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

(B) When a soda bottle is opened,the pressure inside decreases,causing dissolved $CO_2$ gas to form bubbles.
According to Archimedes' principle,any object submerged in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced by it.
The buoyant force $(F_b)$ acting on the bubble is given by $F_b = V \rho_{liquid} g$,where $V$ is the volume of the bubble,$\rho_{liquid}$ is the density of the liquid,and $g$ is the acceleration due to gravity.
The weight of the bubble $(W)$ is $W = V \rho_{gas} g$.
Since the density of the gas $(\rho_{gas})$ is much smaller than the density of the liquid $(\rho_{liquid})$,the buoyant force is significantly greater than the weight of the bubble $(F_b > W)$.
This net upward force causes the bubbles to accelerate and rise to the surface.

(N/A) Buoyancy is the upward force exerted by a fluid on an object immersed in it. This force is known as the buoyant force. The property of a fluid to exert this upward force is called buoyancy.

(N/A) According to Archimedes' principle,an object floats if the weight of the liquid it displaces is equal to its own weight.
$1$. $A$ piece of iron has a small volume,so the weight of the water it displaces is less than its own weight,causing it to sink.
$2$. $A$ ship is designed with a large hollow structure,which increases its volume significantly. This allows the ship to displace a volume of water whose weight is equal to the total weight of the ship,satisfying the condition for flotation.

(N/A) No,the water level in the lake will not change.
According to Archimedes' principle,the weight of the water displaced by the boat is equal to the weight of the boat and the person.
When the person takes a bucket of water from the lake,the weight of the water removed is equal to the weight of the water displaced by the bucket when it was submerged.
When the bucket is placed inside the boat,the total weight of the boat increases by the weight of the water in the bucket.
This additional weight causes the boat to sink slightly deeper,displacing an amount of water equal to the weight of the water in the bucket.
Since the volume of water added to the boat is exactly equal to the volume of water removed from the lake,the net change in the water level of the lake is zero.

(B) No, the container does not obey the principle of Archimedes in free fall.
Archimedes' principle states that the buoyant force is equal to the weight of the fluid displaced, which is given by $F_{b} = V \rho g$.
In a state of free fall, the effective acceleration due to gravity $(g_{eff})$ acting on the fluid is $g - a$. Since $a = g$ in free fall, $g_{eff} = 0$.
Consequently, the buoyant force $F_{b} = V \rho (0) = 0$.
Since the buoyant force becomes zero, the principle of Archimedes is not applicable.

(B) The bigger cork will reach the surface faster. According to Archimedes' principle,the buoyant force acting on an object is equal to the weight of the fluid displaced by it. Since the bigger cork has a larger volume,it displaces a greater volume of water compared to the smaller cork. Consequently,the buoyant force acting on the bigger cork is significantly higher. Since the buoyant force is the upward force responsible for bringing the cork to the surface,the larger cork experiences a greater net upward acceleration,causing it to reach the surface faster.

(N/A) It is easier to swim in sea water than in river water because sea water contains dissolved salts,which increases the density of the water. According to Archimedes' principle,the upthrust (buoyant force) acting on an object is given by $F = V \rho g$,where $V$ is the volume of the displaced fluid,$\rho$ is the density of the fluid,and $g$ is the acceleration due to gravity. Since the density of sea water is higher than that of river water,the upthrust exerted by sea water is greater. This increased buoyant force helps the body float more easily,requiring less effort to stay afloat.

(B) The density of ice is $\rho_{\text{ice}} = 0.917 \, g/cm^3$.
The density of water is $\rho_{w} = 1 \, g/cm^3$.
Let the total volume of the iceberg be $V$.
Let the volume of the iceberg submerged in water be $V^{\prime}$.
For a body floating in equilibrium,the weight of the body must be equal to the weight of the liquid displaced by the submerged part of the body (Archimedes' Principle).
Weight of iceberg = Weight of water displaced
$V \cdot \rho_{\text{ice}} \cdot g = V^{\prime} \cdot \rho_{w} \cdot g$
Dividing both sides by $V \cdot \rho_{w} \cdot g$,we get the fraction of submerged volume:
$\frac{V^{\prime}}{V} = \frac{\rho_{\text{ice}}}{\rho_{w}} = \frac{0.917}{1} = 0.917$.

(N/A) Consider the system consisting of the vessel and the water. The scale is initially adjusted to zero.
When the block is submerged in the water,it experiences an upward buoyant force (upthrust) $F_B$ from the water.
According to Newton's third law,the block exerts an equal and opposite downward force on the water.
Therefore,the reading of the scale increases by an amount equal to the upthrust experienced by the block.
Upthrust $F_B = V_{sub} \rho_w g$,where $V_{sub}$ is the submerged volume of the block and $\rho_w$ is the density of water.
If the entire block of mass $M$ and density $\rho$ is submerged,then $V_{sub} = V = \frac{M}{\rho}$.
Thus,the reading of the scale is $F_B = \left( \frac{M}{\rho} \right) \rho_w g = M g \left( \frac{\rho_w}{\rho} \right)$.

(A) Let the density of water be $\rho_{w}$. The block of height $L$ floats on it. Let $x$ be the height of the block submerged in water.
Volume of the block $V = L^{3}$.
Mass of the block $m = V \rho = L^{3} \rho$.
Weight of the block $W = mg = L^{3} \rho g$.
Case $1$: When the elevator is at rest (or moving with constant velocity),the weight of the block is balanced by the buoyant force.
Weight of the block = Weight of the water displaced.
$L^{3} \rho g = (x L^{2}) \rho_{w} g$.
Therefore,the fraction submerged is $\frac{x}{L} = \frac{\rho}{\rho_{w}}$.
Case $2$: When the elevator is accelerating upward with acceleration $a$,the effective acceleration becomes $g' = (g + a)$.
In this frame,the weight of the block becomes $W' = m(g + a) = L^{3} \rho (g + a)$.
The buoyant force also changes because the effective gravity acting on the fluid changes: $F_{B}' = V_{submerged} \rho_{w} (g + a) = (x' L^{2}) \rho_{w} (g + a)$,where $x'$ is the new submerged height.
Equating the two: $L^{3} \rho (g + a) = x' L^{2} \rho_{w} (g + a)$.
Canceling $(g + a)$ from both sides,we get $L^{3} \rho = x' L^{2} \rho_{w}$.
Therefore,the new fraction submerged is $\frac{x'}{L} = \frac{\rho}{\rho_{w}}$.
Conclusion: The fraction of the block submerged remains unchanged regardless of the acceleration of the elevator.

(C) The volume of the air bubble is $V = \frac{4}{3} \pi r^3 = \frac{4}{3} \times \pi \times (1)^3 \approx 4.1888\,cm^3 \approx 4.19\,cm^3$.
The forces acting on the bubble are the buoyant force $B$ (upward) and the weight $mg$ (downward).
According to Newton's second law,the net force is $B - mg = ma$.
Here,$B = V \rho_w g$,where $\rho_w = 1\,g\,cm^{-3}$ is the density of water.
Substituting $B$ into the equation: $V \rho_w g - mg = ma$.
Rearranging to solve for mass $m$: $m(g + a) = V \rho_w g$.
$m = \frac{V \rho_w g}{g + a} = \frac{V \rho_w}{1 + \frac{a}{g}}$.
Substituting the values: $m = \frac{4.19 \times 1}{1 + \frac{9.8}{980}} = \frac{4.19}{1 + 0.01} = \frac{4.19}{1.01} \approx 4.1485\,g \approx 4.15\,g$.

(B) For the shell to float just submerged,the weight of the shell must be equal to the weight of the water displaced.
Weight of shell = $V_{material} \times \rho_{material} \times g = \frac{4}{3} \pi (R^3 - r^3) \rho_{material} g$.
Weight of displaced water = $V_{total} \times \rho_{water} \times g = \frac{4}{3} \pi R^3 \rho_{water} g$.
Equating the two: $\frac{4}{3} \pi (R^3 - r^3) \rho_{material} g = \frac{4}{3} \pi R^3 \rho_{water} g$.
$(R^3 - r^3) \rho_{material} = R^3 \rho_{water}$.
Given specific gravity $\frac{\rho_{material}}{\rho_{water}} = \frac{27}{8}$.
So,$(R^3 - r^3) \frac{27}{8} = R^3$.
$R^3 - r^3 = \frac{8}{27} R^3$.
$r^3 = R^3 - \frac{8}{27} R^3 = \frac{19}{27} R^3$.
$r = R \left( \frac{19}{27} \right)^{1/3} = \frac{R}{3} (19)^{1/3}$.
Since $(19)^{1/3} \approx 2.668$,$r \approx \frac{2.668}{3} R \approx 0.889 R$.

(B) Case $A$: Here,only the force acting on the weighing pan is the weight of the water. So,$w_{A} = mg$.
Case $B$: In this case,the downward forces on the pan are the weight of the water and the reaction force equal to the buoyant force $(F_{B})$ exerted by the water on the ball. So,$w_{B} = mg + F_{B}$.
Case $C$: In this case,the downward acting forces are the weight of the water and the reaction of the buoyant force,which is the same as in case $B$ because the balls are of the same size. So,$w_{C} = mg + F_{B}$.
Case $D$: In this case,the forces acting on the bottom of the beaker are the weight of the water $(mg)$,the weight of the ball $(m'g)$,and the tension $(T)$ in the string pulling upwards. The buoyant force $(F_{B})$ acts upwards on the ball. For the ball to be in equilibrium,$T + F_{B} = m'g$,so $T = m'g - F_{B}$. The total force on the pan is the weight of the water plus the weight of the ball minus the upward tension: $w_{D} = mg + m'g - T = mg + m'g - (m'g - F_{B}) = mg + F_{B}$.
Wait,let's re-evaluate Case $D$: The total downward force on the pan is the weight of the water $(mg)$ plus the weight of the ball $(m'g)$ minus the tension $(T)$ in the string. Since the ball is submerged,$F_{B} + T = m'g$,so $T = m'g - F_{B}$. Thus,$w_{D} = mg + m'g - (m'g - F_{B}) = mg + F_{B}$.
Therefore,$w_{B} = w_{C} = w_{D} > w_{A}$. However,if the $TT$ ball is very light $(m' \approx 0)$,then $w_{D} \approx mg + F_{B}$. If the lead ball is heavy,$w_{B} = mg + F_{B}$. The correct relationship is $w_{B} = w_{C} = w_{D} > w_{A}$.

(B) Let $m$ be the mass of each coin. The initial submerged depth is $h_0 = 3 \, cm$ for $40$ coins.
By Archimedes' principle,the weight of the container equals the weight of the displaced water: $(40m)g = A \cdot h_0 \cdot \rho_w \cdot g$,where $A$ is the cross-sectional area and $\rho_w$ is the density of water.
Thus,$A \rho_w = \frac{40m}{3}$.
When $N$ additional coins are added,the total mass is $(40+N)m$. The new submerged depth $h'$ is given by $(40+N)m = A \cdot h' \cdot \rho_w$,so $h' = \frac{(40+N)m}{A \rho_w} = \frac{(40+N)m}{40m/3} = \frac{3(40+N)}{40} \, cm$.
The centre of mass $(CM)$ of the system (measured from the base) is $CM = \frac{40m(0) + Nm(9)}{(40+N)m} = \frac{9N}{40+N} \, cm$.
The geometric centre $(GC)$ of the submerged portion is at a distance $h'/2$ from the base: $GC = \frac{h'}{2} = \frac{3(40+N)}{80} \, cm$.
For the transition from stable to unstable equilibrium,the $CM$ must coincide with the $GC$ $(CM = GC)$:
$\frac{9N}{40+N} = \frac{3(40+N)}{80}$
$720N = 3(40+N)^2$
$240N = 1600 + 80N + N^2$
$N^2 - 160N + 1600 = 0$
Using the quadratic formula: $N = \frac{160 \pm \sqrt{160^2 - 4(1600)}}{2} = \frac{160 \pm \sqrt{25600 - 6400}}{2} = \frac{160 \pm \sqrt{19200}}{2} = 80 \pm 40\sqrt{3} \approx 80 \pm 69.28$.
Since $N < 40$ (the container height is $9 \, cm$),we take $N = 80 - 69.28 = 10.72$.
The value of $N$ is closest to $10$.

(A) For a floating block,the weight of the block equals the weight of the displaced water: $Mg = V_{sub} \rho_{water} g$,where $V_{sub}$ is the submerged volume.
Since the total volume of the block $V = V_{sub} + V_0$ is constant (ignoring thermal expansion of wood),we have $V_0 = V - V_{sub} = V - \frac{M}{\rho_{water}}$.
As the temperature of water increases from $0^{\circ} C$ to $10^{\circ} C$,the density of water $\rho_{water}$ initially increases,reaching a maximum at $4^{\circ} C$,and then decreases.
Since $V_0 = V - \frac{M}{\rho_{water}}$,as $\rho_{water}$ increases,the term $\frac{M}{\rho_{water}}$ decreases,causing $V_0$ to increase until $4^{\circ} C$.
After $4^{\circ} C$,as $\rho_{water}$ decreases,the term $\frac{M}{\rho_{water}}$ increases,causing $V_0$ to decrease.
Therefore,the volume of the block above water $V_0$ increases up to $4^{\circ} C$ and then decreases from $4^{\circ} C$ to $10^{\circ} C$,which corresponds to the graph shown in option $A$.

(C) The fraction of the thickness of the ice block that remains above the water surface is given by $x = 1 - (\rho_{\text{ice}} / \rho_{\text{water}})$.
Taking the density of ice $\rho_{\text{ice}} \approx 0.9 \, \text{g/cm}^3$ and density of water $\rho_{\text{water}} = 1 \, \text{g/cm}^3$,we get $x = 1 - 0.9 = 0.1$.
The thickness of the ice block is $H = 8 \, m$.
The height of the ice block above the water level is $h = H \times x = 8 \times 0.1 = 0.8 \, m$.
Since the person is standing on the ice,the distance to the water surface is the height of the ice above the water.
Therefore,the minimum length of rope required is approximately $0.8 \, m$.
Comparing this with the given options,the nearest value is $0.9 \, m$.

(D) The reading of a spring balance measures the tension in the spring,which is equal to the apparent weight of the object.
When the box is in the atmosphere,it experiences an upward buoyant force $F_b$ due to the displaced air.
The reading $R$ in the atmosphere is given by $R = W - F_b$,where $W$ is the true weight of the box.
In an evacuated chamber,the buoyant force $F_b$ becomes zero.
Therefore,the new reading $R'$ will be equal to the true weight $W$.
Since $R' = W$ and $R = W - F_b$,it follows that $R' > R$.
Thus,the reading will be more than $50 \,kg$ because the atmospheric buoyancy force is absent.

(D) For a soap bubble to float in air,the gravitational force must be equal to the buoyant force.
Gravitational force = Buoyant force
$g \times (\text{mass of helium} + \text{mass of soap film}) = \text{Weight of air displaced by bubble}$
Let $r$ be the inner radius of the soap bubble and $t$ be the thickness of the film.
The volume of helium is $V_{He} = \frac{4}{3} \pi r^3$.
The volume of the soap film is $V_{film} \approx 4 \pi r^2 t$ (since $t \ll r$).
The equation of equilibrium is:
$\frac{4}{3} \pi r^3 \rho_{He} g + (4 \pi r^2 t) \rho_{soap} g = \frac{4}{3} \pi r^3 \rho_{air} g$
Dividing by $g$ and rearranging:
$4 \pi r^2 t \rho_{soap} = \frac{4}{3} \pi r^3 (\rho_{air} - \rho_{He})$
$t = \frac{r}{3} \frac{(\rho_{air} - \rho_{He})}{\rho_{soap}}$
Given $r = 1 \, cm = 10^{-2} \, m$,$\rho_{air} = 1.23 \, kg \, m^{-3}$,$\rho_{He} = 0.18 \, kg \, m^{-3}$,and $\rho_{soap} = 1000 \, kg \, m^{-3}$.
$t = \frac{10^{-2}}{3} \times \frac{(1.23 - 0.18)}{1000}$
$t = \frac{10^{-2}}{3} \times \frac{1.05}{1000} = \frac{1.05 \times 10^{-5}}{3} = 0.35 \times 10^{-5} \, m = 3.5 \times 10^{-6} \, m$.
Since $1 \, \mu m = 10^{-6} \, m$,the thickness $t = 3.50 \, \mu m$.

(B) Given,surface area of cube $=$ surface area of sphere.
$6a^2 = 4\pi r^2 \Rightarrow \frac{a}{r} = \sqrt{\frac{4\pi}{6}} = \sqrt{\frac{2\pi}{3}}$.
Buoyant force $F_B = V \cdot \rho_f \cdot g$. Since both are completely submerged in the same fluid,the ratio of buoyant forces is equal to the ratio of their volumes:
$\frac{(F_B)_{\text{cube}}}{(F_B)_{\text{sphere}}} = \frac{V_{\text{cube}}}{V_{\text{sphere}}} = \frac{a^3}{\frac{4}{3}\pi r^3} = \frac{3}{4\pi} \left( \frac{a}{r} \right)^3$.
Substituting $\frac{a}{r} = \sqrt{\frac{2\pi}{3}}$:
$\frac{(F_B)_{\text{cube}}}{(F_B)_{\text{sphere}}} = \frac{3}{4\pi} \left( \sqrt{\frac{2\pi}{3}} \right)^3 = \frac{3}{4\pi} \cdot \frac{2\pi}{3} \cdot \sqrt{\frac{2\pi}{3}} = \frac{1}{2} \sqrt{\frac{2\pi}{3}} = \sqrt{\frac{2\pi}{3 \cdot 4}} = \sqrt{\frac{\pi}{6}}$.
Since $\pi \approx 3.14$,$\frac{\pi}{6} < 1$,therefore $\sqrt{\frac{\pi}{6}} < 1$.
This implies $(F_B)_{\text{cube}} < (F_B)_{\text{sphere}}$.
Thus,the buoyant force is greater for the sphere.

(B) For the equilibrium of the block hung from the spring,the sum of the spring force and the buoyant force must equal the weight of the block.
Let $V$ be the volume of the object and $k$ be the spring constant.
The weight of the object is $W = \rho V g$.
The buoyant force in a liquid of density $\rho_f$ is $F_B = \rho_f V g$.
The spring force is $F_s = kx$.
For equilibrium: $kx + \rho_f V g = \rho V g$.
In the first liquid (density $\rho_1$): $k x_1 + \rho_1 V g = \rho V g \quad \dots(i)$
In the second liquid (density $\rho_2$): $k x_2 + \rho_2 V g = \rho V g \quad \dots(ii)$
From $(i)$,$k = \frac{(\rho - \rho_1) V g}{x_1}$.
From $(ii)$,$k = \frac{(\rho - \rho_2) V g}{x_2}$.
Equating the two expressions for $k$:
$\frac{(\rho - \rho_1) V g}{x_1} = \frac{(\rho - \rho_2) V g}{x_2}$
$(\rho - \rho_1) x_2 = (\rho - \rho_2) x_1$
$\rho x_2 - \rho_1 x_2 = \rho x_1 - \rho_2 x_1$
$\rho (x_2 - x_1) = \rho_1 x_2 - \rho_2 x_1$
$\rho = \frac{\rho_1 x_2 - \rho_2 x_1}{x_2 - x_1}$

(C) When the block is immersed in the liquid,it experiences an upward buoyant force $F_b$ due to the liquid.
For balance $A$,the reading corresponds to the tension $T$ in the spring. Initially,$T = mg = 2 \,kg \times g$. When immersed,$T' = mg - F_b$. Since $F_b > 0$,the new reading $T'$ will be less than $2 \,kg$.
For balance $B$,the initial reading is the weight of the beaker plus the liquid. When the block is immersed,by Newton's third law,the block exerts an equal and opposite downward force $F_b$ on the liquid. Therefore,the new reading on balance $B$ will be the initial weight plus the buoyant force $F_b$. Since $F_b > 0$,the new reading will be greater than $3 \,kg$. However,the total weight of the system (beaker + liquid + block) is $2 \,kg + 3 \,kg = 5 \,kg$. Since the block is not resting on the bottom of the beaker but is supported by spring $A$,the reading on $B$ will be less than the total weight of $5 \,kg$.
Thus,balance $A$ reads less than $2 \,kg$ and balance $B$ reads between $3 \,kg$ and $5 \,kg$.

(C) For the cube to be in equilibrium,the weight of the cube must be equal to the total buoyant force exerted by the two liquids.
Weight of the cube $W = m g$,where $m$ is the mass of the cube.
Buoyant force $F_B = F_{B,A} + F_{B,B} = V_A \rho_A g + V_B \rho_B g$.
Here,the edge length of the cube is $L = 10 \, cm = 0.1 \, m$. The cross-sectional area of the cube is $A_{cube} = L^2 = 100 \, cm^2 = 0.01 \, m^2$.
The volume of the cube submerged in liquid $A$ is $V_A = A_{cube} \times h_A = 100 \, cm^2 \times 4 \, cm = 400 \, cm^3 = 400 \times 10^{-6} \, m^3$.
The volume of the cube submerged in liquid $B$ is $V_B = A_{cube} \times h_B = 100 \, cm^2 \times 6 \, cm = 600 \, cm^3 = 600 \times 10^{-6} \, m^3$.
The densities of the liquids are $\rho_A = 0.6 \times 1000 \, kg/m^3 = 600 \, kg/m^3$ and $\rho_B = 0.4 \times 1000 \, kg/m^3 = 400 \, kg/m^3$.
Equating weight and buoyant force: $m g = (V_A \rho_A + V_B \rho_B) g$.
$m = V_A \rho_A + V_B \rho_B = (400 \times 10^{-6} \, m^3 \times 600 \, kg/m^3) + (600 \times 10^{-6} \, m^3 \times 400 \, kg/m^3)$.
$m = (240,000 \times 10^{-6}) + (240,000 \times 10^{-6}) \, kg = 0.24 \, kg + 0.24 \, kg = 0.48 \, kg$.
Converting to grams,$m = 0.48 \times 1000 \, g = 480 \, g$.

(B) When the boat is floating with iron pieces,the weight of the water displaced by the boat is equal to the total weight of the boat plus the iron pieces $(W_{boat} + W_{iron})$.
According to Archimedes' principle,the buoyant force equals the weight of the displaced water.
When the iron pieces are thrown into the water,they sink because the density of iron is greater than the density of water.
In this case,the iron pieces displace a volume of water equal to their own volume.
Since the density of iron is greater than water,the weight of the water displaced by the iron pieces when submerged is less than the weight of the water displaced when they were floating in the boat.
Therefore,the total volume of water displaced decreases,causing the water level in the pond to decrease.

(A) According to Archimedes' principle,the weight of the floating ice cube is equal to the weight of the water displaced by it.
Let $M$ be the mass of the ice and $V_{air}$ be the volume of the air bubble.
The total volume of the ice cube is $V_{ice} + V_{air}$.
The weight of the ice cube is $W = Mg$.
Since it is floating,the weight of the displaced water is equal to the weight of the ice cube: $W_{displaced} = Mg$.
The volume of the displaced water is $V_{disp} = \frac{Mg}{\rho_w g} = \frac{M}{\rho_w}$,where $\rho_w$ is the density of water.
When the ice melts,it turns into water of mass $M$,occupying a volume $V_{melted} = \frac{M}{\rho_w}$.
However,the air bubble was previously displacing water equal to its own volume $V_{air}$ to keep the cube afloat.
When the ice melts,the air bubble is released,and the water level drops because the volume previously occupied by the air bubble $(V_{air})$ is no longer displacing water.
Therefore,the water level will fall.

(A) For a floating object,the buoyant force must balance the weight of the object.
Let $V$ be the total volume of the ice block and $V_{imm}$ be the volume of ice immersed in water.
Let $\rho_{ice} = 0.9 \, g/cm^3$ be the density of ice and $\rho_{water} = 1.0 \, g/cm^3$ be the density of water.
The weight of the ice block is $W = V \cdot \rho_{ice} \cdot g$.
The buoyant force is $F_B = V_{imm} \cdot \rho_{water} \cdot g$.
Equating the two: $V \cdot \rho_{ice} \cdot g = V_{imm} \cdot \rho_{water} \cdot g$.
$\frac{V_{imm}}{V} = \frac{\rho_{ice}}{\rho_{water}} = \frac{0.9}{1.0} = 0.9$.
This means $90\%$ of the ice is submerged.
The fraction of the volume floating outside the water is $1 - 0.9 = 0.1$.
Converting to percentage: $0.1 \times 100\% = 10\%$.

(A) Let $V$ be the volume of the block,$\rho$ be the density of the block,and $\rho_w$ be the density of water.
In air,the weight of the block is $W_{air} = V \rho g = 60 \, N$.
When submerged in water,the apparent weight is $W_{water} = V \rho g - V \rho_w g = 40 \, N$.
Substituting $V \rho g = 60$,we get $60 - V \rho_w g = 40$,which implies the buoyant force $F_B = V \rho_w g = 20 \, N$.
Now,the specific gravity of the block is defined as $SG = \frac{\rho}{\rho_w} = \frac{V \rho g}{V \rho_w g}$.
Substituting the values,$SG = \frac{60}{20} = 3$.
Therefore,the specific gravity of the block is $3$.

(B) Let the weight of the body in air be $W_a = V \rho_b g$,where $V$ is the volume of the body and $\rho_b$ is its density.
The weight of the body in water is given as $W_w = \frac{1}{3} W_a$.
According to Archimedes' principle,the buoyant force $F_B$ is equal to the weight of the displaced water: $F_B = W_a - W_w = W_a - \frac{1}{3} W_a = \frac{2}{3} W_a$.
Also,$F_B = V \rho_w g$,where $\rho_w$ is the density of water $(1 \ g/cm^3)$.
Equating the two expressions for $F_B$:
$V \rho_w g = \frac{2}{3} (V \rho_b g)$
$\rho_w = \frac{2}{3} \rho_b$
$\rho_b = \frac{3}{2} \rho_w = 1.5 \times 1 \ g/cm^3 = 1.5 \ g/cm^3$.
Thus,the density of the body is $1.5 \ g/cm^3$.

(C) Let $L$ be the side of the cube in $cm$. The density of water is $\rho_w = 1 \,g/cm^3$.
In the first case,the cube with the $200 \,g$ mass is just floating,meaning the total weight is balanced by the buoyant force on the entire volume of the cube:
$(M_{cube} + 200)g = (L^3 \times \rho_w)g$
In the second case,the cube alone is floating,meaning its weight is balanced by the buoyant force on the submerged volume:
$M_{cube}g = (L^2 \times (L - 2) \times \rho_w)g$
Subtracting the second equation from the first:
$(M_{cube} + 200)g - M_{cube}g = (L^3 \times \rho_w)g - (L^2(L - 2) \times \rho_w)g$
$200 = L^3 - L^2(L - 2)$
$200 = L^3 - L^3 + 2L^2$
$200 = 2L^2$
$L^2 = 100$
$L = 10 \,cm$.

(A) The apparent weight of an object submerged in a fluid is given by the formula: $W_{app} = W_{actual} - F_B$,where $F_B$ is the buoyant force.
$W_{app} = V_b \rho_s g - V_b \rho_w g = V_b g (\rho_s - \rho_w)$.
Given:
Volume of the block $V_b = 5 \times 5 \times 5 \, cm^3 = 125 \, cm^3$.
Relative density of steel $\text{RD}_s = \frac{\rho_s}{\rho_w} = 7$,so $\rho_s = 7 \, g/cm^3$ and $\rho_w = 1 \, g/cm^3$.
Substituting the values:
$W_{app} = (5 \times 5 \times 5) \times (7 - 1) \, g wt$
$W_{app} = 6 \times 5 \times 5 \times 5 \, g wt$.
Thus,the correct option is $A$.

(B) Let $V_b$ be the volume of the block and $\rho_b$ be its density.
When the block floats in water,the weight of the block equals the weight of the displaced water:
$V_b \cdot \rho_b \cdot g = (\frac{4}{5} V_b) \cdot \rho_w \cdot g$
Given $\rho_w = 1000 \, kg/m^3$,we have:
$\rho_b = \frac{4}{5} \times 1000 = 800 \, kg/m^3$.
When the block 'just floats' in another liquid of density $\rho_l$,it means the entire volume of the block is submerged:
$V_b \cdot \rho_b \cdot g = V_b \cdot \rho_l \cdot g$
Therefore,$\rho_l = \rho_b = 800 \, kg/m^3$.

(A) Let $V$ be the total volume of the cubical block and $\rho_b$ be its density. Let $\rho_l$ be the density of the liquid.
Initially,the block is in equilibrium,so the weight of the block equals the buoyant force:
$V \rho_b g = V_{immersed} \rho_l g$
Given $V_{immersed} = V / 4$,we have:
$V \rho_b g = (V / 4) \rho_l g \implies \rho_b = \rho_l / 4$
When the system accelerates upward with acceleration $a = g / 4$,the effective gravity becomes $g' = g + a = g + g / 4 = 5g / 4$.
The new buoyant force $F_B'$ must balance the effective weight $W'$ of the block:
$W' = V \rho_b g' = V (\rho_l / 4) (5g / 4)$
$F_B' = V_{immersed}' \rho_l g' = V_{immersed}' \rho_l (5g / 4)$
Equating $W' = F_B'$:
$V (\rho_l / 4) (5g / 4) = V_{immersed}' \rho_l (5g / 4)$
$V / 4 = V_{immersed}'$
Thus,the fraction of volume immersed remains $1 / 4$.

(C) Let $V$ be the volume of the body. When the body is dropped from height $h$,its velocity $v$ just before entering the water is given by $v = \sqrt{2gh}$.
As the body sinks to a maximum depth $h'$,its final velocity becomes zero. We can apply the work-energy theorem: the work done by gravity plus the work done by the buoyant force equals the change in kinetic energy.
Work done by gravity = $(mg)h' = (V \rho g)h'$
Work done by buoyant force = $-(F_B)h' = -(V \sigma g)h'$
Change in kinetic energy = $K_f - K_i = 0 - \frac{1}{2}mv^2 = -\frac{1}{2}(V \rho)(2gh) = -V \rho gh$
Equating the work done to the change in kinetic energy:
$(V \rho g)h' - (V \sigma g)h' = -V \rho gh$
$V g h' (\rho - \sigma) = -V \rho gh$
$h' (\sigma - \rho) = h \rho$
$h' = \frac{h \rho}{\sigma - \rho}$

(C) The correct answer is $C$.
When the stones are in the boat,the boat floats by displacing a volume of water whose weight is equal to the total weight of the boat and the stones (Archimedes' Principle).
Let $M_b$ be the mass of the boat and $M_s$ be the mass of the stones. The total weight is $(M_b + M_s)g$. The volume of water displaced by the boat is $V_1 = \frac{(M_b + M_s)}{\rho_w}$,where $\rho_w$ is the density of water.
When the stones are thrown into the water,they sink to the bottom. The boat now only displaces water equal to its own weight,$V_2 = \frac{M_b}{\rho_w}$. The stones,being submerged,displace a volume of water equal to their own volume,$V_s = \frac{M_s}{\rho_s}$,where $\rho_s$ is the density of the stones.
Since the stones are denser than water $(\rho_s > \rho_w)$,the volume of water displaced by the stones when submerged is less than the volume of water displaced by them when they were floating in the boat $(V_s < \frac{M_s}{\rho_w})$.
Therefore,the total volume of water displaced decreases,causing the water level in the tank to fall.

(D) The forces acting on the sphere are the gravitational force $(Mg)$ acting downwards,the buoyant force $(F_B)$ acting upwards,and the tension $(T)$ in the string acting upwards.
For equilibrium,$T + F_B = Mg$,so $T = Mg - F_B$.
The mass of the sphere is $M = 3 \,kg$. The relative density of the metal is $\rho_{rel, M} = 10$,so the density of the metal is $\rho_M = 10 \times 1000 \,kg/m^3$.
The volume of the sphere is $V = \frac{M}{\rho_M} = \frac{3}{10 \times 1000} \,m^3 = 3 \times 10^{-4} \,m^3$.
The density of the liquid is $\rho_L = 0.8 \times 1000 \,kg/m^3 = 800 \,kg/m^3$.
The buoyant force is $F_B = \rho_L V g = 800 \times (3 \times 10^{-4}) \times 10 = 2.4 \,N$.
The weight of the sphere in air is $Mg = 3 \times 10 = 30 \,N$.
Therefore,the tension in the string is $T = 30 - 2.4 = 27.6 \,N$.

(C) According to the Law of Floatation,for a floating body,the weight of the body is equal to the weight of the displaced liquid.
Since the mass of the block remains constant,the weight of the block remains constant.
Therefore,the weight of the displaced water must also remain constant.
Since the weight of the displaced water is equal to $V_{displaced} \times \rho_{water} \times g$,and both $\rho_{water}$ and $g$ are constant,the volume of displaced water $(V_{displaced})$ must remain the same.
Because the volume of the displaced water does not change,the level of the water in the container will remain the same.

(A) According to the law of floatation,for a body floating in a liquid,the weight of the body is equal to the weight of the liquid displaced.
Let $V$ be the volume of each cubical block and $\rho_w$ be the density of water.
For the $1$st block: Weight of block = Weight of displaced water $\implies V \cdot \rho_1 \cdot g = (V/2) \cdot \rho_w \cdot g \implies \rho_1 = \rho_w / 2$.
For the $2$nd block: Weight of block = Weight of displaced water $\implies V \cdot \rho_2 \cdot g = (3V/4) \cdot \rho_w \cdot g \implies \rho_2 = 3\rho_w / 4$.
The ratio of densities of the blocks is $\rho_1 / \rho_2 = (\rho_w / 2) / (3\rho_w / 4) = (1/2) \times (4/3) = 2/3$.

(D) The apparent weight of the object in water is given by $W_{\text{app}} = W_{\text{air}} - F_B$,where $F_B$ is the buoyant force.
Given $W_{\text{air}} = 10 \,g$ and $W_{\text{app}} = 9 \,g$,the buoyant force is $F_B = 10 - 9 = 1 \,g \text{ (force equivalent)} = 1 \,g \times g$.
According to Archimedes' principle,the buoyant force is equal to the weight of the water displaced: $F_B = V_{\text{total}} \times \rho_w \times g$.
Since $\rho_w = 1 \,g/cm^3$,we have $V_{\text{total}} = 1 \,cm^3$.
The total volume $V_{\text{total}}$ is the sum of the volume of the gold $V_g$ and the volume of the cavity $V_c$: $V_{\text{total}} = V_g + V_c$.
The volume of the gold is $V_g = \frac{\text{mass}}{\rho_g} = \frac{10}{19.3} \approx 0.518 \,cm^3$.
Substituting these values: $1 = 0.518 + V_c$.
Therefore,$V_c = 1 - 0.518 = 0.482 \,cm^3$.

(B) According to Archimedes' principle,the weight of the displaced oil is equal to the weight of the floating ice block.
Let $V_{ice}$ be the volume of the ice and $\rho_{ice}$ be its density. The weight of the ice is $W = V_{ice} \rho_{ice} g$.
The weight of the displaced oil is $W_{oil} = V_{displaced} \rho_{oil} g$.
Since the ice is floating,$W = W_{oil}$,so $V_{displaced} = V_{ice} (\rho_{ice} / \rho_{oil})$.
When the ice melts,it turns into water with volume $V_{water} = V_{ice} (\rho_{ice} / \rho_{water})$.
Since the density of oil $\rho_{oil}$ is generally less than the density of water $\rho_{water}$,the volume of the melted water $V_{water}$ is less than the volume of the displaced oil $V_{displaced}$.
Therefore,the level of the oil in the vessel will go down.

(A) Let the density of the object be $\sigma$ and the density of the liquid be $\rho$. Let the volume of the object be $V$.
The elongation $\Delta L$ in the wire is proportional to the tension $T$ in the wire,i.e.,$\Delta L \propto T$.
Case $1$: When the object is in air,the tension $T_1 = Mg = V\sigma g$. The elongation is $\Delta L_1 = 10 \,mm$.
Case $2$: When the object is submerged in the liquid,the tension $T_2 = Mg - F_B = V\sigma g - V\rho g$,where $F_B$ is the buoyant force. The elongation is $\Delta L_2 = 10 - \frac{10}{3} = \frac{20}{3} \,mm$.
Taking the ratio of the elongations:
$\frac{\Delta L_1}{\Delta L_2} = \frac{T_1}{T_2} = \frac{V\sigma g}{V\sigma g - V\rho g} = \frac{\sigma}{\sigma - \rho}$.
Substituting the values:
$\frac{10}{20/3} = \frac{\sigma}{\sigma - \rho} \Rightarrow \frac{3}{2} = \frac{\sigma}{\sigma - \rho}$.
Cross-multiplying:
$3(\sigma - \rho) = 2\sigma \Rightarrow 3\sigma - 3\rho = 2\sigma \Rightarrow \sigma = 3\rho$.
Therefore,the ratio of the density of the object to the density of the liquid is $\frac{\sigma}{\rho} = 3: 1$.

(A) The weight of the lead lump in air is $W_1 = mg = 200 \,gF$.
The volume of the lead lump is $V = \frac{m}{\rho_{lead}}$,where $\rho_{lead}$ is the density of lead.
When the lead is immersed with half of its volume in brine,the buoyant force $F_B$ acting on it is equal to the weight of the displaced brine:
$F_B = \text{Volume immersed} \times \rho_{brine} \times g = \frac{V}{2} \times \rho_{brine} \times g$.
The new reading of the spring balance $W_2$ is the apparent weight:
$W_2 = W_1 - F_B = mg - \frac{V}{2} \rho_{brine} g$.
Substituting $V = \frac{m}{\rho_{lead}}$:
$W_2 = mg - \frac{m}{2 \rho_{lead}} \rho_{brine} g = mg \left(1 - \frac{\rho_{brine}}{2 \rho_{lead}}\right)$.
Given the specific gravity $\sigma = \frac{\rho_{substance}}{\rho_{water}}$,we have $\frac{\rho_{brine}}{\rho_{lead}} = \frac{1.1}{11.4}$.
$W_2 = 200 \times \left(1 - \frac{1.1}{2 \times 11.4}\right) = 200 \times \left(1 - \frac{1.1}{22.8}\right) = 200 \times \left(1 - 0.048245\right) = 200 \times 0.951755 = 190.351 \,gF$.
Rounding to one decimal place,the reading is approximately $190.4 \,gF$.

(B) For a floating object,the weight of the displaced liquid equals the weight of the object: $V_{sub} \rho_{liq} g = V_{cube} \rho_{cube} g$,where $V_{sub}$ is the submerged volume of the cube.
This implies $V_{sub} \rho_{liq} = V_{cube} \rho_{cube}$.
Let the temperature increase by $\Delta T$. The new volumes and densities are:
$V'_{cube} = V_{cube}(1 + 3\alpha \Delta T)$
$V'_{liq} = V_{liq}(1 + \gamma \Delta T) \implies \rho'_{liq} = \frac{\rho_{liq}}{1 + \gamma \Delta T} \approx \rho_{liq}(1 - \gamma \Delta T)$
For the cube to sink,the fraction of the submerged volume must increase,meaning the density of the liquid must decrease more rapidly than the density of the cube,or effectively,the liquid must expand more than the cube.
The condition for sinking is that the liquid expands more than the cube,i.e.,the volume expansion coefficient of the liquid must be greater than the volume expansion coefficient of the solid cube.
The volume expansion coefficient of the cube is $3\alpha$.
Therefore,the cube will sink if $\gamma > 3\alpha$.

(B) According to Archimedes' principle,the weight of the displaced liquid is equal to the weight of the floating ice block.
Let $M$ be the mass of the ice block. The weight of the ice is $W = Mg$.
The volume of the liquid displaced by the ice is $V_{disp} = \frac{M}{\rho_L}$,where $\rho_L$ is the density of the liquid.
When the ice melts,it turns into water of mass $M$. The volume of this water is $V_{water} = \frac{M}{\rho_W}$,where $\rho_W$ is the density of water.
Since the density of the liquid is less than the density of water $(\rho_L < \rho_W)$,it follows that $\frac{M}{\rho_L} > \frac{M}{\rho_W}$.
Therefore,$V_{disp} > V_{water}$.
Since the volume of the liquid displaced by the ice is greater than the volume of the water formed after melting,the liquid level will go down.

(D) Let $V_w$ be the volume of ice immersed in water and $V_k$ be the volume of ice immersed in kerosene oil.
According to the principle of floatation, the weight of the ice cube is equal to the total buoyant force exerted by both liquids.
Weight of ice $= (V_w + V_k) \rho_{ice} g$
Buoyant force $= V_w \rho_w g + V_k \rho_k g$
Equating the two: $(V_w + V_k) \rho_{ice} g = V_w \rho_w g + V_k \rho_k g$
Dividing by $\rho_w g$ (where $\rho_w = 1 \text{ g/cm}^3$):
$(V_w + V_k) \times 0.9 = V_w \times 1 + V_k \times 0.8$
$0.9 V_w + 0.9 V_k = V_w + 0.8 V_k$
$0.9 V_k - 0.8 V_k = V_w - 0.9 V_w$
$0.1 V_k = 0.1 V_w$
Therefore, $V_w / V_k = 1 / 1$ or $1: 1$.

(A) The weight of the sphere is given by $W = V_{material} \cdot \rho_{sphere} \cdot g = \frac{4}{3} \pi \left( \frac{D^3 - d^3}{8} \right) \sigma \rho_w g$.
The buoyant force is equal to the weight of the water displaced by the total volume of the sphere: $F_b = V_{total} \cdot \rho_w \cdot g = \frac{4}{3} \pi \left( \frac{D^3}{8} \right) \rho_w g$.
For the sphere to just float,the weight must equal the buoyant force: $W = F_b$.
Substituting the expressions: $\frac{4}{3} \pi \left( \frac{D^3 - d^3}{8} \right) \sigma \rho_w g = \frac{4}{3} \pi \left( \frac{D^3}{8} \right) \rho_w g$.
Simplifying the equation: $(D^3 - d^3) \sigma = D^3$.
Dividing by $D^3 \sigma$: $1 - \frac{d^3}{D^3} = \frac{1}{\sigma}$.
Rearranging for the ratio: $\frac{d^3}{D^3} = 1 - \frac{1}{\sigma} = \frac{\sigma - 1}{\sigma}$.
Taking the reciprocal and cube root: $\frac{D}{d} = \left( \frac{\sigma}{\sigma - 1} \right)^{\frac{1}{3}}$.

(B) At equilibrium,the buoyant force equals the weight of the balloon: $Mg = V \rho(h) g$,which simplifies to $M = V \rho(h)$.
For the initial state at $h_1 = 100 m$: $480 = V \rho_0 e^{-\frac{100}{6000}}$.
For the final state at $h_2 = 150 m$ after removing $N$ sandbags: $(480 - N) = V \rho_0 e^{-\frac{150}{6000}}$.
Dividing the two equations: $\frac{480 - N}{480} = \frac{e^{-\frac{150}{6000}}}{e^{-\frac{100}{6000}}} = e^{-\frac{50}{6000}}$.
Using the approximation $e^{-x} \approx 1 - x$ for small $x$: $1 - \frac{N}{480} \approx 1 - \frac{50}{6000}$.
Therefore,$\frac{N}{480} = \frac{50}{6000} = \frac{1}{120}$.
$N = \frac{480}{120} = 4$.

(A) Let $V_0$ be the outer volume of the shell and $V_i$ be the inner volume of the shell.
Let $V$ be the volume of water inside the shell.
Let $\rho_c$ be the relative density of the shell material.
For the shell to float in a half-submerged state,the total weight of the shell and the water inside must equal the buoyant force exerted by the displaced water.
The weight of the shell is $W_s = \rho_c (V_0 - V_i) g$.
The weight of the water inside is $W_w = V g$.
The buoyant force is $F_B = \frac{V_0}{2} g$ (since it is half-submerged).
Equating weight and buoyant force: $\rho_c (V_0 - V_i) g + V g = \frac{V_0}{2} g$.
Simplifying,we get: $\rho_c (V_0 - V_i) = \frac{V_0}{2} - V$.
Thus,$\rho_c = \frac{V_0/2 - V}{V_0 - V_i}$.
If $\rho_c < 0.5$,then $\frac{V_0/2 - V}{V_0 - V_i} < \frac{1}{2}$.
Multiplying both sides by $2(V_0 - V_i)$ (assuming $V_0 > V_i$): $V_0 - 2V < V_0 - V_i$.
This simplifies to $-2V < -V_i$,or $V > V_i / 2$.
Since $V$ is the volume of water and $V_i$ is the total inner volume,$V > V_i / 2$ means the shell is more than half filled.

(C) Let $V = \frac{4}{3} \pi R^3$ be the volume of each sphere.
For the upper sphere (density $\rho$): The forces acting are gravity ($mg = V\rho g$ downwards),spring force ($kx$ upwards,as it is being pulled down by the lower sphere),and buoyant force ($F_{B1} = V(2\rho)g$ upwards). At equilibrium: $V\rho g + kx = V(2\rho)g \implies kx = V\rho g = \frac{4}{3} \pi R^3 \rho g$. Thus,$x = \frac{4 \pi R^3 \rho g}{3 k}$. This confirms statement $(A)$ is correct.
For the lower sphere (density $3\rho$): The forces are gravity ($mg = V(3\rho)g$ downwards),spring force ($kx$ downwards),and buoyant force ($F_{B2} = V(2\rho)g$ upwards). At equilibrium: $V(3\rho)g = V(2\rho)g + kx \implies kx = V\rho g = \frac{4}{3} \pi R^3 \rho g$. This confirms the elongation is indeed $\frac{4 \pi R^3 \rho g}{3 k}$.
Since the buoyant force on the upper sphere $(2V\rho g)$ is greater than its weight $(V\rho g)$,it must be fully submerged to maintain equilibrium,as the spring provides the necessary downward force to balance the net upward buoyant force. Thus,the light sphere is completely submerged. Statement $(D)$ is correct.

(C) $(1)$ For the test-tube to sink,its average density must equal the density of water. The volume of the glass is $V_{\text{glass}} = \frac{\text{mass}}{\text{density}} = \frac{5 \text{ g}}{2.5 \text{ g/cc}} = 2 \text{ cc}$.
Let $V_{\text{gas}}$ be the volume of the trapped air when it starts to sink. The total volume of the test-tube system is $V_{\text{total}} = V_{\text{glass}} + V_{\text{gas}} = 2 + V_{\text{gas}}$.
For the system to sink,the buoyant force must equal the weight: $\rho_{\text{water}} V_{\text{total}} g = m_{\text{total}} g$.
$1 \times (2 + V_{\text{gas}}) = 5 \implies V_{\text{gas}} = 3 \text{ cc}$.
The change in volume is $\Delta V = V_0 - V_{\text{gas}} = 3.3 - 3 = 0.3 \text{ cc}$.
Since $\Delta V = X \text{ cc}$,$X = 0.3$. Note: The options provided seem to imply $X$ in units of $0.1 \text{ cc}$ or a typo in the question's expected format. Based on standard physics problems of this type,$X=0.3$ is the correct physical value.
$(2)$ Using the isothermal process for the trapped air: $P_1 V_1 = P_2 V_2$.
$10^5 \times 3.3 = P_2 \times 3$.
$P_2 = 1.1 \times 10^5 \text{ Pa}$.
$\Delta P = P_2 - P_1 = 1.1 \times 10^5 - 10^5 = 0.1 \times 10^5 = 10 \times 10^3 \text{ Pa}$.
Given $\Delta P = Y \times 10^3 \text{ Pa}$,we get $Y = 10$.

(D) The mass of the cube is $M = 400 \ g = 0.4 \ kg$. The edge length is $L = 10 \ cm = 0.1 \ m$. The total volume of the cube is $V_{total} = L^3 = (0.1 \ m)^3 = 0.001 \ m^3 = 1000 \ cm^3$.
For a floating object,the weight of the object equals the buoyant force: $Mg = \rho_{water} V_{displaced} g$.
$0.4 = 1000 \times V_{displaced}$.
$V_{displaced} = 0.4 / 1000 = 0.0004 \ m^3 = 400 \ cm^3$.
The volume outside the water is $V_{outside} = V_{total} - V_{displaced} = 1000 \ cm^3 - 400 \ cm^3 = 600 \ cm^3$.

(A) Given: Side of the cube $a = 10 \ cm = 0.1 \ m$. Volume of the cube $V = a^3 = (0.1)^3 = 10^{-3} \ m^3$.
Let the mass of the cube be $m$ and its density be $\rho$. Thus,$m = \rho V = \rho \times 10^{-3} \ kg$.
The distance of the unknown mass from the wedge is $2 \ cm = 0.02 \ m$,and the distance of the $200 \ g$ $(0.2 \ kg)$ mass from the wedge is $25 \ cm = 0.25 \ m$.
When half the volume of the cube is submerged,the buoyant force $F_B$ acting on it is given by $F_B = \rho_{water} \times V_{submerged} \times g = 1000 \ kg/m^3 \times (0.5 \times 10^{-3} \ m^3) \times 10 \ m/s^2 = 5 \ N$.
For the rod to be in rotational equilibrium about the wedge point $O$,the net torque must be zero:
$\tau_{net} = (mg - F_B) \times 0.02 \ m - (0.2 \ kg \times 10 \ m/s^2) \times 0.25 \ m = 0$
$(m \times 10 - 5) \times 0.02 = 2 \times 0.25$
$(10m - 5) \times 0.02 = 0.5$
$10m - 5 = 0.5 / 0.02 = 25$
$10m = 30$
$m = 3 \ kg$.