Hydrostatics

On the bottom $\rho g a^3$, on the side face $(1/2) \rho g a^3$.

The pressure $p$ in a liquid at depth $h$ is $\rho g h$, i.e., it varies linearly with height (see figure). Therefore, the total pressure force on the entire side face of the cube is equal in magnitude to the average pressure (i.e., the pressure at depth $a/2$) multiplied by the area of the face.

If the container narrows upwards, the weight and mercury will not detach the bottom, but the oil will. If the container narrows downwards, the opposite is true.

The pressure force of the liquid acts on the bottom from the inside: $\rho g h S$ (where $\rho$ is density, $h$ is the liquid level height, $S$ is the bottom area). If the container narrows upwards, this force is greater than the weight of the liquid poured into the cone by the amount of the weight of the liquid occupying the shaded volume (figure, a). Therefore, mercury and the weight will not detach the bottom. Oil will detach the bottom, because in the case of a cone narrowing upwards, the product $ph$ for oil is greater than for water (oil will occupy a larger volume than the same amount of water. This relative increase in level height $h$ is greater than the relative decrease in density $\rho$). If the container narrows downwards, the pressure force of the liquid poured into the container on the bottom is less than its weight by the amount of the weight of the liquid in the shaded volume (figure, b). Therefore, in this case, oil will not detach the bottom, but the weight and mercury will.

$\Delta h = 4$ cm.

Apply the equilibrium condition for non-uniform liquids in communicating vessels.

$S_2/S_1 = n h m g / A \approx 500$ times.

The water levels in both cases rise by $\Delta h = \frac{m}{\frac{\pi}{4} (D_1^2 + D_2^2) \rho_0}$.

$h_2 = 40$ cm. Alcohol will flow over. The tube should be positioned inclined, descending towards the arm with water.

See problem 3. By plotting the graph of pressure versus height (initial level - mercury level) for both arms (see the following figure), it is easy to see that at the height where the tube is located, the pressure in the arm with alcohol is greater than in the arm with water. Therefore, alcohol will flow over. For equilibrium not to be disturbed when the tap is opened, the tube must be positioned inclined. The difference in heights of the tube connection points can be found using the graph.

$A = gSH^2 (\rho_w - \rho_i)^2 / (2\rho_w) \approx 7.84$ J.

During uniform immersion of the ice floe, the buoyant force $Q$ acting on it increases uniformly, and therefore the force $F = Q - mg = (\rho_w - \rho_i) gSh$ ($m$ is the mass of the ice floe, $\rho_w$ is the density of water, $\rho_i$ is the density of ice, $h$ is the depth of immersion) changes linearly with the immersion height from zero (when the ice floats) to the value $F_H = (\rho_w - \rho_i) gSH.$ Therefore, the work required to immerse the ice floe is equal in magnitude to the work of the average force $(1/2)F_H$ over the path equal to the thickness of the part of the ice floe protruding above the water.

Ice displaces water whose weight is equal to the weight of the ice. When the ice melts, the same amount of water is formed, so the level will not change.

1. The presence of an air bubble in the piece of ice does not change the previous reasoning (the mass of air is negligible), therefore, in this case too, the water level in the vessel will remain the same.
2. If the ice contains a piece of lead, then the volume of water formed when the ice melts, together with the volume of the piece of lead, will be less than in the case of a piece of pure ice of the same weight. Consequently, the level will decrease.

The bottle with water will sink, but the one with mercury will not.

The weight of the bottle with water is greater than the weight of water whose volume is equal to the external volume of the bottle, while the bottle with mercury weighs less than the same volume of mercury.

$h = H - m/(\rho_0 S) = 4$ cm, where $\rho_0$ is the density of water.

0.5 kg.

Less.

When the weight is removed from the beaker, the volume of water displaced by the beaker will decrease by $V_1 = m/\rho_1$ ($m$ is the mass of the weight, $\rho_1$ is the density of water). When the weight is immersed in water, the volume of displaced water will increase by the volume of the weight $V_2 = m/\rho_2$ ($\rho_2$ is the density of the weight). Since $\rho_2 > \rho_1$, then $V_2 < V_1$, which means $H_2 < H_1$.

$\Delta h = m/(\rho S)$.

1st method. When the body is placed in the vessel, the pressure force on the bottom increases by $\Delta F = mg.$ But the body itself does not touch the bottom of the vessel, so the pressure force changes due to the change in water pressure. If the water level in the vessel rises by $\Delta h$, then $\Delta p = \rho g \Delta h$ and $\Delta F = \rho g \Delta h S.$ Therefore $\rho g \Delta h S = mg$. Hence $\Delta h = m/(\rho S).$

2nd method. A body of mass $m$, floating on the surface of a liquid, according to Archimedes' law, will displace (occupy) a volume $\Delta V$ in it, which would be occupied by the same liquid of mass $m$. Thus, this immersion is equivalent (in the sense of changing the liquid level in the vessel) to adding liquid of mass $m$ to the vessel. Then $\Delta h = \Delta V / S = m / (\rho S)$.

$\Delta h = m/(2\rho S)$.

See problem 13. The changes in liquid levels in the arms are the same.

$g \frac{\pi}{4} \{ \rho_0 d^2 (H-h) + (\rho - \rho_0)(D^2 - d^2)h + \rho d^2 h_1 \}$.

The pressure force of the structure on the bottom consists of the pressure force of water on the upper surface of the part, the weight of the structure, and the pressure force of water on the lower surface, directed upwards and equal to $\frac{\pi}{4} (D^2 - d^2)[H - h_1 + h] \rho_0 g.$

No, because the buoyant (Archimedean) force does not arise - hydrostatic pressure is not transmitted to the lower face of the cube due to the absence of liquid under it.

$p = p_0 \{ 1 + 2(S_1/S_2)^{1/2} - 3S_1/S_2 \}$.

The buoyant (Archimedean) force is maximum when the water reaches the top of the plug. At this point, it is equal to the weight of the water in the shaded volume of the plug (see the following figure).

$h = p_0 (V_0 - m/\rho) / (mg) \approx 2.6$ m.

At the sought depth $h$, the water pressure is $p_1 = p_0 + \rho g h$, where $p_0$ is the air pressure above the water surface; $\rho$ is its density. For the air in the bottle, which compresses upon immersion, Boyle's law can be written: $p_0 V_0 = p_1 V_1,$ where $V_0$ and $V_1$ are the volumes of air in the bottle before and after immersion, respectively. Considering, finally, that the bottle at depth $h$ is in equilibrium, supported by the Archimedean force $mg = \rho V_1 g,$ we can obtain $p_1 = p_0 V_0 \rho / m.$

$\rho = R_1^3 (\rho_2 - \rho_1) + \dfrac{R_2^3 \rho_1}{R_1^3}$.

The sphere will be in suspension inside the liquid if its mass (weight) is equal to the mass (weight) of the liquid whose volume is equal to the volume of the sphere.

$V = m (\frac{1}{\rho} - \frac{1}{\rho_w}) \approx 9350 \text{ cm}^3$.

45 g; on the other pan.

In accordance with Newton's third law, a force equal in magnitude to the Archimedean (buoyant) force acting on the block, and directed downwards, will act on the vessel with water from the side of the block. This force arises due to the increase in hydrostatic pressure on the bottom (and walls) of the vessel as a result of the rise in the water level in the vessel after immersing the block (see also problem 13).

In both cases, it will be disturbed; the iron ball will outweigh.

See the previous problem. In the second case, the arms are not equal and the moments of the buoyant forces are also not equal.

$V_x = 25.1 \text{ cm}^3$.

Since the weighed piece of iron is in equilibrium, we can write: $m_i g = P + m_w g$, where $m_i$ and $m_w$ are the masses of iron and water in volumes equal to $V_x$. The right side of the equality is the sum of the forces applied to the body: Archimedean force and the reaction force of the support (suspension), equal to the weight of the body in water $P$. Expressing the masses of iron and water through the volume $V_x$ and the corresponding densities, we get $V_x \rho_i g = P + V_x \rho_w g,$ whence $V_x = \frac{P/g}{\rho_i - \rho_w}.$

1.5 g/cm$^3$.

1st method. See the previous problem.

2nd method. Assume that the weight of the body in air is three units, then in water, according to the condition, the weight is one unit, i.e., the loss of weight in water is two units, which is equal to the weight of water in the volume of the body. Hence the density is $3/2$ g/cm$^3 = 1.5$ g/cm$^3$.

It will decrease, as the liquid pressure on the lower face of the block increases.

It will not change if the compressibility of the body is the same as that of water; it will increase if the compressibility is greater than that of water, and decrease otherwise.

$F \approx 40$ N.

It will decrease by the amount of the weight of water displaced by the walls and bottom of the bucket.

The dynamometer reading will not change.

The water displaced by the weight will spill out of the vessel. Thus, the weight of the water in the vessel will decrease, which exactly compensates for the force arising from the weight on the vessel with water, which is equal to the buoyant (Archimedean) force acting on the weight. (See also problem 10.21).

Place a weight of mass $M = 2m \rho_w / \rho_m \approx 24$ g on the right pan of the balance.

When the weight is immersed, the force acting on the left arm of the balance beam increases, and on the right - decreases by an amount equal to the buoyant force $F = V \rho_w g = m \rho_w g / \rho_m$ (see problem 10.21).

0.75 g/cm$^3$.

Since the stick is thin, we can assume that the buoyant force $Q$ is applied at the midpoint of the part of the stick submerged in water (see figure). By virtue of the equilibrium of the stick, the moments of forces $mg$ and $Q$ relative to point A must be equal, i.e., $mgx = Q \cdot (3/2)x$, whence $Q = (2/3)mg$. Substituting $m = \rho V$ and $Q = (1/2) \rho_w V g$ here ($V$ is the volume of the stick, $\rho$ is the density of the stick material and $\rho_w$ is the density of water), we find $\rho = (3/4) \rho_w$.

At a distance $x = l \frac{\rho_2 - \rho}{(\rho_2 - \rho) + (\rho_1 - \rho)(r_1/r_2)^3}$ from the sphere with density $\rho_1$.

The differences between the forces of gravity and Archimedean forces acting on the weights are respectively: $F_1 = \frac{4}{3} \pi r_1^3 (\rho_1 - \rho)g$ and $F_2 = \frac{4}{3} \pi r_2^3 (\rho_2 - \rho)g$. Since the rod with the weights is in equilibrium, $F_1 x = F_2 (l - x)$ ($x$ is the distance from the weight with density $\rho_1$ to the suspension point of the rod).

$F > \dfrac{\rho(H-h)\pi r^3 - M(R-2r)}{2(R-r)} g$.

The plate will rotate around point B (see the following figure). The water pressures from both sides of the plate balance each other everywhere except for the area above the tube. Here the water pressure force is $F = \rho g (H-h) \pi r^2$. Its moment relative to point B is $\rho g (H-h) \pi r^3$. The plate will turn over if $\rho g (H-h) \pi r^3 < Mg(R-2r) + F(2R-2r);$ $F > \frac{\rho(H-h)\pi r^3 - M(R-2r)}{2(R-r)} g.$

The answer depends on the ratio of the thermal expansion coefficients of the body and the liquid. If the thermal coefficient of the body is less than that of the liquid (as is usually the case), the body will outweigh.

$V_1/V = (\rho - \rho_2)/(\rho_1 - \rho_2)$.

The mass of the body must be equal to the mass of the displaced liquids: $\rho V = \rho_1 V_1 + \rho_2 V_2$, where $V_1 + V_2 = V$. Solving these equations together, we find $V_1 = V \frac{\rho - \rho_2}{\rho_1 - \rho_2}$.

$m \approx 0.85$ kg; $F_1 \approx 1.96$ N; $F_2 \approx 10.29$ N. Does not change.

See the solution to the previous problem.

$H \approx 4.6$ cm; $p \approx 7650$ Pa.

0.2 g/cm$^3$.

Let: mass of the cork $m_1$, its density $\rho_1$, mass of lead $m_2$, its density $\rho_2$. Then the volume of the cork $V_1 = m_1/\rho_1$, volume of lead $V_2 = m_2/\rho_2$, total volume $V = V_1 + V_2 = m_1/\rho_1 + m_2/\rho_2$. According to Archimedes' law, the buoyant force $F$ acts on the body immersed in the liquid [in our case $F = (m_1 + m_2)g - Q$, where $Q$ is the reading of the scales], equal to the weight of the displaced liquid in the volume of the immersed body $F = V \rho_3 g$, where $\rho_3$ is the density of kerosene. Hence $(m_1 + m_2)g - Q = (m_1/\rho_1 + m_2/\rho_2) \rho_3 g$. Solving the obtained equation for the unknown quantity $\rho_1$, we get $\rho_1 = \frac{m_1 g \rho_2 \rho_3}{[(m_1 + m_2)g - Q]\rho_2 - m_2 g \rho_3}$.

When the beaker is immersed in the inverted position, water, compressing the air inside it, partially fills its internal volume. With the same immersion of the beaker in water, in the first case, the displaced volume of water is less than in the second case, and therefore the work is less.

The thin rubber balloon will rise to a great height. At the same height above the Earth, the volume of the thin rubber balloon, and hence the buoyant force acting on it, will be greater than for the fabric balloon.

Decreases by a factor of 1.08.

The lift force $F_L$ of a certain volume of gas is equal to the difference between the weight $m_{air}g$ of the air displaced by the gas and the weight $m_g g$ of the gas itself: $F_L = (m_{air} - m_g)g$. The ratio of the lift forces $(m_{air} - m_{water}) / (m_{air} - m_{helium})$ is equal to the ratio of the differences in molar masses, since the pressure $p$, volume $V$, and temperature $T$ are the same ($pV = mRT/\mu$). Consequently, $F_1/F_2 = (\mu_{air} - \mu_{H2}) / (\mu_{air} - \mu_{He}) = (29 - 2) / (29 - 4) = 27/25 = 1.08$.

The dynamometer shows the sum of the weight of the tube and the force of atmospheric pressure on the tube, equal to the weight of mercury in the tube.

$m = p \cdot 4\pi R^2 / g \approx 5 \times 10^{18}$ kg, where $p$ is atmospheric pressure; $R$ is the radius of the Earth.

The pressures of air and water at level I-I (see figure) are respectively equal: $p_1 = p_0 - \rho_{air} g h$, $p_2 = p_0 - \rho_{water} g h$, where $p_0$ is the pressure at level 0-0; $\rho_{air}$ and $\rho_{water}$ are the densities of air and water. Since $\rho_{water} > \rho_{air}$, then $p_1 > p_2$, and the film will be bent outwards.

It will increase, because the speed of water in the narrow part of the pipe is greater, and the pressure is correspondingly less than in the wide part of the pipe.

$A = (m - \rho V)gh + mgH \approx 150$ J.

To move the body in water, a force $F = mg - F_{buoy} = mg - \rho V g$ ($\rho$ is the density of water) must be applied to it. Therefore, to lift the body from depth $h$ to height $H$ above the water surface, work must be done $A = (m - \rho V)gh + mgH \approx 150$ J. This work is less than the change in potential energy of the body $\Delta W_p = mg(h+H)$ by the amount $\rho V gh \approx 50$ J. The found difference is precisely the amount by which the potential energy of the water decreased when the body was lifted - volume $V$ of water descends from the surface to depth $h$.

$A = \frac{pS}{2} (2H - \frac{p}{\rho_0 g}) \approx 9800$ J, where $\rho_0$ is the density of water.

The water pressure directly under the pump piston is equal to $p - \rho_0 g h$ ($h$ is the piston lift height). The force lifting the piston is equal to the difference between the pressure forces above and below the piston, i.e., the value $F = [p_0 - (p_0 - \rho_0 g h)]S = \rho_0 g h S$. When the piston is lifted to a height $h = p_0 / (\rho_0 g) = 10$ m, this force increases linearly to the value $p_0 S$. Further, this force remains constant. Therefore, the work done in lifting the piston will be equal to: $A = \frac{p_0 S}{2} \frac{p_0}{\rho_0 g} + p_0 S (H - \frac{p_0}{\rho_0 g}) = \frac{p_0 S}{2} (2H - \frac{p_0}{\rho_0 g})$.

$v \approx 44.3$ m/s; $V = 50$ m$^3$.

Applying Torricelli's formula for the outflow velocity from the opening $v^2 = 2gh$, we find $v = \sqrt{2gh}$. In time $t$, a volume of water $V = \pi d^2 v t / 4$ will enter the boat.

$S_1 \approx 4.37$ cm$^2$.

The speed of water at the end of the fire hose is $v_0 = Q/S_0$. Flying out of the fire hose, the water moves with uniform deceleration with acceleration $g$, therefore at height $h$ the speed of water $v_h = \sqrt{v_0^2 - 2gh}$. From the continuity equation of the stream $v_0 S_0 = v_h S_h$, which expresses that the mass of liquid flowing through any cross-section of the stream per unit time is the same, we get $S_h = \frac{v_0 S_0}{v_h} = \frac{Q}{\sqrt{(Q/S_0)^2 - 2gh}}$.

During the time the bullet is in the glass, the water level in it does not have time to change. The resulting high-pressure region upon compression of the water spreads in all directions with the speed of sound, reaches the walls of the glass, and it shatters. This experiment illustrates the low compressibility of liquids.