Buoyancy and Archimedes' Principle Questions in English

(B) The upward force exerted by a liquid on an object immersed in it is known as buoyancy.
When an object is immersed in a fluid,the fluid exerts an upward force on the object.
This force is called the buoyant force or upthrust.
As the object is pushed deeper into the liquid,the volume of the displaced liquid increases,which leads to an increase in the buoyant force.

(C) Whether an object floats or sinks depends on the relative density of the object compared to the liquid.
If the density of an object is greater than the density of the liquid,the weight of the object (force of gravity) exceeds the buoyant force,causing it to sink.
Conversely,if the density of the object is less than the density of the liquid,the buoyant force acting on the object is greater than the force of gravity,causing it to float on the surface.

(A) The bag of cotton is heavier than the iron bar.
This is because the volume of the cotton bag is much larger than that of the iron bar.
According to Archimedes' principle,an object immersed in a fluid (air) experiences an upward buoyant force equal to the weight of the fluid displaced.
Since the cotton bag occupies more space,it displaces a larger volume of air,resulting in a greater buoyant force acting on it compared to the iron bar.
This buoyant force reduces the apparent weight of the cotton bag on the weighing machine.
Therefore,for the weighing machine to show $100\, kg$ for both,the actual mass of the cotton bag must be greater than that of the iron bar to compensate for the higher buoyant force.

(B) When an object is immersed in a liquid,the liquid exerts an upward force on the object due to the pressure difference between the top and bottom surfaces of the object. This upward force is known as the buoyant force or upthrust.

(B) Two forces act on an object immersed in water.
One is the gravitational force,which pulls the object downwards.
The other is the buoyant force,which pushes the object upwards.
If the upward buoyant force is greater than the downward gravitational force,the object will rise to the surface.
Since the density of plastic is less than the density of water,the buoyant force exerted by the water on the plastic block is greater than the gravitational force acting on it.
Therefore,the block of plastic rises to the surface when released under water.

(B) If the density of an object is more than the density of a liquid,then it sinks in the liquid. On the other hand,if the density of an object is less than the density of a liquid,then it floats on the surface of the liquid.
Here,the density of the substance $= \frac{\text{Mass of the substance}}{\text{Volume of the substance}} = \frac{50}{20} = 2.5\, g\, cm^{-3}$.
The density of the substance $(2.5\, g\, cm^{-3})$ is greater than the density of water $(1\, g\, cm^{-3})$. Therefore,the substance will sink in water.

(N/A) Density of the packet = $\frac{\text{Mass}}{\text{Volume}} = \frac{500\, g}{350\, cm^3} \approx 1.43\, g/cm^3$.
Since the density of the packet $(1.43\, g/cm^3)$ is greater than the density of water $(1\, g/cm^3)$,the packet will sink in water.
According to Archimedes' principle,the volume of water displaced by a fully submerged object is equal to the volume of the object.
Mass of displaced water = $\text{Volume of packet} \times \text{Density of water} = 350\, cm^3 \times 1\, g/cm^3 = 350\, g$.

$(A)$ According to the law of flotation, the weight of the object is equal to the weight of the displaced liquid.
Let $V$ be the total volume of the object and $d$ be the density of the object.
Weight of object = $V \cdot d \cdot g$.
If a fraction $f$ of the volume is outside the liquid, then the volume inside the liquid is $V_{in} = V(1 - f)$.
Weight of displaced liquid = $V(1 - f) \cdot d_{liquid} \cdot g$.
Equating the two: $V \cdot d \cdot g = V(1 - f) \cdot d_{liquid} \cdot g$, which simplifies to $d = (1 - f) \cdot d_{liquid}$, or $d_{liquid} = \frac{d}{1 - f}$.
For $d_1$, $f_1 = \frac{1}{9}$, so $d_1 = \frac{d}{1 - 1/9} = \frac{d}{8/9} = 1.125d$.
For $d_2$, $f_2 = \frac{2}{11}$, so $d_2 = \frac{d}{1 - 2/11} = \frac{d}{9/11} = 1.222d$.
For $d_3$, $f_3 = \frac{3}{7}$, so $d_3 = \frac{d}{1 - 3/7} = \frac{d}{4/7} = 1.75d$.
Comparing the values, we get $1.125d < 1.222d < 1.75d$, which implies $d_1 < d_2 < d_3$.

(A) The weight of the liquid displaced by an object is equal to the loss in weight of the object when immersed in the liquid.
Loss in weight = Weight in air - Weight in water.
Loss in weight = $10 \, N - 8 \, N = 2 \, N$.
According to Archimedes' principle,the buoyant force acting on an object is equal to the weight of the liquid displaced by the object,which is equal to the apparent loss in weight.

(N/A) $(i)$ The cube will experience a greater buoyant force in the saturated salt solution because the buoyant force is directly proportional to the density of the fluid $(F_b = \rho V g)$. Since the density of the salt solution is greater than that of water,the buoyant force in the salt solution is higher.
$(ii)$ The smaller cube ($4\, cm$ side) will experience a lesser buoyant force because the buoyant force is directly proportional to the volume of the displaced liquid. Since the volume of the smaller cube is less than the initial cube,the volume of water displaced is less,resulting in a lower buoyant force.
$(b)$ The volume of the ball $V = \frac{\text{mass}}{\text{density}} = \frac{4\, kg}{4000\, kg\, m^{-3}} = 10^{-3}\, m^3$.
Buoyant force $(F_b)$ = weight of the liquid displaced = $\text{density of water} \times \text{volume of ball} \times g$.
$F_b = 1000\, kg\, m^{-3} \times 10^{-3}\, m^3 \times 10\, m\, s^{-2} = 10\, N$.

(N/A) The buoyant force depends on the following two factors:
$1$. The volume of the object submerged in the fluid.
$2$. The density of the fluid.
$(b)$ According to Archimedes' principle, the buoyant force acting on an object immersed in a fluid is equal to the weight of the fluid displaced by the object. Therefore, $\text{Buoyant force} = \text{Weight of the fluid displaced}$.

(N/A) lactometer is used to determine the purity of milk by measuring its relative density. It works on the principle of Archimedes' principle,which states that a body immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced by it.
$(b)$ The values shown by the milkman are honesty,transparency,social responsibility,and awareness.

(N/A) The person was making use of the $Archimedes'$ principle,which relates the buoyant force on an object to the weight of the fluid displaced by it. This allows for the determination of the density of the object.
$(b)$ The density of the gold is determined using the relation:
$\text{Relative Density} = \frac{\text{Weight of the body in air}}{\text{Loss of weight in water}}$
Since the density of water is $1 \text{ g/cm}^3$,the relative density is numerically equal to the density of the gold in $\text{g/cm}^3$.
$(c)$ The values Kanchan imbibed include:
$(i)$ Awareness regarding the quality of products.
$(ii)$ Scientific temper and critical thinking.
$(iii)$ Seeking evidence-based verification for claims.

(N/A) It is difficult to drown in the Dead Sea because its water has a much higher density than ordinary water due to high salt concentration. Consequently,the buoyant force (upthrust) exerted by the water is greater than the weight of the lady,allowing her to float easily.
$(b)$ Raman exhibits values such as scientific temper,critical thinking,and the ability to educate others by sharing his knowledge.

(A) The mass of the balloon filled with helium is less than the mass of the air displaced by it. Hence,the upthrust acting on the balloon is more than its weight. As a result,the balloon experiences a net upthrust which makes it rise. As the balloon rises,it experiences less and less upthrust due to the fact that the density of air decreases with height. At a certain point,the weight of the balloon is completely balanced by the upthrust acting on it. Thereafter,the balloon stops rising.
$(b)$ Value shown by Avdesh: Respect for the teacher and knowledge of the subject.
Value shown by Gautam: Logical reasoning.

(N/A) An iceberg has a smaller density $(0.917 \text{ g cm}^{-3})$ than sea water at the same temperature. The density of sea water is about $1.018 \text{ g cm}^{-3}$,hence an iceberg floats with nearly $10\%$ of its volume outside the water. This means that only a small part of the iceberg is visible above the water surface,while the massive bulk remains hidden beneath,making it dangerous for ships.
$(b)$ John: Caring nature,awareness,and scientific knowledge.
John's father: Open-mindedness and respect for knowledge.