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

(B) Let the total volume of the cube be $V$ and the volume immersed in water be $V_1$. The volume immersed in the other liquid is $(V - V_1)$.
For the cube to be in equilibrium,the weight of the cube must be equal to the sum of the buoyant forces from both liquids.
$Weight = F_{B,liquid} + F_{B,water}$
$V \rho_{cube} g = (V - V_1) \rho_{liquid} g + V_1 \rho_{water} g$
Dividing by $g$ and substituting the given values:
$V \times 0.9 = (V - V_1) \times 0.7 + V_1 \times 1.0$
$0.9 V = 0.7 V - 0.7 V_1 + 1.0 V_1$
$0.9 V - 0.7 V = 0.3 V_1$
$0.2 V = 0.3 V_1$
$\frac{V_1}{V} = \frac{0.2}{0.3} = \frac{2}{3}$
Thus,$2/3$ of the cube is immersed in water.

(A) Let $V$ be the total volume of the iceberg and $V_{\text{sub}}$ be the volume submerged in sea water.
According to the law of floatation, the weight of the iceberg is equal to the weight of the displaced sea water.
$V \cdot \rho_{\text{ice}} \cdot g = V_{\text{sub}} \cdot \rho_{\text{sea water}} \cdot g$
$V \cdot 0.9 = V_{\text{sub}} \cdot 1.1$
$V_{\text{sub}} = \frac{0.9}{1.1} V = \frac{9}{11} V$
The volume unsubmerged is $V_{\text{unsub}} = V - V_{\text{sub}} = V - \frac{9}{11} V = \frac{2}{11} V$.
The percentage fraction of the iceberg which is unsubmerged is $\left( \frac{V_{\text{unsub}}}{V} \right) \times 100 = \left( \frac{2}{11} \right) \times 100 \approx 18.18 \%$.

(D) Let the sphere have mass $m$,volume $V$,and density $\varrho$. Thus,$m = V \varrho$.
When the sphere is dropped from height $h$,its velocity $v$ just before hitting the liquid surface is given by $v^2 = 2gh$.
Inside the liquid,the forces acting on the sphere are its weight $mg$ (downward) and the buoyant force $F_B = V \sigma g$ (upward).
The net force on the sphere inside the liquid is $F_{net} = F_B - mg = V \sigma g - V \varrho g = Vg(\sigma - \varrho)$.
The acceleration $a$ of the sphere inside the liquid is $a = \frac{F_{net}}{m} = \frac{Vg(\sigma - \varrho)}{V \varrho} = g \frac{(\sigma - \varrho)}{\varrho}$.
Since the buoyant force is greater than the weight (as $\sigma > \varrho$),the sphere will decelerate.
Let $d$ be the maximum depth reached. Using the equation of motion $v_f^2 = v_i^2 + 2ad$,where $v_f = 0$ (at maximum depth),$v_i^2 = 2gh$,and $a = -g \frac{(\sigma - \varrho)}{\varrho}$:
$0 = 2gh - 2 \left( g \frac{(\sigma - \varrho)}{\varrho} \right) d$.
Solving for $d$: $d = \frac{h \varrho}{(\sigma - \varrho)}$.

(B) For the cylinder to float in equilibrium, the total weight of the cylinder must be equal to the total buoyant force exerted by the two liquids.
Let $A$ be the cross-sectional area of the cylinder and $d$ be its density.
Weight of the cylinder $W = (\text{Volume}) \times (\text{Density}) \times g = (A \ell) d g$.
Buoyant force from the upper liquid (density $\varrho$): $F_1 = (\text{Volume submerged in upper liquid}) \times \varrho \times g = A (\ell - \ell/4) \varrho g = A (3\ell/4) \varrho g$.
Buoyant force from the lower liquid (density $3\varrho$): $F_2 = (\text{Volume submerged in lower liquid}) \times (3\varrho) \times g = A (\ell/4) (3\varrho) g = A (3\ell/4) \varrho g$.
Equating weight to the total buoyant force: $A \ell d g = A (3\ell/4) \varrho g + A (3\ell/4) \varrho g$.
$d \ell = (3\ell/4) \varrho + (3\ell/4) \varrho = (6\ell/4) \varrho = (3/2) \varrho \ell$.
Therefore, $d = \frac{3}{2} \varrho$.

(D) The force exerted by the liquid on the bottom surface of the cylinder is equal to the weight of the liquid displaced by the cylinder plus the force due to the pressure at the depth of the cylinder's top surface.
By Archimedes' principle,the buoyant force $F_B = \rho V g$.
The downward force due to the liquid column above the cylinder is $F_{top} = P_{top} \times A = (\rho g h) \times (\pi R^2)$.
The total upward force exerted by the liquid on the cylinder is $F_{net} = F_B + F_{top} = \rho V g + \rho g h \pi R^2$.
Thus,the force on the bottom of the cylinder is $F_{bottom} = \rho g (V + \pi R^2 h)$.

(A) The ball is moving with constant velocity. Therefore,the net force acting on it is zero.
Let the density of the ball be $\rho_b$ and the density of the liquid be $\rho_l = 3\rho_b$.
The weight of the ball is $W = V \rho_b g$,where $V$ is the volume of the ball.
As the ball is rising up,the buoyant force $F_B$ acts upwards and the viscous force (friction) $F_v$ acts downwards along with the weight $W$.
The buoyant force is $F_B = V \rho_l g = V (3\rho_b) g = 3 V \rho_b g$.
Since the velocity is constant,the net force is zero:
$F_B = W + F_v$
$F_v = F_B - W = 3 V \rho_b g - V \rho_b g = 2 V \rho_b g$.
The ratio of the force of friction to the weight is:
$\frac{F_v}{W} = \frac{2 V \rho_b g}{V \rho_b g} = \frac{2}{1}$.

(A) Let $V$ be the volume of the body. When the body hits the surface of the lake,its velocity $v_0$ is given by $v_0^2 = 2gh$.
Inside the water,the forces acting on the body are the weight $W = \rho V g$ (downward) and the buoyant force $F_B = \sigma V g$ (upward).
The net upward force is $F_{net} = F_B - W = Vg(\sigma - \rho)$.
The acceleration $a$ of the body inside the water is $a = \frac{F_{net}}{m} = \frac{Vg(\sigma - \rho)}{\rho V} = \frac{g(\sigma - \rho)}{\rho}$.
Since the force is upward,the acceleration is directed upward (opposing the motion),so $a = -\frac{g(\sigma - \rho)}{\rho}$.
Let $H$ be the maximum depth reached. At this depth,the final velocity $v = 0$.
Using the equation of motion $v^2 - u^2 = 2aH$,we have:
$0 - (\sqrt{2gh})^2 = 2 \left( -\frac{g(\sigma - \rho)}{\rho} \right) H$
$-2gh = -\frac{2g(\sigma - \rho)}{\rho} H$
$h = \frac{(\sigma - \rho)}{\rho} H$
$H = \frac{h \rho}{(\sigma - \rho)}$.

(D) Let $V$ be the volume of the body. When the body falls from height $h$,its velocity $v$ just before entering the water is given by $v^2 = 2gh$.
When the body enters the water,it experiences an upward buoyant force $F_B = V \delta g$ and a downward gravitational force $W = V \rho g$.
The net upward force (retarding force) is $F_{net} = F_B - W = Vg(\delta - \rho)$.
The retardation $a$ of the body in the water is $a = \frac{F_{net}}{m} = \frac{Vg(\delta - \rho)}{V \rho} = g \left( \frac{\delta - \rho}{\rho} \right)$.
Let $d$ be the maximum depth reached. Using the equation of motion $v^2 = 2ad$ (where $v$ is the velocity at the surface and $a$ is the retardation),we have $2gh = 2 \left[ g \left( \frac{\delta - \rho}{\rho} \right) \right] d$.
Solving for $d$,we get $d = \frac{h \rho}{(\delta - \rho)}$.

(A) When the block is displaced vertically by a small distance $x$ from its equilibrium position,the additional volume of liquid displaced is $V = A x$.
The additional buoyant force acting on the block is $F = \rho V g = \rho (A x) g$.
This force acts as a restoring force,so $F_{\text{restoring}} = -\rho A g x$.
According to Newton's second law,$F = M a$,where $M$ is the mass of the block.
Thus,$M a = -\rho A g x$,which gives $a = -(\frac{\rho A g}{M}) x$.
Comparing this with the standard $SHM$ equation $a = -\omega^2 x$,we get $\omega^2 = \frac{\rho A g}{M}$,so $\omega = \sqrt{\frac{\rho A g}{M}}$.
Since frequency $n = \frac{\omega}{2 \pi}$,we have $n = \frac{1}{2 \pi} \sqrt{\frac{\rho A g}{M}}$.
Therefore,$n \propto \sqrt{A}$.

(A) According to the Law of Floatation,for an object floating in a liquid,the weight of the object is equal to the weight of the liquid displaced.
Let $V$ be the total volume of the iceberg and $V_{sub}$ be the volume submerged in water.
Weight of the iceberg = $V \cdot \rho_{i} \cdot g$
Weight of the water displaced = $V_{sub} \cdot \rho_{w} \cdot g$
Equating the two: $V \cdot \rho_{i} \cdot g = V_{sub} \cdot \rho_{w} \cdot g$
The fraction of the volume submerged is $\frac{V_{sub}}{V} = \frac{\rho_{i}}{\rho_{w}}$.
Given $\rho_{i} = 0.917 \ g \ cm^{-3}$ and $\rho_{w} = 1.00 \ g \ cm^{-3}$.
Therefore,$\frac{V_{sub}}{V} = \frac{0.917}{1.00} = 0.917$.

(B) The buoyant force $F_b$ acting on a submerged object is given by Archimedes' principle: $F_b = V \rho g$,where $V$ is the volume of the displaced liquid,$\rho$ is the density of the liquid,and $g$ is the acceleration due to gravity.
The density of water is maximum at $4 \ ^\circ C$ $(\rho_4 \approx 1000 \ kg/m^3)$ and is lower at $0 \ ^\circ C$ $(\rho_0 \approx 999.8 \ kg/m^3)$.
Since the volume of the aluminium sphere $V$ remains constant,the buoyancy at $0 \ ^\circ C$ is $F_b = V \rho_0 g$ and at $4 \ ^\circ C$ is $F'_b = V \rho_4 g$.
Comparing the two,since $\rho_0 < \rho_4$,it follows that $F_b < F'_b$.
Therefore,the buoyancy is less in water at $0 \ ^\circ C$ than in water at $4 \ ^\circ C$.

(B) According to the law of floatation,for a body floating in equilibrium,the weight of the body is equal to the buoyant force (weight of the displaced liquid).
Let $V$ be the total volume of each solid,$\rho_P$ and $\rho_Q$ be their densities,and $\rho_w$ be the density of water.
For solid $P$: $V \rho_P g = (V/2) \rho_w g \Rightarrow \rho_P = \frac{1}{2} \rho_w$.
For solid $Q$: $V \rho_Q g = (2V/3) \rho_w g \Rightarrow \rho_Q = \frac{2}{3} \rho_w$.
The ratio of densities is $\frac{\rho_P}{\rho_Q} = \frac{\frac{1}{2} \rho_w}{\frac{2}{3} \rho_w} = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4}$.

(A) According to Archimedes' principle,the apparent loss of weight of an object immersed in a fluid is equal to the weight of the fluid displaced by it.
Loss of weight $w = V \rho_{w} g$,where $V$ is the volume of the object and $\rho_{w}$ is the density of water.
Since the mass $m$ of both spheres is the same,we have $m = V_{iron} \rho_{iron} = V_{lead} \rho_{lead}$.
Thus,the volume $V = \frac{m}{\rho}$.
Substituting this into the expression for loss of weight: $w = \frac{m}{\rho} \rho_{w} g = m g \frac{\rho_{w}}{\rho}$.
For the iron sphere,$w_{1} = m g \frac{\rho_{w}}{\rho_{iron}}$.
For the lead sphere,$w_{2} = m g \frac{\rho_{w}}{\rho_{lead}}$.
Taking the ratio: $\frac{w_{1}}{w_{2}} = \frac{\rho_{lead}}{\rho_{iron}}$.
Since the density of lead is greater than the density of iron $(\rho_{lead} > \rho_{iron})$,it follows that $\frac{w_{1}}{w_{2}} > 1$.

(D) Let $W_s$ and $W_a$ be the weights of steel and aluminium in air,and $V_s$ and $V_a$ be their respective volumes. The density of water is $\rho_w$. When immersed in water,the apparent weight is $W_{app} = W - V\rho_w g$. Given that the apparent weights are equal: $W_s - V_s \rho_w g = W_a - V_a \rho_w g$. Since $W = V \rho g$,we have $V = W / (\rho g)$. Substituting this,$W_s - (W_s / \rho_s) \rho_w g = W_a - (W_a / \rho_a) \rho_w g$,which simplifies to $W_s(1 - \rho_w / \rho_s) = W_a(1 - \rho_w / \rho_a)$. Since the density of steel $\rho_s \approx 7800 \ kg/m^3$ is much greater than the density of aluminium $\rho_a \approx 2700 \ kg/m^3$,the term $(1 - \rho_w / \rho_s)$ is greater than $(1 - \rho_w / \rho_a)$. Therefore,for the equality to hold,$W_s$ must be less than $W_a$. Thus,the aluminium piece weighs more in air.

(B) The upthrust on a body is given by the weight of the liquid displaced. The ratio of upthrusts is equal to the ratio of the densities (or specific gravities) of the liquids.
$\frac{\text{Upthrust in liquid}}{\text{Upthrust in water}} = \frac{\text{Specific gravity of liquid}}{\text{Specific gravity of water}}$
Upthrust in water $= 50 \,g - 40 \,g = 10 \,g$.
Let the upthrust in the liquid be $x$.
$\frac{x}{10 \,g} = \frac{1.5}{1}$
$x = 15 \,g$.
The weight of the body in the liquid is given by: $\text{Weight in air} - \text{Upthrust in liquid}$.
Weight $= 50 \,g - 15 \,g = 35 \,g$.

(B) Given: Mass of the object $m = 10 \ kg$,distance $S = 2 \ m$,time $t = 1 \ s$,initial velocity $u = 0$,and acceleration due to gravity $g = 10 \ m/s^2$.
Using the equation of motion $S = ut + \frac{1}{2}at^2$:
$2 = 0 \times 1 + \frac{1}{2} \times a \times (1)^2$
$2 = \frac{1}{2}a \Rightarrow a = 4 \ m/s^2$.
Now,applying Newton's second law for the sinking object:
$mg - F_B = ma$
Where $F_B$ is the buoyant force.
$F_B = m(g - a) = 10 \times (10 - 4) = 10 \times 6 = 60 \ N$.
According to Archimedes' principle,the buoyant force $F_B$ is equal to the weight of the displaced liquid:
$F_B = m_{liquid} \times g$
$60 = m_{liquid} \times 10$
$m_{liquid} = 6 \ kg$.

(D) For a floating object, the weight of the object equals the weight of the displaced fluid. This is given by the principle: $\rho_s V g = \rho_f V_{sub} g$, where $\rho_s$ is the density of the solid, $\rho_f$ is the density of the fluid, and $V_{sub}$ is the submerged volume.
$1$. In water: $\rho_s V g = \rho_w (0.5 V) g$.
Given $\rho_w = 1000 \,kg \,m^{-3}$, we have $\rho_s = 0.5 \times 1000 = 500 \,kg \,m^{-3}$.
$2$. In oil: $\rho_s V g = \rho_{oil} (0.8 V) g$.
Substituting $\rho_s = 500 \,kg \,m^{-3}$, we get $500 = 0.8 \times \rho_{oil}$.
Therefore, $\rho_{oil} = \frac{500}{0.8} = 625 \,kg \,m^{-3}$.

(C) According to the law of floatation,the weight of the body is equal to the weight of the liquid displaced.
Let $\rho$ be the density of the body and $\rho_w$ be the density of water $(1000 \ kg/m^3)$.
When floating on water,the volume submerged is $V_{sub} = V - \frac{1}{3}V = \frac{2}{3}V$.
Weight of body = Weight of displaced water $\Rightarrow V \rho g = V_{sub} \rho_w g$.
$V \rho = \frac{2}{3}V \rho_w \Rightarrow \rho = \frac{2}{3} \rho_w = \frac{2}{3} \times 1000 = \frac{2000}{3} \ kg/m^3$.
Now,the body floats on a liquid with specific gravity $1.5$,so the density of the liquid is $\rho_l = 1.5 \times 1000 = 1500 \ kg/m^3$.
Let $V'_{sub}$ be the volume submerged in this liquid.
$V \rho g = V'_{sub} \rho_l g \Rightarrow V \times \frac{2000}{3} = V'_{sub} \times 1500$.
$V'_{sub} = V \times \frac{2000}{3 \times 1500} = V \times \frac{2000}{4500} = \frac{4}{9}V$.
The volume above the surface is $V_{above} = V - V'_{sub} = V - \frac{4}{9}V = \frac{5}{9}V$.

(A) According to the principle of floatation,the weight of the flask plus the weight of the water inside it must equal the weight of the water displaced by the flask.
Let $V_{ext}$ be the external volume of the flask (volume of water displaced).
Mass of flask $m_f = 390 \ g$.
Volume of water inside $V_w = 250 \ cm^3$.
Mass of water inside $m_w = \rho_w \times V_w = 1 \times 250 = 250 \ g$.
Total weight of the system = $(390 + 250) \ g = 640 \ g$.
Since the flask just floats,the buoyant force equals the total weight,so the mass of water displaced is $640 \ g$.
Thus,the external volume of the flask $V_{ext} = 640 \ cm^3$.
The volume of the glass material $V_{glass} = V_{ext} - V_{internal} = 640 - 500 = 140 \ cm^3$.
The density of the glass $\rho_{glass} = \frac{m_f}{V_{glass}} = \frac{390}{140} \approx 2.785 \ g/cm^3$.
Specific gravity = $\frac{\rho_{glass}}{\rho_w} = \frac{2.785}{1} \approx 2.8$.

(B) Let $V$ be the total volume of the wooden block and $\sigma$ be its density. Let $\rho$ be the density of water.
Initially,the block floats in water with $(4/5)V$ submerged. By the law of floatation,the weight of the block equals the buoyant force:
$V \sigma g = (4/5)V \rho g$
$\sigma = (4/5) \rho$ ...$(i)$
When oil is poured,the block is submerged such that half its volume $(V/2)$ is in water and half $(V/2)$ is in oil. Let $\rho_o$ be the density of oil.
The total buoyant force equals the weight of the block:
$V \sigma g = (V/2) \rho g + (V/2) \rho_o g$
Substituting $\sigma = (4/5) \rho$ from $(i)$:
$(4/5) V \rho = (1/2) V \rho + (1/2) V \rho_o$
Dividing by $V$ and multiplying by $2$:
$(8/5) \rho = \rho + \rho_o$
$\rho_o = (8/5 - 1) \rho = (3/5) \rho$
Thus,the relative density of oil with respect to water is $3/5$.

(C) Given that,$(1/3)$rd part of the body is above the water surface.
Therefore,the volume of the body outside the water is $V_o = (1/3)V$,where $V$ is the total volume of the body.
Therefore,the volume of the body inside the water is $V_i = V - V_o = V - (V/3) = (2V/3)$.
Let the density of the body be $\sigma$ and the density of water be $\rho = 10^3 \ kg \ m^{-3}$.
According to the law of floatation,the weight of the body is balanced by the buoyant force $(F_B)$.
$W = F_B$
$Mg = V_i \rho g$
Since mass $M = V \sigma$,we have:
$V \sigma g = V_i \rho g$
$\sigma = (V_i / V) \rho$
Substituting the values:
$\sigma = \frac{(2V/3)}{V} \times 10^3 = \frac{2}{3} \times 10^3 = \frac{2000}{3} \ kg \ m^{-3}$.
Thus,the density of the solid is $\frac{2000}{3} \ kg \ m^{-3}$.

(D) The apparent weight of an object submerged in a fluid is given by the formula: $W_{app} = W_{actual} - F_B$, where $W_{actual}$ is the actual weight and $F_B$ is the buoyant force.
For an object of volume $V$ and density $\rho_{obj}$ submerged in a fluid of density $\rho_{fluid}$, the buoyant force is $F_B = V \rho_{fluid} g$.
The actual weight of the bag is $W_{actual} = m g = V \rho_{obj} g$.
Since the bag is filled with water, the density of the object (water) is equal to the density of the surrounding fluid (water), i.e., $\rho_{obj} = \rho_{fluid}$.
Therefore, $W_{app} = V \rho_{obj} g - V \rho_{fluid} g = 0$.
Thus, the reading of the spring balance is $0 \,kg$.

(D) Mass of the block,$m = 0.5 \,kg$.
Density of brass,$\rho = 8 \times 10^3 \,kg \,m^{-3}$.
Volume of the block,$V = \frac{m}{\rho} = \frac{0.5}{8 \times 10^3} = 6.25 \times 10^{-5} \,m^3$.
When the block is fully immersed in water,the upthrust (buoyant) force $F_b$ is given by the weight of the displaced water:
$F_b = V \cdot \rho_w \cdot g = (6.25 \times 10^{-5} \,m^3) \times (10^3 \,kg \,m^{-3}) \times (10 \,ms^{-2}) = 0.625 \,N$.
The tension $T$ in the string is the weight of the block minus the upthrust force:
$T = mg - F_b = (0.5 \,kg \times 10 \,ms^{-2}) - 0.625 \,N = 5 \,N - 0.625 \,N = 4.375 \,N$.
Converting to fraction: $4.375 = \frac{4375}{1000} = \frac{35}{8} \,N$.

(D) Density of the metallic block,$\rho = 5 \text{ g cm}^{-3}$.
Volume of the block,$V = 5 \text{ cm} \times 5 \text{ cm} \times 5 \text{ cm} = 125 \text{ cm}^3$.
Mass of the block,$m = \rho \times V = 5 \times 125 = 625 \text{ g}$.
Weight of the block in air,$W = m \times g = 625 \text{ gf}$.
Upthrust force on the metallic block in water,$F_u = \text{Volume of displaced water} \times \text{density of water} \times g$.
Since $\rho_w = 1 \text{ g cm}^{-3}$,$F_u = 125 \text{ cm}^3 \times 1 \text{ g cm}^{-3} = 125 \text{ gf}$.
Apparent weight = Weight in air - Upthrust force.
Apparent weight $= 625 \text{ gf} - 125 \text{ gf} = 500 \text{ gf}$.
Expressing this in the given format: $500 = 5 \times 5 \times 5 \times 4 \text{ gf}$.

(D) Archimedes' upward thrust (buoyant force) is defined as the weight of the fluid displaced by an object immersed in it.
Mathematically, the buoyant force $F_B = V \rho g$, where $V$ is the volume of the displaced fluid, $\rho$ is the density of the fluid, and $g$ is the acceleration due to gravity.
If there were no gravity, $g = 0$, which implies that the buoyant force $F_B = 0$.
Viscosity, surface tension, and pressure (in a fluid at rest due to intermolecular forces or external applied pressure) do not depend on gravity for their existence.
Therefore, Archimedes' upward thrust will not exist in the absence of gravity.

(C) Let $V$ be the volume of the body,$\rho_b$ be the density of the body,$\rho_w$ be the density of water $(1000 \,kg/m^3)$,and $\rho_l$ be the density of the liquid.
In air: $T_1 = V \rho_b g = 40.2 \,N$.
In water: $T_2 = V(\rho_b - \rho_w)g = 28.4 \,N$.
Buoyant force in water: $F_{Bw} = T_1 - T_2 = 40.2 - 28.4 = 11.8 \,N$.
Since $F_{Bw} = V \rho_w g$,we have $V g = 11.8 / 1000 = 0.0118 \,m^3 \cdot kg/m^3 \cdot m/s^2$.
In liquid: $T_3 = V(\rho_b - \rho_l)g = 16.6 \,N$.
Buoyant force in liquid: $F_{Bl} = T_1 - T_3 = 40.2 - 16.6 = 23.6 \,N$.
Since $F_{Bl} = V \rho_l g$,we have $V \rho_l g = 23.6 \,N$.
Dividing the two buoyant force equations: $\frac{V \rho_l g}{V \rho_w g} = \frac{23.6}{11.8} = 2$.
Therefore,$\rho_l = 2 \rho_w = 2 \times 1000 \,kg/m^3 = 2000 \,kg/m^3$.

(D) Let $R$ be the radius of the sphere and $r$ be the radius of the cavity.
Given: $R = 2 \ cm$,relative density of material $\rho_s = 8$,density of water $\rho_w = 1 \ g/cm^3$.
The sphere just sinks in water,which means its weight equals the buoyant force.
Weight of sphere = $\text{Volume of material} \times \rho_s \times g = \frac{4}{3} \pi (R^3 - r^3) \times 8 \times g$.
Buoyant force = $\text{Total volume of sphere} \times \rho_w \times g = \frac{4}{3} \pi R^3 \times 1 \times g$.
Equating the two: $\frac{4}{3} \pi (R^3 - r^3) \times 8 = \frac{4}{3} \pi R^3$.
$8(R^3 - r^3) = R^3$.
$8R^3 - 8r^3 = R^3 \Rightarrow 7R^3 = 8r^3$.
$r^3 = \frac{7}{8} R^3 = \frac{7}{8} \times (2)^3 = \frac{7}{8} \times 8 = 7 \ cm^3$.
Volume of cavity $V_c = \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 7 = \frac{88}{3} \ cm^3$.

(A) Let $V$ be the volume of the body and $\rho_b$ be its density. Let $\rho_1$ and $\rho_2$ be the densities of the liquid at $20^{\circ} C$ and $100^{\circ} C$ respectively.
By the law of flotation,the weight of the body equals the weight of the displaced liquid.
At $20^{\circ} C$: $V \rho_b g = (\frac{2}{3} V) \rho_1 g \implies \rho_b = \frac{2}{3} \rho_1$.
At $100^{\circ} C$: $V \rho_b g = (\frac{3}{4} V) \rho_2 g \implies \rho_b = \frac{3}{4} \rho_2$.
Equating the two expressions for $\rho_b$: $\frac{2}{3} \rho_1 = \frac{3}{4} \rho_2 \implies \frac{\rho_1}{\rho_2} = \frac{9}{8}$.
Since density $\rho = \frac{\rho_0}{1 + \gamma \Delta t}$,we have $\frac{\rho_1}{\rho_2} = \frac{1 + \gamma \Delta t_2}{1 + \gamma \Delta t_1} = 1 + \gamma \Delta T$ (where $\Delta T = 80^{\circ} C$).
$1 + \gamma (80) = \frac{9}{8} \implies 80 \gamma = \frac{1}{8} \implies \gamma = \frac{1}{640} = 0.0015625 {}^{\circ} C^{-1}$.
$\gamma = 15.6 \times 10^{-4} {}^{\circ} C^{-1}$.

(D) Mass of block, $m = 20 \,g = 0.02 \,kg$.
Initial elongation, $x = 1 \,cm = 0.01 \,m$.
At equilibrium, the spring force equals the weight: $kx = mg$.
Spring constant $k = \frac{mg}{x} = \frac{0.02 \times 10}{0.01} = 20 \,N/m$.
When the block is immersed in water, it experiences an upward buoyant force $F_B = \rho_w V g$, where $V = \frac{m}{\rho_c}$.
The new equilibrium condition is $kx' + F_B = mg$, where $x'$ is the new elongation.
$kx' = mg - \rho_w \left(\frac{m}{\rho_c}\right) g = mg \left(1 - \frac{\rho_w}{\rho_c}\right)$.
$x' = \frac{mg}{k} \left(1 - \frac{\rho_w}{\rho_c}\right) = x \left(1 - \frac{1000}{9000}\right)$.
$x' = 0.01 \times \left(1 - \frac{1}{9}\right) = 0.01 \times \frac{8}{9} \approx 0.00888 \,m$.
$x' \approx 0.89 \,cm$.

(C) Let the outer volume of the block be $V = 1000 \text{ ml} = 10^{-3} \text{ m}^3$.
The specific gravity of the wood is $\rho_w / \rho_{water} = 0.75$.
Thus,the density of the wood material is $\rho_w = 0.75 \times 1000 \text{ kg/m}^3 = 750 \text{ kg/m}^3$.
Let $V_c$ be the volume of the cavity. The volume of the actual wooden material is $V_m = V - V_c$.
The mass of the block is $M = \rho_w \times V_m = 750 \times (10^{-3} - V_c)$.
According to the law of floatation,the weight of the block equals the weight of the water displaced.
The block floats with half its volume immersed,so $V_{displaced} = \frac{V}{2} = 500 \text{ ml} = 5 \times 10^{-4} \text{ m}^3$.
Weight of displaced water = $\rho_{water} \times V_{displaced} \times g = 1000 \times 5 \times 10^{-4} \times g = 0.5g$.
Weight of the block = $M \times g = 750 \times (10^{-3} - V_c) \times g$.
Equating the two: $0.5g = 750 \times (10^{-3} - V_c) \times g$.
$0.5 = 0.75 - 750 \times V_c$.
$750 \times V_c = 0.25$.
$V_c = \frac{0.25}{750} = \frac{1}{3000} \text{ m}^3$.
$V_c = \frac{1}{3000} \times 10^6 \text{ ml} = 333.33 \text{ ml}$.

(C) Let the side of the cube be $a = 40 \ cm = 0.4 \ m$. The volume of the cube is $V = a^3 = (0.4)^3 = 0.064 \ m^3$.
Density of water $\rho_w = 1000 \ kg/m^3$.
Initially,the cube floats with $\frac{1}{4}$ of its volume immersed. By the law of floatation,the weight of the cube equals the weight of the displaced water:
$m_{cube} \cdot g = \rho_w \cdot V_{immersed} \cdot g$
$m_{cube} = 1000 \cdot (\frac{1}{4} \cdot 0.064) = 1000 \cdot 0.016 = 16 \ kg$.
When the disc of mass $M$ is placed on the cube,the total weight is $(m_{cube} + M)g$. The new volume immersed is $\frac{2}{5}$ of $V$:
$(m_{cube} + M)g = \rho_w \cdot (\frac{2}{5} \cdot V) \cdot g$
$16 + M = 1000 \cdot (\frac{2}{5} \cdot 0.064)$
$16 + M = 1000 \cdot 0.0256 = 25.6$
$M = 25.6 - 16 = 9.6 \ kg$.

(D) Given: Side of the cube $a = 10 \ cm$,so the area of each face $A = a^2 = 100 \ cm^2$.
Depth of the cube in water $x = 3 \ cm$.
Depth of the cube in oil $h = a - x = 10 - 3 = 7 \ cm$.
Density of oil $\delta_1 = 0.9 \ g \ cm^{-3}$.
Density of water $\delta_2 = 1 \ g \ cm^{-3}$.
By the law of flotation,the weight of the cube is equal to the total buoyant force exerted by the two liquids.
$W = F_{B1} + F_{B2}$
$mg = (\delta_1 \cdot V_{oil} \cdot g) + (\delta_2 \cdot V_{water} \cdot g)$
$m = \delta_1 \cdot A \cdot h + \delta_2 \cdot A \cdot x$
$m = 0.9 \times 100 \times 7 + 1 \times 100 \times 3$
$m = 630 + 300 = 930 \ g$
Therefore,the mass of the wooden cube is $930 \ g$.

(B) The volume of the material of the hollow sphere is $V_S = \frac{4}{3} \pi (R^3 - r^3)$,where $R = 4 \ cm$ and $r = 2 \ cm$.
For a floating body,the buoyant force equals the weight of the body.
$B = W$
$\rho_L V_{sub} g = \rho_S V_S g$
Since the sphere is half submerged,the submerged volume $V_{sub} = \frac{1}{2} \times ( \frac{4}{3} \pi R^3 ) = \frac{2}{3} \pi R^3$.
Substituting the values:
$2.0 \times \frac{2}{3} \pi (4)^3 = \rho_S \times \frac{4}{3} \pi (4^3 - 2^3)$
$2.0 \times 64 = \rho_S \times (64 - 8)$
$128 = \rho_S \times 56$
$\rho_S = \frac{128}{56} \approx 2.28 \ g \ cm^{-3}$.
Wait,re-evaluating the calculation: $V_S = \frac{4}{3} \pi (64 - 8) = \frac{4}{3} \pi (56) = \frac{224}{3} \pi$.
$V_{sub} = \frac{1}{2} \times \frac{4}{3} \pi (4^3) = \frac{128}{3} \pi$.
Equating forces: $2.0 \times (\frac{128}{3} \pi) = \rho_S \times (\frac{224}{3} \pi)$.
$\rho_S = \frac{256}{224} = 1.1428 \approx 1.14 \ g \ cm^{-3}$.

(B) Volume of the wood block $V_w = \frac{\text{Mass}}{\text{Density}} = \frac{160}{0.8 \times 1} = 200 \,cm^3$.
Let the mass of the metal piece be $x \,g$.
Volume of the metal piece $V_m = \frac{\text{Mass}}{\text{Density}} = \frac{x}{10 \times 1} = \frac{x}{10} \,cm^3$.
For the system to float in water, the total weight of the system must be equal to the buoyant force (upthrust) exerted by the water when the entire system is submerged.
Total weight $W = (m_w + m_m)g = (160 + x)g$.
Total volume of the system $V_{total} = V_w + V_m = (200 + \frac{x}{10}) \,cm^3$.
Buoyant force $F_B = V_{total} \times \rho_{water} \times g = (200 + \frac{x}{10}) \times 1 \times g$.
Equating weight and buoyant force: $(160 + x)g = (200 + \frac{x}{10})g$.
$160 + x = 200 + 0.1x$.
$0.9x = 40$.
$x = \frac{400}{9} \approx 44.4 \,g$.

(C) Given: Mass of iron ball, $m = 100 \,kg$. Density of iron, $\rho_{\text{iron}} = 8 \times 10^3 \,kg/m^3$. Density of sea water, $\rho_{\text{sea water}} = 1000 \,kg/m^3$. Acceleration due to gravity, $g = 10 \,m/s^2$.
When the ball is submerged in sea water, an upthrust (buoyant force) acts on it in the upward direction.
According to Archimedes' principle, the apparent weight of the ball is equal to its actual weight minus the upthrust force.
The tension $T$ in the cable is equal to the apparent weight of the ball.
$T = W - F_B = mg - V \rho_{\text{sea water}} g = mg \left(1 - \frac{\rho_{\text{sea water}}}{\rho_{\text{iron}}}\right)$.
Substituting the values:
$T = 100 \times 10 \left(1 - \frac{1000}{8 \times 10^3}\right) = 1000 \left(1 - \frac{1}{8}\right) = 1000 \times \frac{7}{8} = 875 \,N$.

(D) When the cylinder is immersed in the liquid,it displaces a volume of liquid equal to its own volume $V$.
According to Archimedes' principle,the buoyant force $F_B$ exerted by the liquid on the cylinder is $F_B = V \sigma g$.
By Newton's third law,the cylinder exerts an equal and opposite force on the liquid,which increases the pressure at the bottom of the vessel.
The volume of the cylinder is $V = \frac{m}{\rho}$.
The increase in pressure $\Delta P$ is given by the force exerted on the liquid divided by the cross-sectional area $A$ of the vessel:
$\Delta P = \frac{F_B}{A} = \frac{V \sigma g}{A}$.
Substituting $V = \frac{m}{\rho}$ into the equation:
$\Delta P = \frac{(m/\rho) \sigma g}{A} = \frac{m \sigma g}{\rho A}$.

(D) Weight of the block in air, $w_{\text{air}} = 6000 \,N$.
Weight of the block in water, $w_{\text{water}} = 4000 \,N$.
The buoyant force (upthrust) acting on the block is equal to the weight of the water displaced: $F_B = w_{\text{air}} - w_{\text{water}} = 6000 - 4000 = 2000 \,N$.
The total volume of the block $(V_{\text{total}})$ is equal to the volume of water displaced: $V_{\text{total}} = \frac{F_B}{\rho_{\text{water}} \cdot g} = \frac{2000}{1000 \cdot 10} = 0.2 \,m^3$.
The actual volume of the iron material $(V_{\text{iron}})$ is calculated from its weight in air: $V_{\text{iron}} = \frac{w_{\text{air}}}{\rho_{\text{iron}} \cdot g} = \frac{6000}{6000 \cdot 10} = 0.1 \,m^3$.
The volume of the cavity $(V_{\text{cavity}})$ is the difference between the total volume and the volume of the iron: $V_{\text{cavity}} = V_{\text{total}} - V_{\text{iron}} = 0.2 \,m^3 - 0.1 \,m^3 = 0.1 \,m^3$.

(B) Density of the material,$\rho = 3 \,g/cc$.
Density of the fluid,$\sigma = 7 \,g/cc$.
Let the total volume of the material be $V$ and the fraction of the volume submerged in the fluid be $n$.
According to the principle of flotation,the weight of the block is equal to the weight of the displaced fluid.
$V \rho g = (n V) \sigma g$.
Dividing both sides by $V g$,we get $\rho = n \sigma$.
Therefore,the submerged fraction $n = \frac{\rho}{\sigma} = \frac{3}{7}$.
The fraction of the volume of the block outside the fluid is $1 - n = 1 - \frac{3}{7} = \frac{4}{7} \approx 0.57$.

(B) At equilibrium,the forces acting on the block are the buoyant force $(F_B)$ acting upwards,the weight of the block $(W)$ acting downwards,and the tension in the string $(T)$ acting downwards.
For equilibrium: $F_B = W + T$.
Thus,$T = F_B - W$.
Given: Density of block $\rho_s = 0.5 \ g/cc = 500 \ kg/m^3$,density of liquid $\rho_l = 1 \ g/cc = 1000 \ kg/m^3$,$T = 20 \ N$,and $g = 10 \ m/s^2$.
$T = \rho_l V g - \rho_s V g = V g (\rho_l - \rho_s)$.
Substituting the values: $20 = V \times 10 \times (1000 - 500)$.
$20 = V \times 10 \times 500$.
$20 = 5000 V$.
$V = \frac{20}{5000} = \frac{1}{250} \ m^3$.
The mass of the block is $m = \rho_s V$.
$m = 500 \times \frac{1}{250} = 2 \ kg$.

(C) The apparent weight loss is equal to the weight of the displaced liquid (Archimedes' Principle).
At $30^{\circ} C$: Loss in weight = $45 - 25 = 20 \ g$.
Volume of metal $V_{30} = \frac{\text{Loss}}{\rho_{30}} = \frac{20 \ g}{1.5 \ g/cm^3} = 13.33 \ cm^3$.
At $40^{\circ} C$: Loss in weight = $45 - 27 = 18 \ g$.
Volume of metal $V_{40} = \frac{\text{Loss}}{\rho_{40}} = \frac{18 \ g}{1.25 \ g/cm^3} = 14.40 \ cm^3$.
The coefficient of volume expansion $\gamma$ is given by $V_{40} = V_{30}(1 + \gamma \Delta T)$.
$\gamma = \frac{V_{40} - V_{30}}{V_{30} \Delta T} = \frac{14.40 - 13.33}{13.33 \times (40 - 30)} = \frac{1.07}{133.3} \approx 8.03 \times 10^{-3} /^{\circ} C$.
The coefficient of linear expansion $\alpha = \frac{\gamma}{3} = \frac{8.03 \times 10^{-3}}{3} \approx 2.67 \times 10^{-3} /^{\circ} C$.

(A) Let $V_1$ be the volume of granite in water and $V_2$ be the volume of granite in mercury.
According to the principle of floatation,the total buoyant force acting on the granite must be equal to its weight.
The buoyant force from water is $F_{B1} = V_1 \rho_1 g$.
The buoyant force from mercury is $F_{B2} = V_2 \rho_2 g$.
The weight of the granite is $W = (V_1 + V_2) \rho g$.
Equating the forces: $V_1 \rho_1 g + V_2 \rho_2 g = (V_1 + V_2) \rho g$.
Dividing by $g$: $V_1 \rho_1 + V_2 \rho_2 = V_1 \rho + V_2 \rho$.
Rearranging the terms to group $V_1$ and $V_2$: $V_1 \rho_1 - V_1 \rho = V_2 \rho - V_2 \rho_2$.
$V_1(\rho_1 - \rho) = V_2(\rho - \rho_2)$.
Therefore,the ratio of the volume of granite in water $(V_1)$ to the volume of granite in mercury $(V_2)$ is:
$\frac{V_1}{V_2} = \frac{\rho - \rho_2}{\rho_1 - \rho} = \frac{\rho_2 - \rho}{\rho - \rho_1}$.

(B) The density of the metal is $\rho$. The density of water is $\rho_w = 1 \text{ g/cm}^3$ (or $1000 \text{ kg/m}^3$).
For the hollow sphere to float,its weight must be less than or equal to the weight of the water displaced by the total volume of the sphere.
The volume of the metal shell is $V_m = 4 \pi R^2 t$ (since $t \ll R$).
The mass of the sphere is $m_s = V_m \times \rho = 4 \pi R^2 t \rho$.
The volume of water displaced by the sphere is $V_w = \frac{4}{3} \pi R^3$.
The mass of the displaced water is $m_w = V_w \times \rho_w = \frac{4}{3} \pi R^3 \times 1$ (assuming $\rho$ is relative to water density).
For floatation,$m_s \leq m_w$:
$4 \pi R^2 t \rho \leq \frac{4}{3} \pi R^3$
Dividing both sides by $4 \pi R^2$:
$t \rho \leq \frac{R}{3}$
$t \leq \frac{R}{3 \rho}$

(C) The buoyant force acting on an immersed body is equal to the weight of the fluid displaced by it when the fluid is at rest. When the fluid is accelerating,the effective gravity $g_{\text{eff}}$ must be considered.
In the first case,the block is in equilibrium: $V_{\text{immersed}} \rho g = V_{\text{total}} \rho_b g$,where $\rho$ is the density of water and $\rho_b$ is the density of the block.
Given $40 \%$ is above the surface,$60 \%$ is immersed,so $V_{\text{immersed}} = 0.6 V_{\text{total}}$.
Thus,$0.6 V \rho g = V \rho_b g \implies \rho_b = 0.6 \rho$.
When the lift accelerates upward with $a = g/2$,the effective gravity is $g_{\text{eff}} = g + a = 1.5 g$.
The buoyant force becomes $F_B = V_{\text{immersed}} \rho (1.5 g)$.
The weight of the block in the accelerating frame is $W_{\text{eff}} = V_{\text{total}} \rho_b (1.5 g)$.
For equilibrium,$F_B = W_{\text{eff}} \implies V_{\text{immersed}} \rho (1.5 g) = V_{\text{total}} \rho_b (1.5 g)$.
Dividing by $1.5 g$,we get $V_{\text{immersed}} \rho = V_{\text{total}} \rho_b$.
Substituting $\rho_b = 0.6 \rho$,we get $V_{\text{immersed}} = 0.6 V_{\text{total}}$.
Therefore,the fraction of the volume immersed remains $60 \%$,and the fraction above the surface remains $40 \%$.

(C) For the rod to be in horizontal equilibrium,the net torque about the mid-point must be zero.
Initially,the torque balance is: $w \cdot l = w_{1} \cdot l_{1}$.
When the metal piece of weight $w$ is immersed in water,it experiences an upward buoyant force $F_{B}$. The effective weight becomes $w' = w - F_{B}$.
The buoyant force is $F_{B} = V \rho_{w} g$,where $V$ is the volume of the metal and $\rho_{w}$ is the density of water. Since $w = V \rho_{metal} g$,we have $F_{B} = w \cdot \frac{\rho_{w}}{\rho_{metal}} = \frac{w}{\sigma}$,where $\sigma$ is the specific gravity of the metal.
Thus,the new effective weight is $w' = w(1 - \frac{1}{\sigma})$.
For the new equilibrium,the torque balance is: $w' \cdot l = w_{1} \cdot l_{2}$.
Substituting $w'$,we get: $w(1 - \frac{1}{\sigma}) l = w_{1} l_{2}$.
From the initial condition,$w = \frac{w_{1} l_{1}}{l}$.
Substituting $w$ into the new equilibrium equation: $\frac{w_{1} l_{1}}{l} (1 - \frac{1}{\sigma}) l = w_{1} l_{2}$.
$l_{1} (1 - \frac{1}{\sigma}) = l_{2}$.
$1 - \frac{1}{\sigma} = \frac{l_{2}}{l_{1}}$.
$\frac{1}{\sigma} = 1 - \frac{l_{2}}{l_{1}} = \frac{l_{1} - l_{2}}{l_{1}}$.
Therefore,$\sigma = \frac{l_{1}}{l_{1} - l_{2}}$.

(A) Let $x$ be the weight of the first metal and $(w_1 - x)$ be the weight of the second metal in air.
Let $v_1$ and $v_2$ be the volumes of the two metals respectively.
$v_1 = \frac{x}{\rho_1}$ and $v_2 = \frac{w_1 - x}{\rho_2}$.
The total volume of the alloy is $V = v_1 + v_2 = \frac{x}{\rho_1} + \frac{w_1 - x}{\rho_2}$.
According to Archimedes' Principle,the loss in weight equals the weight of the displaced water:
$w_1 - w_2 = V \rho_w = \left( \frac{x}{\rho_1} + \frac{w_1 - x}{\rho_2} \right) \rho_w$.
Multiplying by $\rho_1 \rho_2$:
$(w_1 - w_2) \rho_1 \rho_2 = (x \rho_2 + (w_1 - x) \rho_1) \rho_w$.
$(w_1 - w_2) \rho_1 \rho_2 = (x \rho_2 + w_1 \rho_1 - x \rho_1) \rho_w$.
$(w_1 - w_2) \rho_1 \rho_2 = x(\rho_2 - \rho_1) \rho_w + w_1 \rho_1 \rho_w$.
$x(\rho_2 - \rho_1) \rho_w = w_1 \rho_1 \rho_2 - w_2 \rho_1 \rho_2 - w_1 \rho_1 \rho_w$.
$x(\rho_2 - \rho_1) \rho_w = w_1 \rho_1(\rho_2 - \rho_w) - w_2 \rho_1 \rho_2$.
$x = \frac{\rho_1}{\rho_w(\rho_2 - \rho_1)} [w_1(\rho_2 - \rho_w) - w_2 \rho_2]$.

(A) Let $V$ be the total volume of the cylindrical block and $\rho$ be its density.
In the first case,the block floats in a liquid of density $\rho_{1}$. According to the law of floatation,the weight of the block equals the buoyant force:
$V \rho g = V x_{1} \rho_{1} g$
$\Rightarrow \rho = x_{1} \rho_{1} \quad ... (1)$
In the second case,the block is fully immersed in two immiscible liquids of densities $\rho_{1}$ and $\rho_{2}$. The volume fraction $x_{2}$ is in liquid $\rho_{1}$ and the remaining fraction $(1 - x_{2})$ is in liquid $\rho_{2}$. The total buoyant force equals the weight of the block:
$V \rho g = V x_{2} \rho_{1} g + V (1 - x_{2}) \rho_{2} g$
$\Rightarrow \rho = x_{2} \rho_{1} + (1 - x_{2}) \rho_{2} \quad ... (2)$
Equating $(1)$ and $(2)$:
$x_{1} \rho_{1} = x_{2} \rho_{1} + (1 - x_{2}) \rho_{2}$
$x_{1} \rho_{1} - x_{2} \rho_{1} = (1 - x_{2}) \rho_{2}$
$\rho_{1} (x_{1} - x_{2}) = \rho_{2} (1 - x_{2})$
$\frac{\rho_{1}}{\rho_{2}} = \frac{1 - x_{2}}{x_{1} - x_{2}}$

(D) Let $m$ be the mass of the body in grams and $V$ be the volume of the body in $\text{cm}^3$. The density of water is $\rho_w = 1 \text{ g/cm}^3$. The apparent weight $W'$ is given by $W' = W_{actual} - F_B$,where $F_B$ is the buoyant force.
For the liquid with specific gravity $1.2$,the density is $\rho_l = 1.2 \text{ g/cm}^3$. The weight is $44 \text{ gwt}$,so:
$44 = m - 1.2V$ $(i)$
For water,the weight is $50 \text{ gwt}$,so:
$50 = m - V$ $(ii)$
Subtracting $(i)$ from $(ii)$:
$(50 - 44) = (m - V) - (m - 1.2V)$
$6 = 0.2V$
$V = 30 \text{ cm}^3$
Substituting $V = 30$ into $(ii)$:
$50 = m - 30$
$m = 80 \text{ g}$

(C) Let $V$ be the volume of the stone,$\sigma$ be the density of the stone,and $\rho$ be the density of water.
The relative density $K$ of the stone is given by $K = \frac{\sigma}{\rho}$,so $\sigma = K\rho$.
The forces acting on the stone as it sinks are:
$1$. Weight of the stone acting downwards: $W = V\sigma g = V(K\rho)g$.
$2$. Buoyant force acting upwards: $F_B = V\rho g$.
According to Newton's second law,the net force $F_{net} = W - F_B = ma$,where $m = V\sigma$ is the mass of the stone.
$V\sigma g - V\rho g = (V\sigma)a$
Dividing by $V\sigma$:
$a = g - \frac{V\rho g}{V\sigma} = g - \frac{\rho}{\sigma}g$
Since $K = \frac{\sigma}{\rho}$,we have $\frac{\rho}{\sigma} = \frac{1}{K}$.
Therefore,the acceleration $a = g - \frac{g}{K} = g\left(1 - \frac{1}{K}\right)$.