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

(B) Let the density of the cork be $d$ and the density of water be $\rho$. The volume of the cork is $V$.
The forces acting on the cork are the buoyant force $F_B = V \rho g$ (upwards),the weight of the cork $W = V d g$ (downwards),and the spring force $F_s$ (downwards).
In the stationary frame,the equilibrium condition is $F_B = W + F_s$,so $F_s = V \rho g - V d g = V(\rho - d)g$.
When the elevator moves downwards with acceleration $a$,the effective acceleration due to gravity becomes $g_{eff} = (g - a)$.
The new spring force is $F_s' = V(\rho - d)(g - a)$.
Since $(g - a) < g$,the new spring force $F_s'$ is less than the original spring force $F_s$. Therefore,the elongation of the spring decreases,and the spring length decreases.

(A) The upthrust (buoyant force) on a body submerged in a liquid is given by the formula $F_B = V \rho g$, where $V$ is the volume of the displaced liquid, $\rho$ is the density of the liquid, and $g$ is the acceleration due to gravity.
When the entire system (beaker and liquid) falls freely under gravity, it experiences an acceleration equal to $g$ downwards. In the frame of reference of the falling beaker, the effective acceleration due to gravity $(g_{eff})$ becomes $g - g = 0$.
Since the upthrust depends on the effective gravitational field, and $g_{eff} = 0$, the upthrust on the body becomes zero.

(C) Initially,the man holds a $1 \,kg$ block in his left hand and a $10 \,kg$ bucket in his right hand.
When the block is dropped into the bucket,the total mass supported by the right hand becomes the sum of the mass of the bucket (with water) and the mass of the block.
Total mass $= 10 \,kg + 1 \,kg = 11 \,kg$.
Therefore,the man carries a load of $11 \,kg$ in his right hand.

(D) When the block $m$ is immersed in the liquid,it experiences an upward buoyant force (upthrust) $F_B$ according to Archimedes' principle.
For balance $A$,the reading is the tension $T$ in the string. Initially,$T = mg = 2 \, kg \cdot g$. When immersed,$T = mg - F_B$. Since $F_B > 0$,the reading of balance $A$ will be less than $2 \, kg$.
For balance $B$,it measures the normal force exerted by the beaker on the pan. Initially,it reads the weight of the beaker plus the liquid. When the block is immersed,the liquid exerts an upward force $F_B$ on the block. By Newton's third law,the block exerts an equal and opposite downward force $F_B$ on the liquid. Thus,the total downward force on the balance $B$ increases by $F_B$. Therefore,the reading of balance $B$ will be more than $5 \, kg$.
Since both statements $(b)$ and $(c)$ are correct,the correct option is $(d)$.

(D) In the compression or extension of a spring,work is done against the restoring force,which increases the elastic potential energy of the spring.
In moving a body against gravity,work is done against the gravitational force of attraction,which increases the gravitational potential energy of the system.
Therefore,in cases $A$,$B$,and $C$,the potential energy of the system increases.
When an air bubble rises in water,it moves in the direction of the buoyant force (upthrust). Since the system performs work in this direction,the potential energy of the system decreases.

(B) The correct answer is $B$. Archimedes' principle states that the buoyant force (uplift) acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. The weight of the displaced fluid is given by $W = m \cdot g$,where $m$ is the mass of the displaced fluid and $g$ is the acceleration due to gravity. If the gravitational effect $(g)$ decreases,the weight of the displaced fluid decreases,thereby reducing the Archimedes uplift force. Viscous forces and electrostatic forces are independent of the gravitational field strength.

(C) The apparent weight of a body submerged in a fluid is given by $W_{app} = V(\rho - \sigma)g = \frac{m}{\rho}(\rho - \sigma)g$,where $m$ is the mass of the body,$\rho$ is the density of the body,and $\sigma$ is the density of the fluid (water,$\sigma = 1 \ g/cm^3$).
Since the two bodies are in equilibrium,their apparent weights must be equal:
$\frac{m_1}{\rho_1}(\rho_1 - \sigma) = \frac{m_2}{\rho_2}(\rho_2 - \sigma)$
Given $m_1 = 36 \ g$,$\rho_1 = 9 \ g/cm^3$,$m_2 = 48 \ g$,and $\sigma = 1 \ g/cm^3$:
$\frac{36}{9}(9 - 1) = \frac{48}{\rho_2}(\rho_2 - 1)$
$4(8) = \frac{48}{\rho_2}(\rho_2 - 1)$
$32 = \frac{48(\rho_2 - 1)}{\rho_2}$
$32\rho_2 = 48\rho_2 - 48$
$16\rho_2 = 48$
$\rho_2 = 3 \ g/cm^3$.

(D) Let $M_0$ be the true mass of the body in vacuum.
When the body is weighed in air,the apparent weight of the body is equal to the apparent weight of the standard weights.
The apparent weight is given by the actual weight minus the upthrust (buoyant force) due to the displaced air.
For the body: Apparent weight $= M_0 g - V_1 d g = M_0 g - (M_0/d_1) d g = M_0 g (1 - d/d_1)$.
For the weights: Apparent weight $= M g - V_2 d g = M g - (M/d_2) d g = M g (1 - d/d_2)$.
Equating the two: $M_0 g (1 - d/d_1) = M g (1 - d/d_2)$.
Therefore,the true mass $M_0 = \frac{M(1 - d/d_2)}{(1 - d/d_1)}$.

(C) Let the total volume of the iceberg be $V$ and its density be $\rho = 900 \ kg/m^3$. The density of water is $\sigma = 1000 \ kg/m^3$.
According to the law of floatation, the weight of the iceberg equals the weight of the water displaced by the submerged part $(V_{in})$.
$V \rho g = V_{in} \sigma g$
$V_{in} = \left( \frac{\rho}{\sigma} \right) V = \left( \frac{900}{1000} \right) V = 0.9 V$
The volume of the iceberg outside the water is $V_{out} = V - V_{in} = V - 0.9 V = 0.1 V$.
The percentage of volume outside the water is $\left( \frac{V_{out}}{V} \right) \times 100 = 0.1 \times 100 = 10\%$.

(A) The volume of the log of wood is given by $V = \frac{\text{mass}}{\text{density}} = \frac{120}{600} = 0.2 \ m^3$.
Let $x$ be the mass that can be placed on the log of wood to make it just sink.
For the log to just sink,the total weight of the log and the added mass must be equal to the buoyant force (weight of the displaced water).
The total weight is $(120 + x)g$.
The buoyant force is $V \rho_{\text{water}} g = 0.2 \times 1000 \times g$.
Equating the two: $(120 + x)g = 0.2 \times 1000 \times g$.
$120 + x = 200$.
$x = 200 - 120 = 80 \ kg$.

(C) The weight of the hemispherical bowl is given by $W = mg = V_{material} \rho g$,where $V_{material} = \frac{2}{3} \pi [ (D/2)^3 - (d/2)^3 ]$ is the volume of the material of the bowl.
Here,$D = 1 \ m$ is the outer diameter and $d$ is the inner diameter. The density of the bowl is $\rho = 2 \times 10^4 \ kg/m^3$.
According to the principle of flotation,the weight of the bowl equals the weight of the liquid displaced by the submerged part of the bowl. Since the bowl is hemispherical and just floats,the volume of the liquid displaced is the volume of the outer hemisphere,$V_{displaced} = \frac{2}{3} \pi (D/2)^3$.
The weight of the displaced liquid is $W_{liquid} = V_{displaced} \sigma g$,where $\sigma = 1.2 \times 10^3 \ kg/m^3$ is the density of the liquid.
Equating the two weights: $\frac{2}{3} \pi (D/2)^3 \sigma g = \frac{2}{3} \pi [ (D/2)^3 - (d/2)^3 ] \rho g$.
Canceling common terms: $(D/2)^3 \sigma = [ (D/2)^3 - (d/2)^3 ] \rho$.
Substituting the values: $(0.5)^3 \times 1.2 \times 10^3 = [ (0.5)^3 - (d/2)^3 ] \times 2 \times 10^4$.
$0.125 \times 1.2 \times 10^3 = [ 0.125 - (d/2)^3 ] \times 20 \times 10^3$.
$0.15 = [ 0.125 - (d/2)^3 ] \times 20$.
$0.0075 = 0.125 - (d/2)^3$.
$(d/2)^3 = 0.125 - 0.0075 = 0.1175$.
$d/2 = (0.1175)^{1/3} \approx 0.49$.
$d = 2 \times 0.49 = 0.98 \ m$.

(B) Let the specific gravities of concrete and sawdust be $\rho_1 = 2.4$ and $\rho_2 = 0.3$ respectively.
According to the principle of floatation,the weight of the whole sphere equals the upthrust (buoyant force) on the sphere.
Weight of sphere = Weight of water displaced
$\frac{4}{3}\pi (R^3 - r^3)\rho_1 g + \frac{4}{3}\pi r^3 \rho_2 g = \frac{4}{3}\pi R^3 \times 1 \times g$
Dividing by $\frac{4}{3}\pi g$,we get:
$(R^3 - r^3)\rho_1 + r^3 \rho_2 = R^3$
$R^3 \rho_1 - r^3 \rho_1 + r^3 \rho_2 = R^3$
$R^3(\rho_1 - 1) = r^3(\rho_1 - \rho_2)$
$\frac{R^3}{r^3} = \frac{\rho_1 - \rho_2}{\rho_1 - 1} = \frac{2.4 - 0.3}{2.4 - 1} = \frac{2.1}{1.4} = 1.5$
Now,the ratio of the mass of concrete $(M_c)$ to the mass of sawdust $(M_s)$ is:
$\frac{M_c}{M_s} = \frac{\frac{4}{3}\pi (R^3 - r^3)\rho_1}{\frac{4}{3}\pi r^3 \rho_2} = \left( \frac{R^3}{r^3} - 1 \right) \frac{\rho_1}{\rho_2}$
Substituting the values:
$\frac{M_c}{M_s} = (1.5 - 1) \times \frac{2.4}{0.3} = 0.5 \times 8 = 4$.

(D) The volume of the block $V = 5 \, cm \times 5 \, cm \times 5 \, cm = 125 \, cm^3$.
The density of the block $\rho = 5 \, g \, cm^{-3}$.
The density of water $\sigma = 1 \, g \, cm^{-3}$.
The apparent weight $W_{app}$ is given by the formula $W_{app} = V(\rho - \sigma)g$.
Substituting the values,$W_{app} = 125 \times (5 - 1) \times g$ (in units of force).
Since $1 \, g$ mass corresponds to $1 \, gf$ weight,the apparent weight in $gf$ is $V(\rho - \sigma) = 125 \times 4 = 500 \, gf$.
Comparing this with the given options,$500 = 4 \times 5 \times 5 \times 5 \, gf$.

(A) Let the volume of the block be $V$, the density of the block be $\rho$, and the density of the liquid be $\sigma$.
Initially, the block is in equilibrium, so the weight of the block equals the buoyant force:
$V \rho g = V_{in} \sigma g$
Given that half the volume is immersed, $V_{in} = V/2$, so:
$V \rho g = (V/2) \sigma g \implies \rho = \sigma/2$.
When the system accelerates upwards with acceleration $a = g/3$, the effective gravity becomes $g' = g + a = g + g/3 = 4g/3$.
The new equilibrium condition is:
$V \rho g' = V'_{in} \sigma g'$
$V \rho (4g/3) = V'_{in} \sigma (4g/3)$
$V \rho = V'_{in} \sigma$
Since $\rho = \sigma/2$, we have $V(\sigma/2) = V'_{in} \sigma$, which gives $V'_{in} = V/2$.
The fraction of volume immersed remains $V'_{in}/V = 0.5$. Thus, the fraction of volume immersed does not change with acceleration.

(B) The apparent weight of an object immersed in a liquid is given by the formula: $W_{app} = V(\rho_s - \rho_l)g$,where $V$ is the volume of the object,$\rho_s$ is the density of the object,and $\rho_l$ is the density of the liquid.
Since the mass $M = V \rho_s$,we can write $V = M / \rho_s$.
Substituting this into the formula: $W_{app} = \frac{M}{\rho_s}(\rho_s - \rho_l)g = M(1 - \frac{\rho_l}{\rho_s})g$.
Given $M = 2.1 \ kg$,$\rho_s = 10.5$,and $\rho_l = 0.8$.
The tension $T$ in the string is equal to the apparent weight in $kg-wt$:
$T = 2.1 \times (1 - \frac{0.8}{10.5}) \ kg-wt$.
$T = 2.1 \times (\frac{10.5 - 0.8}{10.5}) \ kg-wt$.
$T = 2.1 \times (\frac{9.7}{10.5}) \ kg-wt$.
$T = 2.1 \times 0.9238 \approx 1.94 \ kg-wt$.

(D) Let the density of the metal be $\rho$ and the density of the liquid be $\sigma$. Let $V$ be the volume of the sample.
According to Archimedes' principle,the apparent weight is the actual weight minus the buoyant force.
In air: $W_{air} = V \rho g = 210 \ g$ (taking $g$ as a unit factor for weight in grams) ... $(i)$
In water: $W_{water} = V(\rho - 1)g = 180 \ g$ ... (ii)
In liquid: $W_{liquid} = V(\rho - \sigma)g = 120 \ g$ ... (iii)
From $(i)$ and (ii): $V \rho g - V(1)g = 180 \implies 210 - Vg = 180 \implies Vg = 30$.
Since $V \rho g = 210$,we have $30 \rho = 210 \implies \rho = 7$.
From $(i)$ and (iii): $V \rho g - V \sigma g = 120 \implies 210 - V \sigma g = 120 \implies V \sigma g = 90$.
Substituting $Vg = 30$: $30 \sigma = 90 \implies \sigma = 3$.
Thus,the relative density of the metal is $7$ and the relative density of the liquid is $3$.

(C) For a body floating in a liquid,the weight of the body is equal to the buoyant force (Archimedes' Principle).
Let $\rho_A$ and $\rho_B$ be the densities of solids $A$ and $B$,and $\rho_w$ be the density of water.
For solid $A$: $V_A \rho_A g = (V_A/2) \rho_w g \implies \rho_A = \frac{1}{2} \rho_w$.
For solid $B$: $V_B \rho_B g = (2/3 V_B) \rho_w g \implies \rho_B = \frac{2}{3} \rho_w$.
Taking the ratio of the densities: $\frac{\rho_A}{\rho_B} = \frac{(1/2) \rho_w}{(2/3) \rho_w} = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4}$.
Thus,the ratio of the densities of $A$ and $B$ is $3:4$.

(C) For an object floating in a liquid, the weight of the object is equal to the buoyant force (Archimedes' Principle).
Weight of the object = $V_0 d_0 g$.
Buoyant force = $V_{in} d g$, where $V_{in}$ is the volume of the object submerged in the liquid.
Equating the two: $V_0 d_0 g = V_{in} d g$.
This gives the submerged volume: $V_{in} = V_0 \frac{d_0}{d}$.
The volume of the object above the surface is $V_{out} = V_0 - V_{in}$.
Substituting $V_{in}$: $V_{out} = V_0 - V_0 \frac{d_0}{d} = V_0 \left( \frac{d - d_0}{d} \right)$.
The fraction of the object above the surface is $\frac{V_{out}}{V_0} = \frac{d - d_0}{d}$.

(A) The volume of the steel block is $V = 5 \text{ cm} \times 5 \text{ cm} \times 5 \text{ cm} = 125 \text{ cm}^3$.
Given the relative density of steel is $\rho_r = 7$,the density of steel is $\rho = 7 \text{ g/cm}^3$.
The density of water is $\sigma = 1 \text{ g/cm}^3$.
The apparent weight of an object submerged in a liquid is given by $W_{app} = V(\rho - \sigma)g$.
Since we are calculating the weight in $\text{gf}$ (gram-force),we use the formula $W_{app} = V(\rho - \sigma) \text{ gf/cm}^3$.
Substituting the values: $W_{app} = (5 \times 5 \times 5) \times (7 - 1) \text{ gf}$.
$W_{app} = 125 \times 6 \text{ gf} = 6 \times 5 \times 5 \times 5 \text{ gf}$.

(B) When a body is floating in a liquid such that its density is equal to the density of the liquid,it is in a state of neutral equilibrium.
If the body is pushed down slightly,the buoyant force and the weight of the body remain equal because the density of the liquid is uniform.
Since the net force on the body remains zero at any depth,the body will remain submerged at the position where it is left.

(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 of the sphere $(mg)$ acting downwards.
$2$. Tension in the string $(T)$ acting downwards.
$3$. Buoyant force $(F_B)$ acting upwards,where $F_B = V \rho_w g$.
For equilibrium,the net force is zero: $F_B = T + mg$.
Therefore,$T = F_B - mg = V \rho_w g - V \rho_s g$.
Substituting $\rho_w = \eta \rho_s$ and $V \rho_s = m$:
$T = V (\eta \rho_s) g - m g = \eta (V \rho_s) g - mg = \eta mg - mg = (\eta - 1) mg$.

(A) Let $\rho$ be the density of the material of the sphere and $\rho_w$ be the density of water.
When the sphere is floating half-immersed,the weight of the sphere is balanced by the buoyant force:
$W_{sphere} = F_{buoyant}$
$V \cdot \rho \cdot g = (V/2) \cdot \rho_w \cdot g$
$\rho = \rho_w / 2$
Now,let $V'$ be the volume of water poured inside the sphere so that it just sinks. For the sphere to sink,the total weight (sphere + water) must equal the buoyant force when the sphere is fully submerged:
$W_{total} = F_{buoyant, total}$
$(V \cdot \rho \cdot g) + (V' \cdot \rho_w \cdot g) = V \cdot \rho_w \cdot g$
Substitute $\rho = \rho_w / 2$:
$V \cdot (\rho_w / 2) \cdot g + V' \cdot \rho_w \cdot g = V \cdot \rho_w \cdot g$
Divide by $\rho_w \cdot g$:
$V/2 + V' = V$
$V' = V - V/2 = V/2$
Therefore,the minimum volume of water to be poured is $V/2$.

(A) According to the Law of Floatation,a floating object displaces a weight of liquid equal to its own weight.
Since the weight of the block remains constant,the weight of the displaced water must also remain constant.
Because the weight of the displaced water is equal to the product of the volume of displaced water,the density of water,and the acceleration due to gravity $(W = V_{disp} \cdot \rho_{water} \cdot g)$,and both the density of water and gravity are constant,the volume of displaced water $(V_{disp})$ must remain constant.
Since the volume of displaced water does not change,the level of water in the container remains unchanged.

(B) Let the density of the ball be $\rho = 0.4 \times 10^3 \ kg/m^3$ and the density of water be $\sigma = 1.0 \times 10^3 \ kg/m^3$.
When the ball falls from a height $h_0 = 9 \ cm$,its velocity $v$ just before entering the water is given by $v^2 = 2gh_0$.
Inside the water,the ball experiences an upward buoyant force $F_B = V\sigma g$ and a downward gravitational force $W = V\rho g$. The net force is $F_{net} = W - F_B = V(\rho - \sigma)g$.
The retardation $a$ inside the water is $a = \frac{F_{net}}{m} = \frac{V(\rho - \sigma)g}{V\rho} = \left( \frac{\rho - \sigma}{\rho} \right)g$.
Substituting the values: $a = \left( \frac{0.4 - 1.0}{0.4} \right)g = -\frac{0.6}{0.4}g = -1.5g$.
The ball sinks to a depth $h$ where its final velocity becomes zero. Using $v_f^2 - v_i^2 = 2ah$:
$0 - (2gh_0) = 2(-1.5g)h$.
$2gh_0 = 3gh$.
$h = \frac{2}{3}h_0 = \frac{2}{3} \times 9 \ cm = 6 \ cm$.

(B) According to the law of floatation,the weight of the body is equal to the weight of the water displaced by the immersed part of the body.
Let $\rho_A$ and $\rho_B$ be the densities of solids $A$ and $B$,respectively,and $\rho_W$ be the density of water.
For solid $A$,the volume immersed is $\frac{1}{2} V_A$. Thus,$\frac{1}{2} V_A \rho_W g = V_A \rho_A g$,which gives $\rho_A = \frac{1}{2} \rho_W$.
For solid $B$,the volume above water is $\frac{1}{4} V_B$,so the volume immersed is $V_B - \frac{1}{4} V_B = \frac{3}{4} V_B$. Thus,$\frac{3}{4} V_B \rho_W g = V_B \rho_B g$,which gives $\rho_B = \frac{3}{4} \rho_W$.
The ratio of the density of $A$ to that of $B$ is $\frac{\rho_A}{\rho_B} = \frac{\frac{1}{2} \rho_W}{\frac{3}{4} \rho_W} = \frac{1}{2} \times \frac{4}{3} = \frac{2}{3}$.

(C) Let $M$ be the mass of the boat and $m$ be the total mass of the steel balls. Let $\rho_{\omega}$ be the density of water and $\sigma_{s}$ be the density of steel.
When the balls are in the boat,the boat floats. According to the law of floatation,the weight of the displaced water is equal to the total weight of the boat and the balls: $W = (M + m)g$.
The volume of water displaced is $V_{1} = \frac{M + m}{\rho_{\omega}}$.
When the steel balls are thrown into the water,they sink because $\sigma_{s} > \rho_{\omega}$.
The boat now only displaces water equal to its own weight: $V_{boat} = \frac{M}{\rho_{\omega}}$.
The steel balls displace water equal to their own volume: $V_{balls} = \frac{m}{\sigma_{s}}$.
The total volume of water displaced is $V_{2} = \frac{M}{\rho_{\omega}} + \frac{m}{\sigma_{s}}$.
Since $\sigma_{s} > \rho_{\omega}$,it follows that $\frac{m}{\sigma_{s}} < \frac{m}{\rho_{\omega}}$.
Therefore,$V_{2} < V_{1}$.
As the volume of displaced water decreases,the level of water in the tank will fall.

(C) According to Archimedes' principle,the upthrust (buoyant force) acting on an object immersed in a fluid is given by the formula: $F = V \rho g$,where $V$ is the volume of the displaced fluid (which equals the volume of the submerged part of the object),$\rho$ is the density of the liquid,and $g$ is the acceleration due to gravity.
Since the upthrust $F$ is equal for both pieces and they are immersed in the same liquid (same $\rho$ and $g$),it follows that $V$ must be equal for both pieces.
Therefore,both pieces must have equal volumes.

(B) For a floating object,the weight of the object is equal to the weight of the displaced fluid (Archimedes' Principle).
Let $V$ be the total volume of the cylinder,$\rho$ be the density of the wood,and $\sigma$ be the density of water.
The weight of the cylinder is $W = V \rho g$.
The volume of water displaced is $V_{disp} = \frac{V}{2}$.
The buoyant force is $F_B = V_{disp} \sigma g = \frac{V}{2} \sigma g$.
Since the cylinder is floating,$W = F_B$.
$V \rho g = \frac{V}{2} \sigma g$.
Therefore,$\rho = \frac{\sigma}{2}$.
This means the density of wood is half the density of water.

(B) Let the density of the candle be $\rho_c$ and the density of the liquid be $\rho_l$. Let the total length of the candle be $L_0$ and the submerged length be $L_s$. According to the principle of floatation,the weight of the candle equals the weight of the displaced liquid:
$\rho_c \cdot A \cdot L_0 = \rho_l \cdot A \cdot L_s$
where $A$ is the cross-sectional area of the candle.
This implies $L_s = (\rho_c / \rho_l) L_0$. Let $k = \rho_c / \rho_l$. Then $L_s = k L_0$.
The height of the top of the candle above the liquid surface is $h = L_0 - L_s = L_0(1 - k)$.
As the candle burns,its total length $L_0$ decreases at a rate $v = dL_0/dt = 2 \text{ cm/hour}$.
The rate of change of the height of the top of the candle is $dh/dt = (dL_0/dt)(1 - k)$.
Assuming the candle is half-submerged (as suggested by the figure),$k = 0.5$.
Therefore,$dh/dt = 2 \text{ cm/hour} \times (1 - 0.5) = 1 \text{ cm/hour}$.
Since $dh/dt$ is positive,the top of the candle falls at a rate of $1 \text{ cm/hour}$.

(B) Let the mass of the ice be $M_i$ and the mass of the glass ball be $M_g$.
Initially,the ice block with the glass ball floats in water. According to the law of floatation,the total weight of the system is equal to the weight of the displaced water.
Total weight $W = (M_i + M_g)g$.
The volume of water displaced initially is $V_1 = \frac{M_i + M_g}{\rho_w}$,where $\rho_w$ is the density of water.
When the ice melts,the ice turns into water with volume $V_{iw} = \frac{M_i}{\rho_w}$. The glass ball sinks to the bottom.
The volume of water displaced by the glass ball is $V_g = \frac{M_g}{\rho_g}$,where $\rho_g$ is the density of the glass ball.
Since the glass ball is denser than water $(\rho_g > \rho_w)$,the volume of water displaced by the submerged glass ball is $V_g = \frac{M_g}{\rho_g} < \frac{M_g}{\rho_w}$.
The final volume of the water level corresponds to $V_2 = V_{iw} + V_g = \frac{M_i}{\rho_w} + \frac{M_g}{\rho_g}$.
Comparing $V_1$ and $V_2$,since $\frac{M_g}{\rho_g} < \frac{M_g}{\rho_w}$,it follows that $V_2 < V_1$.
Therefore,the level of water falls.

(D) The correct answer is $None$ of these. The floating of a ship and the sinking of a needle are governed by the principle of flotation and the concept of density.
According to the law of flotation,a body floats if the weight of the fluid displaced by the submerged part of the body is equal to the weight of the body itself.
$A$ ship is designed such that its average density is less than the density of water,allowing it to displace a weight of water equal to its own weight even when only a small portion is submerged.
Conversely,the density of steel is much greater than the density of water. $A$ needle,being solid steel,has an average density greater than water. Therefore,the weight of the water displaced by the needle is less than the weight of the needle,causing it to sink.
This phenomenon is fundamentally explained by the relationship between the weight of the object and the buoyant force (Archimedes' principle),which depends on the object's density relative to the fluid.

(A) The construction and operation of a submarine are based on Archimedes' principle.
According to this principle,any object,wholly or partially immersed in a fluid,is buoyed up by a force equal to the weight of the fluid displaced by the object.
$A$ submarine has ballast tanks that can be filled with water or air to change its overall density.
By adjusting the amount of water in these tanks,the submarine can control its buoyancy to submerge,surface,or maintain a constant depth.

(D) Let $V$ be the volume of the body. The velocity of the body just before entering the liquid is $v = \sqrt{2gh}$.
When the body enters the liquid,the forces acting on it are its weight $W = Vdg$ (downwards) and the buoyant force $F_B = VDg$ (upwards).
The net force $F_{net}$ acting on the body inside the liquid is $F_B - W = VDg - Vdg = V(D - d)g$.
The retardation $a$ of the body is given by $a = \frac{F_{net}}{m} = \frac{V(D - d)g}{Vd} = \left( \frac{D - d}{d} \right)g$.
The body comes to rest when its velocity becomes zero. Using $v = at$,we get $t = \frac{v}{a}$.
Substituting the values,$t = \frac{\sqrt{2gh}}{\left( \frac{D - d}{d} \right)g} = \frac{\sqrt{2gh}}{g} \cdot \frac{d}{D - d} = \sqrt{\frac{2h}{g}} \cdot \frac{d}{D - d}$.

(A) For the cylinder to float in equilibrium,its weight must be equal to the total upthrust (buoyant force) exerted by the two liquids.
Weight of the cylinder = $V \times D \times g = (A/5) \times L \times D \times g$.
Upthrust due to the upper liquid (density $d$): $F_1 = (A/5) \times (3L/4) \times d \times g$.
Upthrust due to the lower liquid (density $2d$): $F_2 = (A/5) \times (L/4) \times 2d \times g$.
Equating weight to total upthrust:
$(A/5) \times L \times D \times g = (A/5) \times (3L/4) \times d \times g + (A/5) \times (L/4) \times 2d \times g$.
Dividing both sides by $(A/5) \times L \times g$:
$D = (3/4)d + (1/4) \times 2d = (3/4)d + (2/4)d = (5/4)d$.
Therefore,the density of the solid is $D = \frac{5}{4}d$.

(D) Initially,the block floats with the coin. According to the law of floatation,the total weight of the block and the coin is balanced by the buoyant force,which is equal to the weight of the water displaced by the submerged part of the block. 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,it sinks (assuming its density is greater than water). The block now only needs to support its own weight $Mg$. The new volume of water displaced by the block is $V'_{disp} = M/\rho_w$. Since $V'_{disp} < V_{disp}$,the submerged depth $l$ of the block decreases.
Regarding $h$,the total volume of water displaced initially was $V_{disp} = (M+m)/\rho_w$. After the coin falls,the block displaces $M/\rho_w$ and the coin displaces its own volume $V_c = m/\rho_c$. The new total volume of water displaced is $V'_{total} = M/\rho_w + m/\rho_c$. Since the density of the coin $\rho_c > \rho_w$,it follows that $m/\rho_c < m/\rho_w$. Therefore,$V'_{total} < V_{disp}$. As the total volume of water displaced decreases,the water level $h$ also decreases. Thus,both $l$ and $h$ decrease.

(C) As the sphere floats in the liquid,its weight is equal to the total upthrust force acting on it.
Let $V$ be the total volume of the sphere and $\rho$ be its density.
Weight of the sphere = $V \rho g$
Upthrust due to oil = (Volume in oil) $\times$ (Density of oil) $\times g = (V/2) \times 0.8 \times g = 0.4 Vg$
Upthrust due to mercury = (Volume in mercury) $\times$ (Density of mercury) $\times g = (V/2) \times 13.6 \times g = 6.8 Vg$
Total upthrust = $0.4 Vg + 6.8 Vg = 7.2 Vg$
Equating weight and total upthrust:
$V \rho g = 7.2 Vg$
$\rho = 7.2 \; g/cm^3$

(A) The upthrust (buoyant force) on a body submerged in a liquid is given by the formula:
$F_B = V \rho_{\text{liquid}} (g - a)$
where $V$ is the volume of the liquid displaced,$\rho_{\text{liquid}}$ is the density of the liquid,$g$ is the acceleration due to gravity,and $a$ is the downward acceleration of the system.
In the case of free fall,the entire system (beaker and liquid) accelerates downwards with an acceleration equal to the acceleration due to gravity,i.e.,$a = g$.
Substituting $a = g$ into the formula:
$F_B = V \rho_{\text{liquid}} (g - g) = V \rho_{\text{liquid}} (0) = 0$.
Therefore,the upthrust on the body is zero.

(A) Let $L$ be the total length of the rod. The rod is uniform,so its weight $W = A L \rho g$ acts at its center of mass,which is at a distance $L/2$ from the bottom end $P$.
Let $l$ be the length of the rod submerged in the liquid. The depth of the liquid is $h = L/2$. From the geometry,$\sin \theta = h/l = (L/2)/l = L/(2l)$,which implies $l = L/(2 \sin \theta)$.
The buoyant force $F_B = A l \rho_0 g$ acts at the center of the submerged portion,which is at a distance $l/2$ from the bottom end $P$.
For rotational equilibrium about the bottom end $P$,the torque due to weight must balance the torque due to the buoyant force:
$W \cdot (L/2) \cos \theta = F_B \cdot (l/2) \cos \theta$
$(A L \rho g) \cdot (L/2) \cos \theta = (A l \rho_0 g) \cdot (l/2) \cos \theta$
$L^2 \rho = l^2 \rho_0$
$(l/L)^2 = \rho / \rho_0$
Substituting $l = L/(2 \sin \theta)$:
$(1 / (2 \sin \theta))^2 = \rho / \rho_0$
$1 / (4 \sin^2 \theta) = \rho / \rho_0$
$\sin^2 \theta = \frac{1}{4} \cdot \frac{\rho_0}{\rho}$
$\sin \theta = \frac{1}{2} \sqrt{\frac{\rho_0}{\rho}}$

(B) Let $M$ be the mass of the ice block.
According to the principle of floatation,the weight of the ice equals the weight of the liquid displaced.
Thus,$M \cdot g = V_D \cdot \sigma_L \cdot g$,where $V_D$ is the volume of liquid displaced and $\sigma_L$ is the density of the liquid.
Therefore,$V_D = \frac{M}{\sigma_L}$.
When the ice melts,it forms water of mass $M$. The volume of this water is $V_F = \frac{M}{\sigma_W}$,where $\sigma_W$ is the density of water $(1 \ g/cm^3)$.
Given $\sigma_L = 1.2 \ g/cm^3$ and $\sigma_W = 1 \ g/cm^3$,we have $\sigma_L > \sigma_W$.
Comparing the volumes: Since $\sigma_L > \sigma_W$,it follows that $\frac{M}{\sigma_L} < \frac{M}{\sigma_W}$,which means $V_D < V_F$.
Since the volume of the water formed $(V_F)$ is greater than the volume of the liquid displaced by the floating ice $(V_D)$,the level of the liquid in the beaker will rise.

(A) The resistance of a rod is given by $R = \rho \frac{L}{A}$. Since the material $(\rho)$ and length $(L)$ are constant,$R \propto \frac{1}{A}$.
Given $\frac{R_1}{R_2} = \frac{1}{2}$,we have $\frac{A_1}{A_2} = \frac{R_2}{R_1} = 2$,which means $A_1 = 2A_2$.
When a body is dipped in water,the loss of weight is equal to the buoyant force,given by $F_B = V \sigma_L g = A L \sigma_L g$,where $V$ is the volume,$\sigma_L$ is the density of the liquid,and $g$ is the acceleration due to gravity.
Since $L$,$\sigma_L$,and $g$ are the same for both rods,the loss of weight is directly proportional to the cross-sectional area $A$.
Therefore,$\frac{\text{Loss of weight}_1}{\text{Loss of weight}_2} = \frac{A_1}{A_2} = 2$.
Since the ratio is $2$,rod $A$ (the first rod) experiences a greater loss of weight.

(A) The resistance of a rod is given by $R = \rho \frac{l}{A}$.
Since the volume $V = A \cdot l$,we can write $A = \frac{V}{l}$.
Substituting this,$R = \rho \frac{l^2}{V}$.
Since density $\sigma = \frac{m}{V}$,we have $V = \frac{m}{\sigma}$.
Thus,$R = \rho \frac{l^2 \sigma}{m}$.
Given that the length $l$ and material (density $\sigma$ and resistivity $\rho$) are the same for both rods,we have $R \propto \frac{1}{m}$.
Given $R_A : R_B = 1 : 2$,it follows that $m_A : m_B = 2 : 1$,which means $m_A > m_B$.
The loss in weight when immersed in water is equal to the weight of the displaced water,which is $W_{loss} = V_{rod} \cdot \sigma_{water} \cdot g = \frac{m}{\sigma_{rod}} \cdot \sigma_{water} \cdot g$.
Since $m_A > m_B$,the loss in weight in rod $A$ will be more than in rod $B$.

(C) The apparent weight of a body immersed in a liquid is given by $W_{app} = V(\rho - \sigma)g$,where $V$ is the volume,$\rho$ is the density of the body,and $\sigma$ is the density of the liquid.
Since $V = m/\rho$,the apparent weight is $W_{app} = \frac{m}{\rho}(\rho - \sigma)g = m(1 - \frac{\sigma}{\rho})g$.
Given that the two bodies are in equilibrium in water,their apparent weights must be equal:
$m_1(1 - \frac{\sigma}{\rho_1}) = m_2(1 - \frac{\sigma}{\rho_2})$
Given: $m_1 = 36 \, g$,$\rho_1 = 9 \, g/cm^{3}$,$m_2 = 48 \, g$,and $\sigma = 1 \, g/cm^{3}$ (density of water).
Substituting the values:
$36(1 - \frac{1}{9}) = 48(1 - \frac{1}{\rho_2})$
$36(\frac{8}{9}) = 48(\frac{\rho_2 - 1}{\rho_2})$
$32 = 48(\frac{\rho_2 - 1}{\rho_2})$
$\frac{32}{48} = \frac{\rho_2 - 1}{\rho_2}$
$\frac{2}{3} = 1 - \frac{1}{\rho_2}$
$\frac{1}{\rho_2} = 1 - \frac{2}{3} = \frac{1}{3}$
$\rho_2 = 3 \, g/cm^{3}$.

(A) For the cylinder to float in equilibrium,the weight of the cylinder must be equal to the total buoyant force exerted by the two liquids.
Let $A_c = A/5$ be the cross-sectional area of the cylinder.
The length of the cylinder submerged in the upper liquid (density $d$) is $3L/4$.
The length of the cylinder submerged in the lower liquid (density $2d$) is $L/4$.
Weight of the cylinder = $V_{total} \times D \times g = (A_c \times L) \times D \times g = (A/5) \times L \times D \times g$.
Buoyant force = (Weight of upper liquid displaced) + (Weight of lower liquid displaced)
Buoyant force = $(A_c \times 3L/4) \times d \times g + (A_c \times L/4) \times 2d \times g$
Buoyant force = $(A/5) \times (3L/4) \times d \times g + (A/5) \times (L/4) \times 2d \times g$
Equating weight and buoyant force:
$(A/5) \times L \times D \times g = (A/5) \times g \times [ (3L/4) \times d + (L/4) \times 2d ]$
$L \times D = (3L/4) \times d + (2L/4) \times d$
$L \times D = (5L/4) \times d$
$D = 5/4 d$

(C) Let $V$ be the total volume of the sphere and $\rho$ be the density of the sphere.
According to the principle of flotation,the weight of the sphere is equal to the total buoyant force exerted by the two liquids.
Weight of the sphere = $V \rho g$
Buoyant force due to oil = $(\frac{V}{2}) \rho_{\text{oil}} g$
Buoyant force due to mercury = $(\frac{V}{2}) \rho_{\text{Hg}} g$
Equating the forces: $V \rho g = \frac{V}{2} \rho_{\text{oil}} g + \frac{V}{2} \rho_{\text{Hg}} g$
Dividing by $Vg$: $\rho = \frac{\rho_{\text{oil}} + \rho_{\text{Hg}}}{2}$
Substituting the values: $\rho = \frac{0.8 + 13.6}{2} = \frac{14.4}{2} = 7.2 \text{ g/cm}^3$.

(C) According to the principle of flotation, the weight of the object is equal to the weight of the liquid displaced by the submerged part of the object.
Let $V_{in}$ be the volume of the object submerged in the liquid.
Weight of the object = $V_0 d_0 g$.
Weight of the displaced liquid = $V_{in} d g$.
Equating the two: $V_0 d_0 g = V_{in} d g$.
This gives $V_{in} = V_0 \frac{d_0}{d}$.
The volume of the object outside the liquid is $V_{out} = V_0 - V_{in}$.
Substituting $V_{in}$: $V_{out} = V_0 - V_0 \frac{d_0}{d} = V_0 \left( 1 - \frac{d_0}{d} \right) = V_0 \left( \frac{d - d_0}{d} \right)$.
The fraction of the volume outside the liquid is $\frac{V_{out}}{V_0} = \frac{d - d_0}{d}$.

(C) Let $d$ be the density of the cylinder and $A$ be the area of the cross-section of the cylinder.
According to the law of floatation,the weight of the cylinder is equal to the total upthrust exerted by the two liquids.
Weight of cylinder = $L \times A \times d \times g$
Upthrust by denser liquid (density $n\rho$) = $(pL \times A) \times n\rho \times g$
Upthrust by lighter liquid (density $\rho$) = $((L - pL) \times A) \times \rho \times g$
Equating weight to total upthrust:
$L \times A \times d \times g = (pL \times A) \times n\rho \times g + ((L - pL) \times A) \times \rho \times g$
Dividing both sides by $A \times L \times g$:
$d = p \times n\rho + (1 - p) \times \rho$
$d = np\rho + \rho - p\rho$
$d = [1 + (n - 1)p]\rho$

(C) The forces acting on the sphere are its weight $(W = Vdg)$ acting downwards and the buoyant force $(F_B = V\rho g)$ acting upwards,where $V$ is the volume of the sphere.
According to Newton's second law,the net force $F_{net} = ma = Vda$.
$Vda = Vdg - V\rho g$
Dividing both sides by $Vd$,we get:
$a = g - \frac{\rho g}{d} = g \left( 1 - \frac{\rho}{d} \right) = g \left( \frac{d - \rho}{d} \right)$.
Thus,the sphere falls vertically with an acceleration of $g \left( \frac{d - \rho}{d} \right)$.

(D) The reading of the spring balance corresponds to the apparent weight of the block when submerged.
Apparent weight = Actual weight - Upthrust (Buoyant force).
The actual weight of the block in air is $W = 12\, N$.
The upthrust is given by $F_B = V_{sub} \cdot \rho_w \cdot g$,where $V_{sub}$ is the submerged volume,$\rho_w$ is the density of water $(1000\, kg/m^3)$,and $g$ is the acceleration due to gravity $(10\, m/s^2)$.
Given $V = 1000\, cm^3 = 10^{-3}\, m^3$,and half of the block is submerged,$V_{sub} = 0.5 \times 10^{-3}\, m^3 = 500 \times 10^{-6}\, m^3$.
$F_B = (500 \times 10^{-6}\, m^3) \times (10^3\, kg/m^3) \times (10\, m/s^2) = 5\, N$.
Apparent weight = $12\, N - 5\, N = 7\, N$.

(A) The volume of the ball is $V = \frac{4}{3}\pi r^3$. The density of the ball is $\sigma = 0.5\rho$.
Initially,the ball floats with half its volume submerged. The buoyant force $F_B = V_{submerged} \rho g = (V/2)\rho g = (2/3)\pi r^3 \rho g$. Since it is in equilibrium,the weight of the ball $W = V \sigma g = V(0.5\rho)g = (2/3)\pi r^3 \rho g$. Thus,$F_B = W$.
To push the ball down until it is fully submerged,we must apply an external force to overcome the net buoyant force. The work done is equal to the change in potential energy of the system.
When the ball is pushed down by a distance $x$,the additional volume submerged is $A(x)dx$. The work done $W = \int_{0}^{r} F_{ext} dx$. The net force required at any depth $y$ (where $y$ is the depth of the center below the equilibrium position) is $F_{ext} = F_{buoyant} - Weight = (V_{submerged}(y) - V_{submerged, initial})\rho g$.
Integrating the force over the displacement,the work done is $W = \frac{5}{12}\pi r^4 \rho g$.