From the fact that the ball floats we can equate its weight and the buoyant force. Since it is submerged exactly halfway we can write:

$$
begin{gather*}
m , g = F_{text{buoyant}} \
rho_{text{ball}} cdot V cdot g = rho_{text{water}} cdot frac{V}{2} cdot g
end{gather*}
$$

From which we deduce:

$$
begin{align*}
rho_{text{ball}} = frac{rho_{text{water}} }{2}
end{align*}
$$

From this we see that the statement A is incorrect.

Obviously statements B and C are incorrect since the forces acting on the ball are balanced.

We conclude that statement D is correct.

First we will assume that the temperature in the lake is roughly constant, so as the air bubble rises the expansion process is isothermal in essence.

Secondly conclude that the gas pressure is always equal to the static pressure in the water surrounding it for that given depth. Knowing this we can write:

$$
begin{align*}
P(h_1) cdot V_1 = P(h_2) cdot V_2
end{align*}
$$

Our two heights are $h_1 = 21text{ m}$ and $h_2 = 0text{ m}$.

Now we find the pressures by increasing the atmospheric pressure for the amount of hydrostatic pressure as follows:

$$
begin{align*}
P(h) = P_{text{atm}} + rho cdot g cdot h
end{align*}
$$

Inserting the numbers we have:

$$
begin{align*}
P(h_1) &= 101.3text{ kPa} + 1000 ; frac{text{kg}}{text{m}^3} cdot 9.81 ; frac{text{m}}{text{s}^2} cdot 21text{ m} = 307.31text{ kPa} \
P(h_2) &= 101.3text{ kPa}
end{align*}
$$

Now we find the volume $V_2$:

$$
begin{align*}
V_2 &= V_1 cdot frac{P(h_1)}{P(h_2)} \
V_2 &= 0.001text{ m}^3 cdot frac{307.31text{ kPa} }{101.3text{ kPa}} \
V_2 &approx 0.003text{ m}^3
end{align*}
$$

The volume of the bubble is closest to C) $V_2 = 0.003text{ m}^3$

We know that the cross sectional area of a fluid flow and the speed of its flow are related by the Equation of Continuity, which can be understood as a continuity of volume, and follows from the incompressibility of the fluid.
It states:

$$
begin{equation*}
A cdot v = text{const}
end{equation*}
$$

Firstly consider this, both the internal energy and average kinetic energy of the particles are proportional to the temperature.

Secondly, we know that the ideal gas equation reads:

$$
begin{equation*}
P cdot V = N cdot k cdot T
end{equation*}
$$

Which means that the pressure is proportional to $T$ when the volume and number of particles are fixed, as they are in our sealed, rigid container.

So it follows that B and C are incorrect.

D is obviously incorrect since $Psim T$

The correct answer is A.

We know from the problem that the water flow is faster in the narrower part of the pipe since $8 ; frac{text{m}}{text{s}} > 4 ; frac{text{m}}{text{s}}$.
This could also have been concluded from the Equation of Continuity.

We know that the static pressure is lower in regions where the speed of the fluid is greater, this stems from Bernoulli’s equation. Notice that the pipe was horizontal, so we didn’t have to take into consideration the gravitational potential term in the equation.

We conclude that the static pressure decreases in the narrow part of the pipe since the speed of the fluid flow is greater there.

### Knowns

– The volume of the container $V = 0.022text{ m}^3$

– The pressure of the container $P = 310 text{ kPa}$

– The temperature of the container $T = 25text{textdegree}text{C} = 298.15text{ K}$

### Calculation

To calculate the number of helium atoms we use the Ideal Gas Equation:

$$
begin{equation*}
PV = N , k , T
end{equation*}
$$

rearranging for number of particles:

$$
begin{equation*}
N = frac{PV}{k , T}
end{equation*}
$$

Plugging in the values we get:

$$
begin{align*}
N = frac{310 cdot 10^{3}text{ Pa} cdot 0.022text{ m}^3 }{1.38 cdot 10^{-23} ; frac{text{J}}{text{K}} cdot 298.15text{ K} } = 1.66 cdot 10^{24}
end{align*}
$$

The number of helium atoms in the balloon is $N = 1.66 cdot 10^{24}$, which is approximately $1.7 cdot 10^{24}$.

The correct solution is:
begin{enumerate}[A)]
item
$N = 1.7 cdot 10^{24}$
end{enumerate}

From the fact that the object floats we can equate its weight and the buoyant force. Since $75%$ of the object is submerged in the fluid we can write:

$$
begin{gather*}
m , g = F_{text{buoyant}} \
rho_{text{object}} cdot V cdot g = rho_{text{water}} cdot 0.75 V cdot g
end{gather*}
$$

From which we deduce:

$$
begin{align*}
rho_{text{object}} = 0.75 , rho_{text{water}}
end{align*}
$$

Now renaming $rho_{text{water}} = rho$

We see that statement B is the correct one.

Statement B is true, $rho_{text{object}} = frac{3}{4} cdot rho$

We need to find the spring constant of a new spring we get by cutting in half a spring of coefficient of stiffness $k$.

First remember, the spring constant is defined as:

$$
begin{equation*}
k = frac{F}{x}
end{equation*}
$$

Where $F$ is the force pulling on the spring, and $x$ is its elongation.

Now consider the following, if we joined together end to end two springs of half length, such as the new one we get by cutting a spring in half, we would construct the initial spring.

When one pulls on such a system of two joined springs he would get that the same force $F$ stretches both springs, and that each of them is extended. Call these elongations $x_1$ and $x_2$, but be aware that $x_1 = x_2$ since the spring are the same.

So the effective coefficient of the system would be:

$$
begin{align*}
k_{_{eff}} = frac{F}{(x_1 + x_2)}
end{align*}
$$

Where you could also write the individual equations:

$$
begin{align*}
k_1 &= frac{F}{x_1} \
k_2 &= frac{F}{x_2}
end{align*}
$$

In our case the situation gets simplified drastically by $k_1 = k_2$ and $x_1 = x_2$ since the springs are the same, so we have:

$$
begin{align*}
k_{text{eff}} = frac{F}{2x_1} = frac{1}{2} cdot k_1
end{align*}
$$

We conclude that the coefficient of stiffness of the total spring is half as big as the one associated with the half spring.

That is, by cutting a spring in half, we increase the stiffness constant of each new half length spring by two

The correct answer is C) $2k$

The correct answer is C) $2k$

When a container is closed by a massive piston, the pressure in the gas can be found by increasing the atmospheric pressure by the amount the massive piston contributes:

$$
begin{equation*}
P = P_{text{atm}} + frac{m , g}{A}
end{equation*}
$$

Where $m$ is the mass of the piston, and $A$ its area.

The initial pressure in the container $P_{text{i}}$ is found by using $m = 2text{ kg}$.

When another mass is added the total mass becomes $m = 2text{ kg} + 1text{ kg} = 3text{ kg}$.

Obviously this will result in a larger pressure $P_{text{f}}$:

$$
begin{align*}
P_{text{f}} = P_{text{i}} + frac{1text{ kg}}{A}
end{align*}
$$

Now for the volume we use the fact the the process is isothermal in essence to rewrite the Ideal gas Equation as follows:

$$
begin{equation*}
P cdot V = text{const}
end{equation*}
$$

Now applying it for the initial and final positions we get:

$$
begin{align*}
P_{text{i}} cdot V_{text{i}} = P_{text{f}} cdot V_{text{f}}
end{align*}
$$

Rearranging:

$$
begin{align*}
V_{text{f}} = V_{text{i}} cdot frac{P_{text{i}} }{P_{text{f}} }
end{align*}
$$

Since we know that the pressure increases, we see that $frac{P_{text{i}} }{P_{text{f}}} < 1$ so $V_{text{f}} < V_{text{i}}$

This could have been reasoned by common sense as well.

When the mass is added the pressure increases $P_{text{f}} > P_{text{i}}$ and the volume decreases $V_{text{f}} < V_{text{i}}$