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

(D) Let $V$ be the volume of the object and $d$ be its density. The apparent weight of the object in a liquid is given by the actual weight minus the buoyant force.
In the first liquid: $m_1 g = V d g - V d_1 g = V g (d - d_1) \implies m_1 = V(d - d_1) \quad (1)$
In the second liquid: $m_2 g = V d g - V d_2 g = V g (d - d_2) \implies m_2 = V(d - d_2) \quad (2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{m_1}{m_2} = \frac{d - d_1}{d - d_2}$
$m_1(d - d_2) = m_2(d - d_1)$
$m_1 d - m_1 d_2 = m_2 d - m_2 d_1$
$d(m_1 - m_2) = m_1 d_2 - m_2 d_1$
$d = \frac{m_1 d_2 - m_2 d_1}{m_1 - m_2}$

(D) Let $V$ be the total volume of the body and $\rho$ be its density.
According to the law of floatation,the weight of the body is equal to the weight of the liquid displaced.
Case $1$: In water (density $\rho_w = 1 \text{ g/cm}^3$):
Volume inside water = $V - 0.4V = 0.6V$.
Weight of body = Weight of water displaced $\implies V \rho g = (0.6V) \rho_w g \implies \rho = 0.6 \rho_w = 0.6 \text{ g/cm}^3$.
Case $2$: In oil (density $\rho_o$):
Volume inside oil = $V - 0.6V = 0.4V$.
Weight of body = Weight of oil displaced $\implies V \rho g = (0.4V) \rho_o g \implies \rho = 0.4 \rho_o$.
Equating the density of the body from both cases: $0.6 \rho_w = 0.4 \rho_o$.
Relative density of oil = $\frac{\rho_o}{\rho_w} = \frac{0.6}{0.4} = 1.5$.

(C) For a body floating in a liquid, the weight of the body equals the weight of the displaced liquid. Let $V$ be the total volume of the body and $d$ be its density.
Case $1$: Floating in water (density $\rho_w = 1 \text{ g/cm}^3$)
Volume immersed $V_{in} = V - 0.4V = 0.6V$.
Weight of body = Weight of displaced water
$V \cdot d \cdot g = (0.6V) \cdot \rho_w \cdot g$
$d = 0.6 \cdot 1 = 0.6 \text{ g/cm}^3$.
Case $2$: Floating in oil (density $\rho_{oil}$)
Volume immersed $V_{in} = V - 0.6V = 0.4V$.
Weight of body = Weight of displaced oil
$V \cdot d \cdot g = (0.4V) \cdot \rho_{oil} \cdot g$
$0.6 = 0.4 \cdot \rho_{oil}$
$\rho_{oil} = \frac{0.6}{0.4} = 1.5$.
Thus, the relative density of the oil is $1.5$.

(D) Let $V$ be the total volume of the body,$d$ be its density,and $\sigma$ be the density of water.
For a floating body,the weight equals the buoyant force: $Vdg = V(1 - \frac{1}{n})\sigma g$.
Thus,$d = (\frac{n-1}{n})\sigma$.
When the body is submerged,the net upward force $F_{net}$ is $F_B - mg = V\sigma g - Vdg = V\sigma g - V(\frac{n-1}{n})\sigma g = V\sigma g (1 - \frac{n-1}{n}) = V\sigma g (\frac{1}{n})$.
The acceleration $a$ is given by $a = \frac{F_{net}}{m} = \frac{V\sigma g / n}{Vd} = \frac{\sigma g / n}{(\frac{n-1}{n})\sigma} = \frac{g}{n-1}$.
Using the kinematic equation $h = \frac{1}{2}at^2$,we have $h = \frac{1}{2} (\frac{g}{n-1}) t^2$.
Solving for $t$,we get $t = \sqrt{\frac{2h(n-1)}{g}}$.
Therefore,$t \propto \sqrt{n-1}$.

(B) Let the density of the body be $\rho_b = 1.2 \times 10^3 \text{ kg/m}^3$, the density of the liquid be $\rho_l = 2.4 \times 10^3 \text{ kg/m}^3$, the height of fall be $h = 1 \text{ m}$, and the maximum depth reached be $d$.
According to the Work-Energy Theorem, the total work done by all forces (gravity and buoyancy) on the body from the point of release to the point of maximum depth is zero (since the change in kinetic energy is zero).
Work done by gravity = $Mg(h + d) = V \rho_b g (h + d)$.
Work done by buoyancy = $-Bd = -V \rho_l g d$.
Equating the work done: $V \rho_b g (h + d) - V \rho_l g d = 0$.
Dividing by $Vg$: $\rho_b (h + d) = \rho_l d$.
Substituting the values: $(1.2 \times 10^3)(1 + d) = (2.4 \times 10^3) d$.
$1.2(1 + d) = 2.4d$.
$1 + d = 2d$.
$d = 1 \text{ m}$.

(D) For a floating object,the weight of the object is equal to the buoyant force exerted by the liquid.
$Mg = F_{b}$
Let $A$ be the area of the base of the cubical block and $h$ be the height of the submerged part.
The mass of the block is $M = \rho_{b} \times A \times H$.
The buoyant force is $F_{b} = \rho_{l} \times A \times h \times g$.
Equating the two: $\rho_{b} \times A \times H \times g = \rho_{l} \times A \times h \times g$.
Simplifying,we get: $\rho_{b} \times H = \rho_{l} \times h$.
Substituting the given values: $600 \times 8.0 = 900 \times h$.
$h = \frac{600 \times 8.0}{900} = \frac{2}{3} \times 8.0 = \frac{16}{3} \approx 5.33 \ cm$.
Rounding to one decimal place,the height of the submerged part is $5.3 \ cm$.

(A) According to Archimedes' principle,when an object floats,the buoyant force is equal to the weight of the object.
When the metal coin is placed on the wooden cube,the system (cube + coin) remains in equilibrium.
The additional buoyant force provided by the extra submerged volume of the cube must balance the weight of the metal coin.
The additional volume of water displaced is $V_{sub} = \text{Area} \times \Delta h = (10 \ \text{cm} \times 10 \ \text{cm}) \times 3.87 \ \text{cm} = 387 \ \text{cm}^3$.
Since the density of water is $\rho_w = 1 \ \text{g/cm}^3$,the mass of the displaced water is $m = \rho_w \times V_{sub} = 1 \ \text{g/cm}^3 \times 387 \ \text{cm}^3 = 387 \ \text{g}$.
Therefore,the mass of the metal coin is $387 \ \text{g}$.