Buoyancy, Archimedes' Principle and Laws of Floatation Questions in English

(D) The forces acting on the cylinder are the gravitational force $(mg)$ acting downwards and the buoyant force $(F_b)$ acting upwards.
According to Archimedes' principle,the buoyant force is equal to the weight of the fluid displaced by the cylinder.
The volume of the cylinder is $V = \pi r^2 h$.
Therefore,the buoyant force is $F_b = \rho Vg = \rho (\pi r^2 h) g = \pi r^2 \rho gh$.
The net force $F_{net}$ on the cylinder is the difference between the buoyant force and the gravitational force.
$F_{net} = F_b - mg = \pi r^2 \rho gh - mg$.
Since we are asked for the magnitude of the net force,we take the absolute value:
$|F_{net}| = |\pi r^2 \rho gh - mg|$.

(B) Initially,the lead mass is submerged,so it displaces a volume of water equal to its own volume $(V_{lead})$. The wood block floats,displacing water equal to its weight $(W_{wood}/\rho_{water}g)$.
When the lead is removed,the water level drops because the volume $V_{lead}$ is no longer displaced.
When the lead is placed on the wood,the total weight supported by the water increases by the weight of the lead $(W_{lead} = m_{lead}g)$.
According to Archimedes' principle,the wood block must displace an additional volume of water equal to the weight of the lead divided by the density of water: $\Delta V = W_{lead} / (\rho_{water}g) = m_{lead} / \rho_{water}$.
Since the density of lead $(\rho_{lead} \approx 11.3 \times 10^3 \ kg/m^3)$ is much greater than the density of water $(\rho_{water} = 1.0 \times 10^3 \ kg/m^3)$,the volume of water displaced by the weight of the lead $(m_{lead}/\rho_{water})$ is significantly larger than the volume of the lead itself $(m_{lead}/\rho_{lead})$.
Therefore,the water level rises higher than it was when the lead was submerged,but since the container was initially full and water spilled out when the wood was first placed,the level will rise back to the brim.

(A) For the sphere to float just submerged in water,the weight of the sphere must be equal to the buoyant force exerted by the water.
Weight of the sphere = Volume of solid material $\times$ density of material $\times g = \frac{4}{3} \pi (R^3 - r^3) \rho g$
Buoyant force = Volume of displaced water $\times$ density of water $\times g = \frac{4}{3} \pi R^3 \rho_w g$
Equating the two: $\frac{4}{3} \pi (R^3 - r^3) \rho g = \frac{4}{3} \pi R^3 \rho_w g$
Given relative density $\frac{\rho}{\rho_w} = 9$,we have:
$9(R^3 - r^3) = R^3$
$9R^3 - 9r^3 = R^3$
$8R^3 = 9r^3$
$r^3 = \frac{8}{9} R^3$

(A) Mass of empty balloon $= 1\, g$.
Mass of balloon filled with water $= 101\, g$.
Mass of water inside the balloon $= 101\, g - 1\, g = 100\, g$.
Since the density of water is $\rho_w$,the volume of the balloon $V = \frac{100\, g}{\rho_w} = 100\, cm^3$.
When the balloon is fully immersed in water,it experiences an upward buoyant force equal to the weight of the displaced water.
Buoyant force $F_B = V \cdot \rho_w \cdot g = 100\, cm^3 \cdot 1\, g/cm^3 \cdot g = 100\, g$ (in terms of weight).
The effective weight of the balloon in water is given by: $W_{\text{eff}} = W_{\text{total}} - F_B$.
$W_{\text{eff}} = 101\, g - 100\, g = 1\, g$.
Therefore,the weight of the balloon in water is $1\, g$.

(B) Let the volume of the ball be $V$,the density of the ball be $\rho_b = 0.4 \times 10^3 \, kg/m^3$,and the density of water be $\rho_w = 1.0 \times 10^3 \, kg/m^3$.
The potential energy of the ball at the surface of the water is $U = mgh = V \rho_b gh$.
When the ball enters the water,it experiences an upward buoyant force (upthrust) $F_B = V \rho_w g$ and a downward gravitational force $W = V \rho_b g$.
The net force acting on the ball while submerged is $F_{net} = F_B - W = Vg(\rho_w - \rho_b)$,which is directed upwards.
By the work-energy theorem,the work done by the net force as the ball sinks to a depth $d$ must equal the initial potential energy of the ball:
$W_{net} = F_{net} \times d = Vg(\rho_w - \rho_b)d$.
Equating the work done to the initial potential energy:
$Vg(\rho_w - \rho_b)d = V \rho_b gh$.
Solving for $d$:
$d = \frac{\rho_b}{\rho_w - \rho_b} h$.
Substituting the given values:
$d = \frac{0.4 \times 10^3}{1.0 \times 10^3 - 0.4 \times 10^3} \times 9 \, cm = \frac{0.4}{0.6} \times 9 \, cm = \frac{2}{3} \times 9 \, cm = 6 \, cm$.

(A) For a floating body, the weight of the body is equal to the total buoyant force exerted by the liquids.
Weight of the cylinder $W = (\text{Volume}) \times (\text{Density of solid}) \times g = (A \times L) \times D \times g$.
Buoyant force $F_B = (\text{Volume submerged in liquid 1}) \times (\text{Density of liquid 1}) \times g + (\text{Volume submerged in liquid 2}) \times (\text{Density of liquid 2}) \times g$.
Given: Volume in lighter liquid (density $d$) is $V_1 = A \times (3L/4)$ and volume in denser liquid (density $2d$) is $V_2 = A \times (L/4)$.
Equating weight and buoyant force: $(A \times L) \times D \times g = (A \times 3L/4) \times d \times g + (A \times L/4) \times 2d \times g$.
Dividing both sides by $A \times L \times g$: $D = (3/4)d + (1/4) \times 2d$.
$D = (3/4)d + (2/4)d = 5d/4$.

(D) The velocity $u$ of the body when it enters the liquid is given by conservation of energy:
$mgh = \frac{1}{2} mu^2 \implies u = \sqrt{2gh}$
Let $V$ be the volume of the body.
Mass of the body $m = Vd$.
Weight of the body $W = Vdg$.
Buoyant force $F_B = VDg$.
Net upward force $F_{net} = F_B - W = Vg(D-d)$.
Retardation $a = \frac{F_{net}}{m} = \frac{Vg(D-d)}{Vd} = \left( \frac{D-d}{d} \right) g$.
Since the body is moving downwards, the acceleration is $a' = -\left( \frac{D-d}{d} \right) g$.
Using the equation of motion $v = u + a't$, where $v = 0$ at the point of instantaneous rest:
$0 = \sqrt{2gh} - \left( \frac{D-d}{d} \right) gt$
$t = \left( \frac{d}{D-d} \right) \sqrt{\frac{2h}{g}}$

(D) The upthrust (buoyant force) on a body submerged in a liquid is given by the formula $F_B = V_{disp} \cdot \rho_{liquid} \cdot g_{eff}$,where $g_{eff}$ is the effective acceleration due to gravity.
When the entire system (beaker and liquid) is falling freely under gravity,the effective acceleration $g_{eff}$ becomes $g - g = 0$.
Since the upthrust depends on the effective gravity,and $g_{eff} = 0$,the upthrust on the body becomes zero.

(D) The apparent loss in weight of the body when immersed in water is equal to the buoyant force (thrust) exerted by the water.
Loss in weight $= 264\, g - 221\, g = 43\, g$.
Since the density of water is $\rho_w = 1\, g/cc$,the volume of the water displaced is equal to the total volume of the copper piece $(V_{total})$.
$V_{total} = \frac{\text{Loss in weight}}{\rho_w} = \frac{43\, g}{1\, g/cc} = 43\, cc$.
The volume of the copper material $(V_{material})$ is calculated using its mass and density:
$V_{material} = \frac{\text{Mass}}{\text{Density of } Cu} = \frac{264\, g}{8.8\, g/cc} = 30\, cc$.
The volume of the internal cavity $(V_{cavity})$ is the difference between the total volume and the volume of the material:
$V_{cavity} = V_{total} - V_{material} = 43\, cc - 30\, cc = 13\, cc$.

(B) Let $a$ be the side of the cube and $\rho_w$ be the density of water $(1 \, g/cm^3)$.
When the mass $m = 200 \, g$ is placed on the cube,it floats. The weight of the mass is supported by the buoyant force equal to the weight of the water displaced by the volume of the cube submerged.
When the mass is removed,the cube rises by $h = 2 \, cm$. This means the volume of water displaced decreases by an amount equal to the volume of the cube's cross-section multiplied by the rise in height.
The weight of the removed mass is equal to the weight of the water displaced by the volume that was previously submerged due to that mass.
Weight of mass = Weight of water displaced by the volume $(a \times a \times h)$.
$m \cdot g = (a^2 \cdot h) \cdot \rho_w \cdot g$.
Substituting the values: $200 = a^2 \times 2 \times 1$.
$a^2 = 100$.
$a = 10 \, cm$.

(C) Let the mass of water consumed by the man be $m$.
Initially,the man and the boat displace a volume of water $V_1$ such that the weight of the displaced water equals the total weight of the man and the boat (Archimedes' Principle).
When the man drinks mass $m$ of water,the total weight of the man and the boat increases by $mg$.
To support this additional weight,the boat must displace an additional volume of water $V_{add} = \frac{m}{\rho}$,where $\rho$ is the density of water.
However,by drinking mass $m$ of water from the pond,the total volume of water in the pond decreases by $V_{removed} = \frac{m}{\rho}$.
Since the additional volume displaced by the boat $(V_{add})$ is exactly equal to the volume of water removed from the pond $(V_{removed})$,the net change in the water level of the pond is zero.
Therefore,the level of water in the pond remains the same.

(B) Let $m_i$ be the mass of ice and $m_f$ be the mass of the iron piece.
Before the ice melts, the total weight of the system is $(m_i + m_f)g$. Since the system is floating, the buoyant force equals the weight of the system: $F_b = (m_i + m_f)g$.
The volume of water displaced by the floating system is $V_{disp} = \frac{F_b}{\rho_w g} = \frac{m_i + m_f}{\rho_w} = \frac{m_i}{\rho_w} + \frac{m_f}{\rho_w}$.
After the ice melts, the ice becomes water with volume $V_{water\_from\_ice} = \frac{m_i}{\rho_w}$. The iron piece sinks to the bottom, displacing its own volume $V_{iron} = \frac{m_f}{\rho_f}$.
The total volume occupied by the water and the iron piece is $V_{total} = \frac{m_i}{\rho_w} + \frac{m_f}{\rho_f}$.
Comparing the displaced volumes: $V_{disp} - V_{total} = (\frac{m_i}{\rho_w} + \frac{m_f}{\rho_w}) - (\frac{m_i}{\rho_w} + \frac{m_f}{\rho_f}) = m_f(\frac{1}{\rho_w} - \frac{1}{\rho_f})$.
Since the density of iron $\rho_f$ is greater than the density of water $\rho_w$, we have $\frac{1}{\rho_w} > \frac{1}{\rho_f}$.
Therefore, $V_{disp} > V_{total}$, which means the water level decreases.

(C) When the block is moved upward by a height $h$ within the liquid,the system consists of the block and the displaced liquid.
The change in potential energy of the block is $\Delta U_b = m_b gh = (\sigma_b V) gh$.
As the block moves up,an equal volume $V$ of liquid moves down by height $h$ to fill the space previously occupied by the block. The change in potential energy of the displaced liquid is $\Delta U_l = -m_l gh = -(\sigma_l V) gh$.
The total change in gravitational potential energy of the system is $\Delta U = \Delta U_b + \Delta U_l$.
$\Delta U = (\sigma_b V) gh - (\sigma_l V) gh = (\sigma_b - \sigma_l) Vgh$.

(B) The volume of the cubical block is $V = l^3$.
Let $h$ be the height of the block above the surface of the mercury.
The depth of the block submerged in mercury is $(l - h)$.
The volume of mercury displaced is $V_{disp} = (l - h)l^2$.
According to the law of floatation,the weight of the displaced mercury must equal the weight of the steel block.
Weight of displaced mercury = $(l - h)l^2 \rho_m g$.
Weight of the steel block = $l^3 \rho_s g$.
Equating the two: $(l - h)l^2 \rho_m g = l^3 \rho_s g$.
Dividing both sides by $l^2 g$: $(l - h) \rho_m = l \rho_s$.
$l - h = l \frac{\rho_s}{\rho_m}$.
$h = l - l \frac{\rho_s}{\rho_m} = l(1 - \frac{\rho_s}{\rho_m})$.

(D) The total mass of the balloon system is $M_{total} = M + m$.
When the balloon is completely immersed in water, it displaces a volume of water equal to its own volume.
Since the balloon contains water of mass $M$ and the balloon material itself has mass $m$, the total volume $V$ of the system is the volume of the water $M$ plus the volume of the balloon material $m$.
However, the density of the water inside is the same as the density of the surrounding water.
According to Archimedes' Principle, the buoyant force $F_B$ acting on the balloon is equal to the weight of the displaced water: $F_B = (M + \rho_{material} \cdot V_{material})g$, where $\rho_{material} \cdot V_{material} = m$.
Thus, $F_B = (M + m)g$.
The apparent weight $W_{app}$ is given by $W_{app} = W_{actual} - F_B = (M + m)g - (M + m)g = 0$.
Wait, if the balloon material has negligible volume, the buoyant force is equal to the weight of the water displaced, which is $M \cdot g$. The total weight is $(M + m)g$. The apparent weight is $(M + m)g - Mg = mg$.
Therefore, the apparent mass is $m$.

(B) The buoyant force $F_B$ acting on an object submerged in a fluid is given by Archimedes' principle: $F_B = \rho_{water} V_{displaced} g$, where $\rho_{water}$ is the density of water, $V_{displaced}$ is the volume of the fluid displaced by the object, and $g$ is the acceleration due to gravity.
As the jar is pushed deeper into the water, the hydrostatic pressure at the depth of the jar increases.
According to Boyle's Law $(PV = \text{constant})$, as the pressure $P$ on the trapped air increases, the volume $V$ of the trapped air decreases.
Since the jar is submerged, the total volume of the system displaced $(V_{displaced})$ is the sum of the volume of the glass jar itself and the volume of the trapped air.
As the volume of the trapped air decreases, the total volume of the system displaced $(V_{displaced})$ decreases.
Consequently, the buoyant force $F_B = \rho_{water} V_{displaced} g$ decreases.

(C) In the first situation,the block is floating in water. By the law of floatation,the weight of the block equals the weight of the displaced water:
$V_b \rho_b g = V_s \rho_w g$
$\frac{V_s}{V_b} = \frac{\rho_b}{\rho_w} = \frac{4}{5} \quad ... (i)$
Here,$V_b$ is the volume of the block,$V_s$ is the submerged volume,$\rho_b$ is the density of the block,and $\rho_w$ is the density of water.
In the second situation,the block is floating in a mixture of oil and water. The total weight of the block is balanced by the buoyant force from both the oil and the water:
$V_b \rho_b g = \left(\frac{V_b}{2}\right) \rho_o g + \left(\frac{V_b}{2}\right) \rho_w g$
Dividing by $V_b g$,we get:
$\rho_b = \frac{\rho_o}{2} + \frac{\rho_w}{2}$
$2 \rho_b = \rho_o + \rho_w$
Substitute $\rho_b = \frac{4}{5} \rho_w$ from equation $(i)$:
$2 \left(\frac{4}{5} \rho_w\right) = \rho_o + \rho_w$
$\frac{8}{5} \rho_w - \rho_w = \rho_o$
$\rho_o = \frac{3}{5} \rho_w = 0.6 \rho_w$
Therefore,the relative density of oil with respect to water is $\frac{\rho_o}{\rho_w} = 0.6$.

(D) Let the side of the cube be $\ell = 0.5\,m$. The volume of the cube is $V = \ell^3 = (0.5)^3 = 0.125\,m^3$.
Initially,$30\%$ of the volume is submerged,so the buoyant force balances the weight of the block:
$V_{sub} \cdot \rho_w \cdot g = V \cdot \rho_{block} \cdot g$
$0.3 \cdot V \cdot \rho_w = V \cdot \rho_{block}$
$\rho_{block} = 0.3 \cdot \rho_w = 0.3 \cdot 1000 = 300\,kg/m^3$.
The mass of the block is $m_{block} = V \cdot \rho_{block} = 0.125 \cdot 300 = 37.5\,kg$.
When an additional mass $M$ is placed on the block,the block is fully submerged,meaning the total weight equals the buoyant force for the full volume $V$:
$(m_{block} + M)g = V \cdot \rho_w \cdot g$
$M = V \cdot \rho_w - m_{block} = (0.125 \cdot 1000) - 37.5 = 125 - 37.5 = 87.5\,kg$.

(C) Let $R$ be the outer radius and $r$ be the radius of the cavity.
The density of the sphere material is $\rho = 27 \times \rho_{water} = 27 \rho_{w}$.
For the sphere to just sink in water,its weight must be equal to the buoyant force acting on it.
Weight of the sphere = $V_{solid} \times \rho \times g = \frac{4}{3} \pi (R^3 - r^3) \times 27 \rho_{w} \times g$.
Buoyant force = $V_{total} \times \rho_{w} \times g = \frac{4}{3} \pi R^3 \times \rho_{w} \times g$.
Equating the two: $\frac{4}{3} \pi (R^3 - r^3) \times 27 \rho_{w} g = \frac{4}{3} \pi R^3 \rho_{w} g$.
$27(R^3 - r^3) = R^3$.
$27R^3 - 27r^3 = R^3$.
$26R^3 = 27r^3$.
$\frac{r^3}{R^3} = \frac{26}{27}$.
Taking the cube root on both sides,$\frac{r}{R} = \frac{(26)^{1/3}}{3}$.

(B) Let the mass of the cube be $m$ and the side length be $a$. The density of water is $\rho_w = 1\,g/cm^3$.
When the $200\,g$ mass is placed on the cube,it floats such that the entire volume is submerged (or at the surface level). The total weight equals the buoyant force:
$(m + 200)g = a^3 \rho_w g$ --- $(i)$
When the mass is removed,the cube floats with $2\,cm$ above the water level. The submerged volume is $a^2(a - 2)$:
$mg = a^2(a - 2) \rho_w g$ --- $(ii)$
Subtracting equation $(ii)$ from equation $(i)$:
$(m + 200)g - mg = a^3 \rho_w g - a^2(a - 2) \rho_w g$
$200 = a^3 \rho_w - (a^3 - 2a^2) \rho_w$
$200 = a^3 \rho_w - a^3 \rho_w + 2a^2 \rho_w$
$200 = 2a^2 \rho_w$
Since $\rho_w = 1\,g/cm^3$:
$200 = 2a^2$
$a^2 = 100$
$a = 10\,cm$.

(D) Let $V$ be the volume of the sphere,$\rho_s$ be the density of the sphere,and $\rho_w$ be the density of water.
Given that the sphere is $\eta$ times lighter than water,we have $\rho_w = \eta \rho_s$.
The mass of the sphere is $m = V \rho_s$,so $V = \frac{m}{\rho_s}$.
The forces acting on the sphere are:
$1$. Weight $(W = mg)$ acting downwards.
$2$. Tension $(T)$ in the string acting downwards.
$3$. Buoyant force $(F_B = V \rho_w g)$ acting upwards.
For equilibrium,the upward force must equal the downward forces:
$F_B = W + T$
$T = F_B - W$
$T = V \rho_w g - mg$
Substitute $V = \frac{m}{\rho_s}$ and $\rho_w = \eta \rho_s$:
$T = \left( \frac{m}{\rho_s} \right) (\eta \rho_s) g - mg$
$T = \eta mg - mg$
$T = (\eta - 1) mg$

(C) Let the mass of wood be $M$ and the mass of concrete be $m$. The density of water is $\rho_w$. The specific gravity is the ratio of the density of the substance to the density of water.
Volume of wood $V_w = \frac{M}{0.5 \rho_w} = \frac{2M}{\rho_w}$.
Volume of concrete $V_c = \frac{m}{2.5 \rho_w} = \frac{m}{2.5 \rho_w} = \frac{0.4m}{\rho_w}$.
For the combination to float with its entire volume submerged,the total weight must equal the buoyant force:
$(M + m)g = (V_w + V_c) \rho_w g$
$M + m = \left(\frac{2M}{\rho_w} + \frac{0.4m}{\rho_w}\right) \rho_w$
$M + m = 2M + 0.4m$
$m - 0.4m = 2M - M$
$0.6m = M$
$\frac{M}{m} = \frac{0.6}{1} = \frac{6}{10} = \frac{3}{5}$.

(A) Initially,the block and the coin float together. The total weight of the block and the coin is balanced by the buoyant force,which is equal to the weight of the displaced water. Let $M$ be the mass of the block and $m$ be the mass of the coin. The total weight is $(M+m)g$. The volume of water displaced is $V_{disp} = (M+m)/\rho_w$,where $\rho_w$ is the density of water.
When the coin falls into the water,the block alone floats. The weight of the block is $Mg$. The volume of water displaced by the block is $V'_{disp} = M/\rho_w$. The coin sinks to the bottom and displaces a volume of water equal to its own volume,$V_{coin} = m/\rho_c$,where $\rho_c$ is the density of the coin.
The total volume of water displaced is now $V_{total} = V'_{disp} + V_{coin} = M/\rho_w + m/\rho_c$. Since the density of the coin $\rho_c$ is much greater than the density of water $\rho_w$,$m/\rho_c < m/\rho_w$. Therefore,the total volume of water displaced decreases $(V_{total} < V_{disp})$,which causes the water level $h$ to decrease.
The submerged depth of the block $l$ is proportional to the weight of the block alone $(Mg)$,which is less than the weight of the block plus the coin $(Mg+mg)$. Thus,the submerged depth $l$ also decreases.

(C) According to Archimedes' principle,the buoyant force $F_B$ is equal to the weight of the displaced liquid,given by $F_B = V_{sub} \rho_{water} g_{eff}$.
When the lift accelerates upwards with acceleration $a$,the effective acceleration due to gravity becomes $g_{eff} = g + a$.
Both the wooden piece and the water experience this same effective gravity $g_{eff}$.
The weight of the wooden piece is $W = V_{total} \rho_{wood} g_{eff}$.
For equilibrium,$W = F_B$,which implies $V_{total} \rho_{wood} g_{eff} = V_{sub} \rho_{water} g_{eff}$.
Since $g_{eff}$ cancels out from both sides,the fraction of the volume submerged $(V_{sub} / V_{total} = \rho_{wood} / \rho_{water})$ remains independent of the acceleration of the lift.
Therefore,the wooden piece will remain half-submerged.
Thus,option $C$ is correct.

(A) According to the principle of floatation, the additional weight of the man must be equal to the weight of the water displaced by the boat due to the sinking.
The volume of water displaced by the boat is $V = \text{length} \times \text{breadth} \times \text{depth}$.
Given: $\text{length} = 3\,m$, $\text{breadth} = 2\,m$, and $\text{depth} = 1\,cm = 0.01\,m$.
$V = 3 \times 2 \times 0.01 = 0.06\,m^3$.
The mass of the displaced water is $m_w = \rho \times V$, where $\rho = 1000\,kg/m^3$.
$m_w = 1000 \times 0.06 = 60\,kg$.
Since the weight of the man equals the weight of the displaced water, the mass of the man is $60\,kg$.

(B) The weight of the ornament in air is $W_{air} = 10\, g$ (mass $m = 10\, g$).
The weight of the ornament in water is $W_{water} = 6\, g$.
The apparent loss in weight is equal to the buoyant force,which equals the weight of the water displaced: $F_{buoyant} = W_{air} - W_{water} = 10\, g - 6\, g = 4\, g$.
Since the density of water is $\rho_{water} = 1\, g/cc$,the volume of water displaced is $V_{total} = \frac{F_{buoyant}}{\rho_{water} \cdot g} = 4\, cc$.
This $V_{total}$ represents the external volume of the ornament (material volume + cavity volume).
The volume of the material alone is $V_{material} = \frac{m}{\rho_{material}} = \frac{10\, g}{20\, g/cc} = 0.5\, cc$.
The volume of the cavity is $V_{cavity} = V_{total} - V_{material} = 4\, cc - 0.5\, cc = 3.5\, cc$.

(C) Let $V$ be the outer volume of the sphere.
Given that the inner volume is half the outer volume,the volume of the material of the sphere is $V_{material} = V - V/2 = V/2$.
According to the principle of floatation,the weight of the sphere is equal to the weight of the water displaced by the submerged part.
Weight of sphere = $V_{material} \times \rho \times g = (V/2) \times \rho \times g$,where $\rho$ is the density of the material.
Weight of displaced water = $V_{submerged} \times \sigma \times g = (4/5)V \times \sigma \times g$,where $\sigma = 10^3 \ kg/m^3$ is the density of water.
Equating the two: $(V/2) \rho g = (4/5) V \sigma g$.
$\rho/2 = 4\sigma/5$.
$\rho = (8/5) \sigma = 1.6 \times 10^3 \ kg/m^3$.

(B) Let the mass of the cube be $m$ and the side length be $a$. The density of water is $\rho_w = 1\,g/cm^3$.
When the $200\,g$ mass is on the cube,it is fully submerged or floating at a specific level. Assuming it floats such that the entire volume is submerged (or the displacement equals the total weight):
$(m + 200)g = a^3 \rho_w g$ --- $(i)$
When the mass is removed,the cube rises by $2\,cm$,meaning the submerged height becomes $(a - 2)$. The weight of the cube is now balanced by the buoyant force on the submerged volume:
$mg = a^2(a - 2) \rho_w g$ --- $(ii)$
Subtracting equation $(ii)$ from $(i)$:
$(m + 200)g - mg = a^3 \rho_w g - a^2(a - 2) \rho_w g$
$200 = a^3 \rho_w - (a^3 - 2a^2) \rho_w$
Since $\rho_w = 1\,g/cm^3$:
$200 = a^3 - a^3 + 2a^2$
$200 = 2a^2$
$a^2 = 100$
$a = 10\,cm$.

(A) Let the weight of the block in air be $W_{actual} = 60\,N$.
Let the apparent weight of the block in water be $W_{app} = 40\,N$.
The buoyant force $F_B$ acting on the block is the difference between the actual weight and the apparent weight:
$F_B = W_{actual} - W_{app} = 60\,N - 40\,N = 20\,N$.
According to Archimedes' principle,the buoyant force is equal to the weight of the water displaced:
$F_B = V \cdot \rho_w \cdot g = 20\,N$,where $V$ is the volume of the block and $\rho_w$ is the density of water.
The actual weight of the block is $W_{actual} = V \cdot \rho_s \cdot g = 60\,N$,where $\rho_s$ is the density of the block.
Dividing the two equations:
$\frac{W_{actual}}{F_B} = \frac{V \cdot \rho_s \cdot g}{V \cdot \rho_w \cdot g} = \frac{\rho_s}{\rho_w} = \text{Specific Density}$.
$\text{Specific Density} = \frac{60}{20} = 3$.

(B) When the stones are in the boat, they displace a volume of water equal to their weight (Archimedes' Principle for floating objects). Let $W_s$ be the weight of the stones and $\rho_w$ be the density of water. The volume of water displaced is $V_1 = W_s / (\rho_w g)$.
When the stones are dropped into the water, they sink and displace a volume of water equal to their own volume. Let $V_s$ be the volume of the stones and $\rho_s$ be the density of the stones. The volume of water displaced is $V_2 = V_s = W_s / (\rho_s g)$.
Since the density of the stones is greater than the density of water $(\rho_s > \rho_w)$, it follows that $V_1 > V_2$.
Because the volume of water displaced decreases when the stones are moved from the boat to the bottom of the tank, the water level in the tank falls.

(A) The relative density $(RD)$ of a solid is defined as the ratio of its weight in air to the loss of weight in water.
$RD = \frac{\text{Weight in air}}{\text{Weight in air} - \text{Weight in water}}$
Given:
Weight in air $(W_a)$ = $15 \ N$
Weight in water $(W_w)$ = $12 \ N$
$RD = \frac{15}{15 - 12} = \frac{15}{3} = 5$
Therefore,the relative density of the block is $5$.

(A) On Earth,a balloon filled with hydrogen rises because the buoyant force (upthrust) is greater than the weight of the balloon.
On the moon,there is no atmosphere,which means there is no air to provide a buoyant force.
Therefore,the balloon will undergo free fall under the influence of the moon's gravity alone.
The acceleration due to gravity on the moon is $\frac{g}{6}$,where $g$ is the acceleration due to gravity on Earth.
Since the balloon is in free fall,its acceleration will be equal to the acceleration due to gravity on the moon,which is $\frac{g}{6}$.
Thus,both the $Assertion$ and $Reason$ are correct,and the $Reason$ correctly explains why the balloon falls with that specific acceleration.

(D) According to the principle of flotation,for the boy and the log to float,the total weight of the system must be equal to the upthrust (buoyant force) exerted by the water.
Let $V$ be the volume of the wooden log.
Mass of the boy = $60\, kg$.
Density of wood $\rho_{wood} = 0.6 \times 1000 = 600\, kg/m^3$.
Density of water $\rho_{water} = 1000\, kg/m^3$.
Total weight of the system = Weight of boy + Weight of log
$= 60g + (V \times 600)g$
Upthrust (Buoyant force) = Weight of water displaced by the log
$= V \times 1000 \times g$
Equating the two:
$60g + 600Vg = 1000Vg$
$60 = 1000V - 600V$
$60 = 400V$
$V = \frac{60}{400} = \frac{3}{20}\, m^3$.

(D) The buoyant force acts at the centre of buoyancy,which is the centre of mass of the displaced fluid. This point coincides with the centre of mass of the object only if the object is homogeneous (uniform density). Since the object is not necessarily homogeneous,the Assertion is incorrect.
The Reason statement is a general principle in mechanics: for a uniform force field (like gravity) acting on a body,the resultant force acts at the centre of mass. However,the buoyant force is not a uniform force field acting on the object itself; it is the resultant of pressure forces acting on the surface of the object. Thus,the Reason is also incorrect.

(D) For the sphere to float with minimum liquid density,it must be completely submerged. In this state,the weight of the sphere equals the buoyant force.
Weight $W = mg = \int \rho(r) g dV = \int_{0}^{R} \rho_{0} \left(1 - \frac{r^{2}}{R^{2}}\right) g (4 \pi r^{2} dr)$.
$W = 4 \pi \rho_{0} g \int_{0}^{R} \left(r^{2} - \frac{r^{4}}{R^{2}}\right) dr$.
$W = 4 \pi \rho_{0} g \left[ \frac{r^{3}}{3} - \frac{r^{5}}{5R^{2}} \right]_{0}^{R} = 4 \pi \rho_{0} g \left( \frac{R^{3}}{3} - \frac{R^{3}}{5} \right) = 4 \pi \rho_{0} g \left( \frac{2R^{3}}{15} \right) = \frac{8}{15} \pi R^{3} \rho_{0} g$.
Buoyant force $B = V_{sphere} \rho_{l} g = \frac{4}{3} \pi R^{3} \rho_{l} g$.
Equating $W = B$: $\frac{8}{15} \pi R^{3} \rho_{0} g = \frac{4}{3} \pi R^{3} \rho_{l} g$.
Solving for $\rho_{l}$: $\rho_{l} = \frac{8}{15} \times \frac{3}{4} \rho_{0} = \frac{2}{5} \rho_{0}$.

(A) Let $A$ be the cross-sectional area of the cylinder and $\rho_0$ and $\rho_4$ be the densities of water at $0^{\circ} C$ and $4^{\circ} C$ respectively.
At $0^{\circ} C,$ the submerged length of the cylinder is $h_0 = 100 \; cm - 20 \; cm = 80 \; cm.$
By the principle of flotation,the weight of the cylinder equals the weight of the displaced water:
$mg = A \times h_0 \times \rho_0 \times g \implies m = A \times 80 \times \rho_0.$
At $4^{\circ} C,$ the submerged length of the cylinder is $h_4 = 100 \; cm - 21 \; cm = 79 \; cm.$
Since the mass of the cylinder remains constant:
$m = A \times 79 \times \rho_4.$
Equating the two expressions for $m$:
$A \times 80 \times \rho_0 = A \times 79 \times \rho_4.$
Therefore,the ratio of the density at $4^{\circ} C$ to the density at $0^{\circ} C$ is:
$\frac{\rho_4}{\rho_0} = \frac{80}{79} \approx 1.0126.$
Rounding to the nearest given option,the value is close to $1.01$.

(N/A) According to the law of atmospheres,the number density is given by:
$n_{2}=n_{1} \exp \left[-m g\left(h_{2}-h_{1}\right) / k_{B} T\right]$
For a particle of mass $m$ and density $\rho$ suspended in a medium of density $\rho^{\prime}$,the effective weight $W_{eff}$ is the actual weight minus the buoyant force (Archimedes' principle).
Let $V$ be the volume of the particle. Then $m = V\rho$.
The buoyant force is equal to the weight of the displaced medium: $F_{B} = V\rho^{\prime}g = (m/\rho)\rho^{\prime}g$.
The effective weight is:
$W_{eff} = mg - F_{B} = mg - (m/\rho)\rho^{\prime}g = mg(1 - \rho^{\prime}/\rho) = mg(\rho - \rho^{\prime})/\rho$.
Substituting this effective weight into the law of atmospheres and using $k_{B} = R/N_{A}$:
$n_{2} = n_{1} \exp \left[ -\frac{mg(\rho - \rho^{\prime})}{\rho} \frac{(h_{2} - h_{1})}{k_{B}T} \right]$
Substituting $k_{B} = R/N_{A}$:
$n_{2} = n_{1} \exp \left[ -\frac{mg(\rho - \rho^{\prime})}{\rho} \frac{(h_{2} - h_{1}) N_{A}}{RT} \right]$
Rearranging the terms,we get:
$n_{2} = n_{1} \exp \left[ -\frac{mg N_{A} (\rho - \rho^{\prime}) (h_{2} - h_{1})}{\rho RT} \right]$

(B) As the air bubble rises in water,it moves against the gravitational force,which might suggest an increase in potential energy. However,the bubble is also expanding due to the decrease in hydrostatic pressure as it moves towards the surface. More importantly,the work done by the buoyant force is positive,and the work done by gravity is negative. Since the bubble is in a fluid,the system's potential energy is defined by the configuration of the fluid and the bubble. As the bubble rises,the water (which has a much higher density) moves downward to occupy the space vacated by the bubble. This results in a net decrease in the gravitational potential energy of the entire system (water + bubble). Therefore,the potential energy of the system decreases.

(A) The buoyant force is the upward force exerted by a fluid on an object that is wholly or partially immersed in it.
This force arises due to the difference in pressure at different depths of the fluid,where the pressure at the bottom of the object is greater than the pressure at the top.
This phenomenon is known as buoyancy.
The buoyant force acts vertically upward through the center of buoyancy,which coincides with the center of mass of the displaced fluid.

(N/A) Archimedes' principle states: "When a body is partially or completely immersed in a fluid, the buoyant force acting on it is equal to the weight of the fluid displaced by the body, and it acts in the upward direction through the center of gravity of the displaced fluid."
Proof:
Consider a solid cube of height $h$ and cross-sectional area $A$ immersed in a liquid of density $\rho$ at a depth $x$ from the surface, as shown in the figure.
The forces acting on the left and right sides of the body are equal and opposite, so they cancel each other out.
The pressure on the upper surface of the body is $P_{1} = x \rho g$.
The pressure on the lower surface of the body is $P_{2} = (x + h) \rho g$.
The force on the upper surface is $F_{1} = P_{1} A = x \rho g A$ (acting downwards).
The force on the lower surface is $F_{2} = P_{2} A = (x + h) \rho g A$ (acting upwards).
The buoyant (resultant) force $F_{b}$ acting on the body is:
$F_{b} = F_{2} - F_{1}$
$F_{b} = (x + h) \rho g A - x \rho g A$
$F_{b} = h \rho g A$
Since $A h = V$ (the volume of the body), and the volume of the body equals the volume of the displaced liquid:
$F_{b} = V \rho g$
Since mass $m = V \rho$, we have:
$F_{b} = m g$
This force acts in the upward direction. Here, $m$ is the mass of the displaced liquid. Thus, the buoyant force is equal to the weight of the displaced liquid. This proves Archimedes' principle.

(N/A) The law of floatation states that a body floats on the surface of a liquid if the weight of the body is equal to the weight of the liquid displaced by the submerged part of the body.
When a body is partially or completely immersed in a liquid,it experiences two forces:
$(1)$ Weight of the body $(W)$: $W = V_s \rho_s g$ (acting downwards),where $V_s$ is the volume of the body and $\rho_s$ is the density of the body.
$(2)$ Buoyant force $(F_b)$: $F_b = V_f \rho_f g$ (acting upwards),where $V_f$ is the volume of the displaced liquid and $\rho_f$ is the density of the liquid.
Cases:
$(a)$ If $W > F_b$: The body sinks in the liquid (e.g.,an iron nail).
$(b)$ If $W = F_b$: The body remains in equilibrium at any depth (e.g.,a submarine).
$(c)$ If $W < F_b$: The body floats on the surface of the liquid (e.g.,a wooden block).

(N/A) When a body floats on the surface of a liquid,the weight of the body is equal to the weight of the liquid displaced by the body.
According to the principle of floatation: $W_{\text{body}} = W_{\text{displaced liquid}}$.
Let $V$ be the total volume of the body and $\rho$ be its density.
Let $V^{\prime}$ be the volume of the part of the body immersed in the liquid and $\rho_{l}$ be the density of the liquid.
The weight of the body is $W = V \rho g$.
The weight of the displaced liquid (buoyant force) is $F_{B} = V^{\prime} \rho_{l} g$.
Equating the two: $V \rho g = V^{\prime} \rho_{l} g$.
Thus,the ratio of the immersed volume to the total volume is given by: $\frac{V^{\prime}}{V} = \frac{\rho}{\rho_{l}}$.
Therefore,the volume of the immersed part is $V^{\prime} = V \left( \frac{\rho}{\rho_{l}} \right)$.

(N/A) Buoyant force is the upward force exerted by a fluid on an object that is wholly or partially immersed in it.
According to Archimedes' principle,the magnitude of this force is equal to the weight of the fluid displaced by the object.
Mathematically,$F_b = \rho \cdot V \cdot g$,where $\rho$ is the density of the fluid,$V$ is the volume of the displaced fluid,and $g$ is the acceleration due to gravity.

(B) The buoyant force is the upward force exerted by a fluid that opposes the weight of an immersed object.
According to Archimedes' principle,this force acts in the vertical direction,specifically in the upward direction,opposite to the force of gravity.
Therefore,the direction of the buoyant force is always upward.

(N/A) Archimedes' principle states that when a body is partially or fully immersed in a fluid,it experiences an upward buoyant force equal to the weight of the fluid displaced by the body.
Mathematically,the buoyant force $F_B$ is given by $F_B = V \rho g$,where $V$ is the volume of the displaced fluid,$\rho$ is the density of the fluid,and $g$ is the acceleration due to gravity.

(N/A) The law of floatation states that a body will float in a liquid if the weight of the liquid displaced by the submerged part of the body is equal to the total weight of the body.
Mathematically, for a body of mass $M$ floating in a liquid of density $\rho$, the condition is:
$Weight \ of \ the \ body = Weight \ of \ the \ displaced \ liquid$
$Mg = V_{submerged} \rho g$
Where $V_{submerged}$ is the volume of the body submerged in the liquid.

(N/A) For a body of total volume $V$ and density $\rho$ floating in a liquid of density $\sigma$,let $V_{sub}$ be the volume of the submerged part.
According to the law of floatation,the weight of the body is equal to the weight of the liquid displaced by the submerged part.
Weight of the body = $V \rho g$
Weight of the liquid displaced = $V_{sub} \sigma g$
Equating the two: $V \rho g = V_{sub} \sigma g$
Therefore,the volume of the submerged part is $V_{sub} = V \left( \frac{\rho}{\sigma} \right)$.

(A) body sinks in a liquid when its weight $(W)$ is greater than the buoyant force $(F_B)$ acting on it.
Mathematically,let $\rho_b$ be the density of the body and $\rho_l$ be the density of the liquid.
The weight of the body is $W = V \rho_b g$,where $V$ is the volume of the body.
The buoyant force is $F_B = V \rho_l g$.
For the body to sink,$W > F_B$.
Substituting the expressions,$V \rho_b g > V \rho_l g$,which simplifies to $\rho_b > \rho_l$.
Therefore,a body sinks if the density of the body is greater than the density of the liquid.

(C) body remains in equilibrium in a liquid at any depth if its density is exactly equal to the density of the liquid.
According to the law of floatation,the weight of the body $(W = V \rho_{body} g)$ must be equal to the buoyant force $(F_B = V \rho_{liquid} g)$.
If $\rho_{body} = \rho_{liquid}$,then $W = F_B$.
Since the buoyant force depends only on the volume of the displaced liquid and the density of the liquid (which is assumed to be uniform),the buoyant force does not change with depth.
Therefore,the net force $(W - F_B)$ remains zero at any depth,allowing the body to remain in neutral equilibrium.

(B) body floats on the surface of a liquid when the weight of the body is exactly balanced by the buoyant force (upthrust) exerted by the liquid.
According to the law of floatation,for a body to float in equilibrium,the weight of the body must be equal to the weight of the liquid displaced by the submerged part of the body.
Mathematically,$W_{body} = F_{buoyant}$.
If $W_{body} > F_{buoyant}$,the body will sink.
If $W_{body} < F_{buoyant}$,the body will rise to the surface and float partially submerged.