that liquid?
The pressures at point- $A$ due to the two fluids must be equal (or the one with the higher pressure would push the lower-pressure fluid away). Therefore,
Pressure due to water $\times$ Pressure due to unknown liquid
$$
\rho_{1} g h_{1}=\rho_{2} g h_{2}
$$
from which $\quad \rho_{2}=\frac{h_{1}}{h_{2}} \rho_{1}=\frac{40}{31}\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)=1290 \mathrm{~kg} / \mathrm{m}^{3}=1.3 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$

Dominique Jan Tan

Dominique Jan Tan

Numerade Educator

Problem 13

The U-tube device connected to the tank in Fig. 13-6 is called a manometer. As you can see, the mercury in the tube stands higher in one side than the other. What is the pressure in the tank if atmospheric pressure is $76 \mathrm{~cm}$ of mercury? The density of mercury is $13.6 \mathrm{~g} / \mathrm{cm}^{3}$.
$$
\begin{aligned}
\text { Pressure at } A_{1} &=\text { Pressure at } A_{2} \\
(P \text { in } \operatorname{tank})+(P \text { due to } 5 \mathrm{~cm} \text { mercury }) &=(P \text { due to atmosphere }) \\
P+(0.05 \mathrm{~m})\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right) &=(0.76 \mathrm{~m})\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)
\end{aligned}
$$
from which $P=95 \mathrm{kPa}$.
Or, more simply perhaps, we could note that the pressure in the tank is $5.0 \mathrm{~cm}$ of mercury lower than atmospheric. So the pressure is $71 \mathrm{~cm}$ of mercury, which is $94.6 \mathrm{kPa}$.

Arpit Gupta

Numerade Educator

Problem 14

The mass of a block of aluminum is $25.0 \mathrm{~g}$. ( $a$ ) What is its volume? $(b)$ What will be the tension in a string that suspends the block when the block is totally submerged in water? The density of aluminum is $2700 \mathrm{~kg} / \mathrm{m}^{3}$.
This problem is basically about buoyant force. (a) Because $\rho=m / V$, we have
$$
V=\frac{m}{\rho}=\frac{0.0250 \mathrm{~kg}}{2700 \mathrm{~kg} / \mathrm{m}^{3}}=9.26 \times 10^{-6} \mathrm{~m}^{3}=9.26 \mathrm{~cm}^{3}
$$
(b) The block displaces $9.26 \times 10^{-6} \mathrm{~m}^{3}$ of water when submerged, so the buoyant force on it is
$$
\begin{aligned}
F_{B} &=\text { Weight of displaced water }=(\text { Volume })(\rho \text { of water })(g) \\
&=\left(9.26 \times 10^{-6} \mathrm{~m}^{3}\right)\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=0.0908 \mathrm{~N}
\end{aligned}
$$
The tension in the supporting cord plus the buoyant force must equal the weight of the block if it is to be in equilibrium (see Fig. 13-7). That is, $F_{T}+F_{B}=m g$, from which
$$
F_{\tau}=m g-F_{p}=(0.0250 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)-0.0908 \mathrm{~N}=0.154 \mathrm{~N}
$$

Arpit Gupta

Numerade Educator

Problem 15

Using a scale, a piece of alloy has a measured mass of $86 \mathrm{~g}$ in air and $73 \mathrm{~g}$ when immersed in water. Find its volume and its density.
The apparent change in measured mass is due to the buoyant force of the water. Figure $13-7$ shows the situation when the object is in water. From the figure, $F_{B}+F_{T}=m g$, so
$$F_{B}=(0.086)(9.81) \mathrm{N}-(0.073)(9.81) \mathrm{N}=(0.013)(9.81) \mathrm{N}$$
But $F_{B}$ must be equal to the weight of the displaced water.
or
$$
\begin{aligned}
F_{B} &=\text { Weight of water }=(\text { Mass of water })(g) \\
&=(\text { Volume of water })(\text { Density of water })(g) \\
(0.013)(9.81) \mathrm{N} &=V\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)
\end{aligned}
$$
from which $V=1.3 \times 10^{-5} \mathrm{~m}^{3}$. This is also the volume of the piece of alloy. Therefore,
$$
\rho \text { of alloy }=\frac{\text { Mass }}{\text { Volume }}=\frac{0.086 \mathrm{~kg}}{1.3 \times 10^{-5} \mathrm{~m}^{3}}=6.6 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}
$$

Arpit Gupta

Numerade Educator

Problem 16

A solid aluminum cylinder with $\rho=2700 \mathrm{~kg} / \mathrm{m}^{3}$ has a measured mass of $67 \mathrm{~g}$ in air and $45 \mathrm{~g}$ when immersed in turpentine. Determine the density of turpentine.
The $F_{B}$ acting on the immersed cylinder is
$$F_{B}=(0.067-0.045)(9.81) \mathrm{N}=(0.022)(9.81) \mathrm{N}$$
This is also the weight of the displaced turpentine.
The volume of the cylinder is, from $\rho=m / V$,
$$
V \text { of cylinder }=\frac{m}{\rho}=\frac{0.067 \mathrm{~kg}}{2700 \mathrm{~kg} / \mathrm{m}^{3}}=2.5 \times 10^{-5} \mathrm{~m}^{3}
$$
This is also the volume of the displaced turpentine. We therefore have, for the turpentine,
$$
\rho=\frac{\text { Mass }}{\text { Volume }}=\frac{(\text { Weight }) / g}{\text { Volume }}=\frac{(0.022)(9.81) /(9.81)}{2.48 \times 10^{-5}} \frac{\mathrm{kg}}{\mathrm{m}^{3}}=8.9 \times 10^{2} \mathrm{~kg} / \mathrm{m}^{3}
$$

Arpit Gupta

Numerade Educator

Problem 17

A glass stopper has a mass of $2.50 \mathrm{~g}$ when measured in air, $1.50 \mathrm{~g}$ in water, and $0.70 \mathrm{~g}$ in sulfuric acid. What is the density of the acid? What is its specific gravity?
The $F_{B}$ on the stopper in water is $(0.00250-0.00150)(9.81) \mathrm{N}$. This is the weight of the displaced water. Since $\rho=m / V$, or $\rho g=F_{W} / V$,
$$\begin{aligned}
\text { Volume of stopper } &=\text { Volume of displaced water }=\frac{\text { weight }}{\rho \mathrm{g}} \\
V &=\frac{(0.00100)(9.81) \mathrm{N}}{\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}=1.00 \times 10^{-6} \mathrm{~m}^{3}
\end{aligned}$$
The buoyant force in acid is
$$\left[(2.50-0.70) \times 10^{-3}\right](9.81) \mathrm{N}=(0.00180)(9.81) \mathrm{N}$$
But this is equal to the weight of displaced acid, $m g$. Since $\rho=m / V$, and since $m=0.00180 \mathrm{~kg}$ and $V=1.00 \times 10^{-6} \mathrm{~m}^{3}$
$$\rho \text { of acid }=\frac{0.00180 \mathrm{~kg}}{1.00 \times 10^{-6} \mathrm{~m}^{3}}=1.8 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$$
Then, for the acid,
$$\text { sp } \mathrm{gr}=\frac{\rho \text { of acid }}{\rho \text { of water }}=\frac{1800}{1000}=1.8$$
Alternative Method
$$\begin{array}{l}
\text { Weight of displaced water }=\left[(2.50-1.50) \times 10^{-3}\right](9.81) \mathrm{N} \\
\text { Weight of displaced acid }=\left[(2.50-0.70) \times 10^{-3}\right](9.81) \mathrm{N}
\end{array}$$
so sp gr of acid $=\frac{\text { Weight of displaced acid }}{\text { Weight of equal volume of displaced water }}=\frac{1.80}{1.00}=1.8$
Then, since sp gr of acid $=(\rho$ of acid $) /(\rho$ of water $)$,
$$
\rho \text { of acid }=(\text { sp gr of acid })(\rho \text { of water })=(1.8)\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)=1.8 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}
$$

Arpit Gupta

Numerade Educator

Problem 18

The density of ice is $917 \mathrm{~kg} / \mathrm{m}^{3}$. What fraction of the volume of a piece of ice will be above the liquid when floating in fresh water?
The piece of ice will float in the water, since its density is less than $1000 \mathrm{~kg} / \mathrm{m}^{3}$, the density of water. As it does,
$F_{B}=$ Weight of displaced water $=$ Weight of piece of ice
But the weight of the ice is $\rho_{\text {ice }} g V$, where $V$ is the volume of the piece. In addition, the weight of the displaced water is $\rho_{w} g V^{\prime}$, where $V^{\prime}$ is the volume of the displaced water. Substituting into the above equation
$$\begin{aligned}
\rho_{\text {ice }} g V &=\rho_{w} g V^{\prime} \\
V^{\prime} &=\frac{\rho_{\text {ice }}}{\rho_{w}} V=\frac{917}{1000} V=0.917 V
\end{aligned}$$
The fraction of the volume that is above water is then
$$
\frac{V-V^{\prime}}{V}=\frac{V-0.917 V}{V}=1-0.917=0.083 \text { or } 8.3 \%
$$

Arpit Gupta

Numerade Educator

Problem 19

A $60-\mathrm{kg}$ rectangular box, open at the top, has base dimensions of $1.0 \mathrm{~m}$ by $0.80 \mathrm{~m}$ and a depth of $0.50 \mathrm{~m} .(a)$ How deep will it sink in fresh water? $(b)$ What weight $F_{W b}$ of ballast will cause it to sink to a depth of $30 \mathrm{~cm}$ ?
(a) Assuming that the box floats,
$$\begin{array}{c}
F_{B}=\text { Weight of displaced water }=\text { Weight of box } \\
\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(1.0 \mathrm{~m} \times 0.80 \mathrm{~m} \times y)=(60 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)
\end{array}$$
where $y$ is the depth the box sinks. Solving yields $y=0.075 \mathrm{~m}$. Because this is smaller than $0.50 \mathrm{~m}$, our assumption is shown to be correct.
(b) $F_{B}=$ weight of box $+$ weight of ballast
But the $F_{B}$ is equal to the weight of the displaced water. Therefore, the above equation becomes
$$\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(1.0 \mathrm{~m} \times 0.80 \mathrm{~m} \times 0.30 \mathrm{~m})=(60)(9.81) \mathrm{N}+F_{W b}$$
from which $F_{W b}=1760 \mathrm{~N}=1.8 \mathrm{kN}$. The ballast must have a mass of $(1760 / 9.81) \mathrm{kg}=180 \mathrm{~kg}$.

Arpit Gupta

Numerade Educator

Problem 20

A foam plastic $\left(\rho_{p}=0.58 \mathrm{~g} / \mathrm{cm}^{3}\right)$ is to be used as a life preserver. What volume of plastic must be used if it is to keep 20 percent (by volume) of an 80 -kg man above water in a lake? The average density of the man is $1.04 \mathrm{~g} / \mathrm{cm}^{3}$.
Keep in mind that a density of $1 \mathrm{~g} / \mathrm{cm}^{3}$ equals $1000 \mathrm{~kg} / \mathrm{m}^{3}$. At equilibrium
$$\begin{array}{l}
F_{B} \text { on man }+F B \text { on plastic }=\text { Weight of man }+\text { Weight of plastic } \\
\quad\left(\rho_{w}\right)\left(0.80 V_{m}\right) g+\rho_{w} V_{p} g=\rho_{m} V_{m} g+\rho_{p} V_{p} g
\end{array}$$
or
$$\left(\rho_{w}-\rho_{p}\right) V_{p}=\left(\rho_{m}-0.80 \rho_{w}\right) V_{m}$$
where subscripts $m, w$, and $p$ refer to man, water, and plastic, respectively.
But $\rho_{m} V_{m}=80 \mathrm{~kg}$ and so $V_{m}=(80 / 1040) \mathrm{m}^{3}$. Substitution gives
$$\left[(1000-580) \mathrm{kg} / \mathrm{m}^{3}\right] V_{p}=\left[(1040-800) \mathrm{kg} / \mathrm{m}^{3}\right]\left[(80 / 1040) \mathrm{m}^{3}\right]$$
from which $V_{p}=0.044 \mathrm{~m}^{3}$.

Arpit Gupta

Numerade Educator

Problem 21

A partly filled beaker of water sits on a scale, and its weight is $2.30 \mathrm{~N}$. When a piece of metal suspended from a thread is totally immersed in the beaker (but not touching bottom), the scale reads $2.75 \mathrm{~N}$. What is the volume of the metal?

The water exerts an upward buoyant force on the metal. According to Newton's Third Law of action and reaction, the metal exerts an equal downward force on the water. It is this force that increases the scale reading from $2.30 \mathrm{~N}$ to $2.75 \mathrm{~N}$. Hence the buoyant force is $2.75-2.30=0.45 \mathrm{~N}$. Then, because
$$F_{B}=\text { weight of displaced water }=\rho_{w} g V=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(V)$$
we have the volume of the displaced water, and of the piece of metal, namely,
$$
V=\frac{0.45 \mathrm{~N}}{9810 \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}^{2}}=46 \times 10^{-6} \mathrm{~m}^{3}=46 \mathrm{~cm}^{3}
$$

Arpit Gupta

Numerade Educator

Problem 22

A piece of pure gold $\left(\rho=19.3 \mathrm{~g} / \mathrm{cm}^{3}\right)$ is suspected to have a hollow center. It has a mass of $38.25 \mathrm{~g}$ when measured in air and $36.22 \mathrm{~g}$ in water. What is the volume of the central hole in the gold?
Remember that you go from a density in $\mathrm{g} / \mathrm{cm}^{3}$ to $\mathrm{kg} / \mathrm{m}^{3}$ by multiplying by 1000 . From $\rho=m / V$,
Actual volume of $38.25 \mathrm{~g}$ of gold $=\frac{0.03825 \mathrm{~kg}}{19300 \mathrm{~kg} / \mathrm{m}^{3}}=1.982 \times 10^{-6} \mathrm{~m}^{3}$
Volume of displaced water $=\frac{(38.25-36.22) \times 10^{-3} \mathrm{~kg}}{1000 \mathrm{~kg} / \mathrm{m}^{3}}=2.030 \times 10^{-6} \mathrm{~m}^{3}$
$$
\text { Volume of hole }=(2.030-1.982) \mathrm{cm}^{3}=0.048 \mathrm{~cm}^{3}
$$

Arpit Gupta

Numerade Educator

Problem 23

A wooden cylinder has a mass $m$ and a base area $A$. It floats in water with its axis vertical. Show that the cylinder undergoes SHM if given a small vertical displacement. Find the frequency of its motion.

When the cylinder is pushed down a distance $y$, it displaces an additional volume $A y$ of water. Because this additional displaced volume has mass $A y \rho_{w}$, an additional buoyant force $A y \rho_{w} g$ acts on the cylinder, where $\rho_{w}$ is the density of water. This is an unbalanced force on the cylinder and is a restoring force. In addition, the force is proportional to the displacement and so is a Hooke's Law force. Therefore, the cylinder will undergo SHM, as described in Chapter $11 .$

Comparing $F_{B}=A \rho_{w} g y$ with Hooke's Law in the form $F=k y$, we see that the elastic constant for the motion is $k=A \rho_{w} g$. This, acting on the cylinder of mass $m$, causes it to have a vibrational frequency of
$$
f=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}=\frac{1}{2 \pi} \sqrt{\frac{A \rho_{w} g}{m}}
$$

Arpit Gupta

Numerade Educator

Problem 24

What must be the volume $V$ of a $5.0$ -kg balloon filled with helium $\left(\rho_{\mathrm{He}}=0.178 \mathrm{~kg} / \mathrm{m}^{3}\right)$ if it is to lift a $30-\mathrm{kg}$ load? Use $\rho_{\text {air }}=1.29 \mathrm{~kg} / \mathrm{m}^{3}$.
The buoyant force, $V \rho_{\mathrm{air}} g$, must lift the weight of the balloon, its load, and the helium within it:
$$
V \rho_{\text {air }} g=(35 \mathrm{~kg})(g)+V \rho_{\mathrm{He}} g
$$
which gives
$$
V=\frac{35 \mathrm{~kg}}{\rho_{\mathrm{air}}-\rho_{\mathrm{He}}}=\frac{35 \mathrm{~kg}}{1.11 \mathrm{~kg} / \mathrm{m}^{3}}=32 \mathrm{~m}^{3}
$$

Arpit Gupta

Numerade Educator

Problem 25

Find the density $\rho$ of a fluid at a depth $h$ in terms of its density $\rho_{0}$ at the surface.
If a mass $m$ of fluid has volume $V_{0}$ at the surface, then it will have volume $V_{0}-\Delta V$ at a depth $h$. The density at depth $h$ is then
$$
\rho=\frac{m}{V_{0}-\Delta V} \quad \text { while } \quad \rho_{0}=\frac{m}{V_{0}}
$$
which gives
$$
\frac{\rho}{\rho_{0}}=\frac{V_{0}}{V_{0}-\Delta V}=\frac{1}{1-\left(\Delta V / V_{0}\right)}
$$
However, from Chapter 12, the bulk modulus is $B=P /\left(\Delta V / V_{0}\right)$ and so $\Delta V / V_{0}=P / B$. Making this substitution, we obtain
$$
\frac{\rho}{\rho_{0}}=\frac{1}{1-P / B}
$$
If we assume that $\rho$ is close to $\rho_{0}$, then the pressure at depth $h$ is approximately $\rho_{0} g h$, and so
$$
\frac{\rho}{\rho_{0}}=\frac{1}{1-\left(\rho_{0} g h / B\right)}
$$

Arpit Gupta

Numerade Educator

Problem 26

A $60-\mathrm{kg}$ performer balances on a cane. The end of the cane in contact with the floor has an area of $0.92 \mathrm{~cm}^{2}$. Find the pressure exerted on the floor by the cane. (Neglect the weight of the cane.)

Prabhat Tyagi

Numerade Educator

Problem 27

A certain town receives its water directly from a water tower. If the top of the water in the tower is $26.0 \mathrm{~m}$ above the water faucet in a house, what should be the water pressure at the faucet? (Neglect the effects of other water users.)

Arpit Gupta

Numerade Educator

Problem 28

At a height of $10 \mathrm{~km}$ ( $33000 \mathrm{ft}$ ) above sea level, atmospheric pressure is about $210 \mathrm{~mm}$ of mercury. What is the net resultant normal force on a $600 \mathrm{~cm}^{2}$ window of an airplane flying at this height? Assume the pressure inside the plane is $760 \mathrm{~mm}$ of mercury. The density of mercury is $13600 \mathrm{~kg} / \mathrm{m}^{3}$.

Arpit Gupta

Numerade Educator

Problem 29

A narrow tube is sealed onto a tank as shown in Fig. $13-8$. The base of the tank has an area of $80 \mathrm{~cm}^{2}$.
(a) Remembering that pressure is determined by the height of the column of liquid, find the force on the bottom of the tank due to oil when the tank and capillary are filled with oil $\left(\rho=0.72 \mathrm{~g} / \mathrm{cm}^{3}\right)$ to the height $h_{1} .(b)$ Repeat for an oil height of $h_{2}$.

Arpit Gupta

Numerade Educator

Problem 30

Repeat Problem $13.29$, but now find the force on the top wall of the tank due to the oil.

Arpit Gupta

Numerade Educator

Problem 31

Compute the pressure required for a water supply system that will raise water $50.0 \mathrm{~m}$ vertically.

Prabhat Tyagi

Numerade Educator

Problem 32

The area of a piston of a force pump is $8.0 \mathrm{~cm}^{2}$. What force must be applied to the piston to raise oil $\left(\rho=0.78 \mathrm{~g} / \mathrm{cm}^{2}\right)$ to a height of $6.0 \mathrm{~m}$ ? Assume the upper end of the oil is open to the atmosphere.

Arpit Gupta

Numerade Educator

Problem 33

The diameter of the large piston of a hydraulic press is $20 \mathrm{~cm}$, and the area of the small piston is $0.50 \mathrm{~cm}^{2}$. If a force of $400 \mathrm{~N}$ is applied to the small piston, $(a)$ what is the resulting force exerted on the large piston? (b) What is the increase in pressure underneath the small piston? ( $c$ ) Underneath the large piston?

Arpit Gupta

Numerade Educator

Problem 34

A metal cube, $2.00 \mathrm{~cm}$ on each side, has a density of $6600 \mathrm{~kg} / \mathrm{m}^{3}$. Find its apparent mass when it is totally submerged in water.

Arpit Gupta

Numerade Educator

Problem 35

A solid wooden cube, $30.0 \mathrm{~cm}$ on each edge, can be totally submerged in water if it is pushed downward with a force of $54.0 \mathrm{~N}$. What is the density of the wood?

Arpit Gupta

Numerade Educator

Problem 36

A metal object "weighs" $26.0 \mathrm{~g}$ in air and $21.48 \mathrm{~g}$ when totally immersed in water. What is the volume of the object? Its mass density?

Arpit Gupta

Numerade Educator

Problem 37

A solid piece of aluminum $\left(\rho=2.70 \mathrm{~g} / \mathrm{cm}^{3}\right)$ has a mass of $8.35 \mathrm{~g}$ when measured in air. If it is hung from a thread and submerged in a vat of oil $\left(\rho=0.75 \mathrm{~g} / \mathrm{cm}^{3}\right)$, what will be the tension in the thread?

Arpit Gupta

Numerade Educator

Problem 38

A beaker contains oil of density $0.80 \mathrm{~g} / \mathrm{cm}^{3}$. A $1.6-\mathrm{cm}$ cube of aluminum $\left(\rho=2.70 \mathrm{~g} / \mathrm{cm}^{3}\right)$ hanging ertically on a thread is submerged in the oil. Find the tension in the thread.

Arpit Gupta

Numerade Educator

Problem 39

A tank containing oil of sp $\mathrm{gr}=0.80$ rests on a scale and weighs $78.6 \mathrm{~N}$. By means of a very fine wire, a $6.0 \mathrm{~cm}$ cube of aluminum, sp $\mathrm{gr}=2.70$, is submerged in the oil. Find $(a)$ the tension in the wire and $(b)$ the scale reading if none of the oil overflows.

Penny Riley

Numerade Educator

Problem 40

Downward forces of $45.0 \mathrm{~N}$ and $15.0 \mathrm{~N}$, respectively, are required to keep a plastic block totally immersed in water and in oil, respectively. If the volume of the block is $8000 \mathrm{~cm}^{3}$, find the density of the oil.

Arpit Gupta

Numerade Educator

Problem 41

Determine the unbalanced force acting on an iron ball $\left(r=1.5 \mathrm{~cm}, \rho=7.8 \mathrm{~g} / \mathrm{cm}^{3}\right)$ when just released while totally immersed in $(a)$ water and $(b)$ mercury $\left(\rho=13.6 \mathrm{~g} / \mathrm{cm}^{3}\right) .$ What will be the initial acceleration of the ball in each case?

Arpit Gupta

Numerade Educator

Problem 42

A $2.0-\mathrm{cm}$ cube of metal is suspended by a fine thread attached to a scale. The cube appears to have a mass of $47.3 \mathrm{~g}$ when measured submerged in water. What will its mass appear to be when submerged in glycerin, sp $\mathrm{gr}=1.26 ?[$ Hint $:$ Find $\rho$ too. $]$

Arpit Gupta

Numerade Educator

Problem 43

A balloon and its gondola have a total (empty) mass of $2.0 \times 10^{2} \mathrm{~kg}$. When filled, the balloon contains $900 \mathrm{~m}^{3}$ of helium at a density of $0.183 \mathrm{~kg} / \mathrm{m}^{3}$. Find the added load, in addition to its own weight, that the balloon can lift. The density of air is $1.29 \mathrm{~kg} / \mathrm{m}^{3}$.

Arpit Gupta

Numerade Educator

Problem 44

A piece of metal has a measured mass of $5.00 \mathrm{~g}$ in air, $3.00 \mathrm{~g}$ in water, and $3.24 \mathrm{~g}$ in benzene. Determine the
mass density of the metal and of the benzene.

Arpit Gupta

Numerade Educator

Problem 45

A spring whose composition is not completely known might be either bronze (sp gr 8.8) or brass (sp gr 8.4). It has a mass of $1.26 \mathrm{~g}$ when measured in air and $1.11 \mathrm{~g}$ in water. Which is it made of?

Arpit Gupta

Numerade Educator

Problem 46

What fraction of the volume of a piece of quartz $\left(\rho=2.65 \mathrm{~g} / \mathrm{cm}^{3}\right)$ will be submerged when it is floating in a container of mercury ( $\rho=13.6 \mathrm{~g} / \mathrm{cm}^{3}$ )

Arpit Gupta

Numerade Educator

Problem 47

A cube of wood floating in water supports a 200-g mass resting on the center of its top face. When the mass is removed, the cube rises $2.00 \mathrm{~cm}$. Determine the volume of the cube.

Arpit Gupta

Numerade Educator

Problem 48

A cork has a measured mass of $5.0 \mathrm{~g}$ in air. A sinker has a measured mass of $86 \mathrm{~g}$ in water. The cork is attached to the sinker and both together have a measured mass of $71 \mathrm{~g}$ when under water. What is the density
of the cork?

Arpit Gupta

Numerade Educator

Problem 49

A glass of water has a $10-\mathrm{cm}^{3}$ ice cube floating in it. The glass is filled to the brim with cold water. By the time the ice cube has completely melted, how much water will have flowed out of the glass? The sp gr of ice is $0.92$.

Arpit Gupta

Numerade Educator

Problem 50

A glass tube is bent into the form of a U. A $50.0-\mathrm{cm}$ height of olive oil in one arm is found to balance $46.0 \mathrm{~cm}$ of water in the other. What is the density of the olive oil?

Arpit Gupta

Numerade Educator

Problem 51

On a day when the pressure of the atmosphere is $1.000 \times 10^{5} \mathrm{~Pa}$, a chemist distills a liquid under slightly reduced pressure. The pressure within the distillation chamber is read by an oil-filled manometer (density of oil $=0.78 \mathrm{~g} / \mathrm{cm}^{3}$ ). The difference in heights on the two sides of the manometer is $27 \mathrm{~cm}$. What is the pressure in the distillation chamber?

Arpit Gupta

Numerade Educator