Tag: upthrust in fluids, archimedes' principle and floatation

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Questions Related to upthrust in fluids, archimedes' principle and floatation

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

A metallic wire of length, "l" is lying horizontally on the surface of liquid of density $ '\rho' $ The maximum radius of wire so that it may not sink will be

  1. $ \sqrt { \frac { 2T }{ \pi \rho g } } $
  2. $ \sqrt { \frac { T }{ \pi \rho g } } $
  3. $ \sqrt { \frac { 2T }{ \rho g } } $
  4. $ \sqrt { \frac { T }{ \rho g } } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a wire of length l and radius r to stay afloat on a liquid surface due to surface tension, the maximum downward gravitational force (weight) must equal the upward force due to surface tension acting on both sides: mg = 2*T*l. Substituting mass as volume times density (pi * r^2 * l * rho) and solving for radius r gives sqrt(2T / (pi * rho * g)).

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

A cube of wood supporting a $200$ gm mass just floats in water. When the mass is removed the cube rises $2$ cm at equilibrium. Find size of the cube.

  1. 10cm

  2. 12cm

  3. 15cm

  4. 4cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The weight of the 200g mass equals the weight of the water displaced by the additional 2cm immersion. 200g = (Area * 2cm) * density_water. Area = 100 cm^2. Side length = 10 cm.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

A cubical box of wood of side $30\, cm$ weighing $21.6\, kg$ floats on water with two faces horizontal. The depth of immersion of box is :

  1. $30\, cm$
  2. $12\, cm$
  3. $6\, cm$
  4. $24\, cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a floating body, the weight of the floating object equals the buoyant force, which is the weight of the displaced water. Using the dimensions of the cubical box, the mass divided by the area of the base gives the volume of water displaced per unit area, yielding the depth of immersion. Calculation: (21.6 kg) / (0.3 m * 0.3 m * 1000 kg/m^3) = 0.24 m or 24 cm.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

A wire of length $L$ metrs, made of a material of specific gravity $8$ is floating horizontally on the surface of water. If it is not wet by water, the maximum diameter of the wire (in mm) up to which it can continue to float is (surface tension of water is) ($T=70\times 10^{-3} \ N/m$)

  1. $1.5$
  2. $1.1$
  3. $0.75$
  4. $0.55$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The weight of the wire (pi * r^2 * L * rho_wire * g) is balanced by surface tension (2 * T * L). Solving for diameter d = 2r: pi * (d/2)^2 * rho_wire * g = 2 * T. Plugging in values gives d approximately 1.5 mm.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

A hollow cylinder of copper of length $25\, cm$ and area of cross-section $15\, cm^2$, floats in water with $3/5$ of its length inside water. Then 

  1. Apparent density of hollow copper cylinder is $0.6\, gcm^{-3}$
  2. Weight of the cylinder is $225\, gf$
  3. Extra force required to completely submerge it in water is $150\, gf$
  4. Extra force required to completely submerge it in water is $225\, gf$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Weight of the cylinder = buoyant force = (3/5 * 25 cm * 15 cm^2) * 1 g/cm^3 = 225 gf. This confirms option B.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

Two solids $A$ and $B$ float in water. It is observed that $A$ floats with half its volume immersed and $B$ floats with $\dfrac{2}{3}$ of its volume immersed. Compare the densities of A and B.

  1. $4:3$
  2. $2:3$
  3. $3:4$
  4. $1:3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a floating object, density_object / density_fluid = fraction_submerged. Density_A / Density_water = 1/2. Density_B / Density_water = 2/3. Ratio A:B = (1/2) / (2/3) = 3/4.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

The weight of the liquid displacement by a body when the body is immersed in it is called 

  1. Apparent weight

  2. Upthrust

  3. Lateral pressure

  4. Relative density of body
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

According to Archimedes principle, when a body is immersed in a liquid, the upward buoyant force acting on it is equal to the weight of the liquid displaced by the body. This upward force is termed upthrust.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

Ice ______ in water, because the weight of water displaced by the immersed part of the ice is _____ its own weight 

  1. sinks, more than

  2. sinks, less than

  3. floats , equal to

  4. floats , less than
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to Archimedes principle, A body immersed in water experiences an upward force equal to the mass of the fluid displaced by the body. If the weight of an object is greater than the weight of displaced fluid, it will float. If the two are equal, it is suspended, neither floating nor sinking. For example, when an object is placed in water, it will displace its own volume of water, and that water will push back against it proportionally, producing an upthrust.
Water has a weight density of $62$ pounds per cubic foot. It an object weighing $62$ pounds has a volume that displaces $2$ cubic feet of water, it will float. The displaced water will weigh $124$ pounds and the pressure of that water would be enough to keep the object floating.
Hence, Ice floats in water, because the weight of water is displaced by the immersed part of the ice is more than its own weight and the statement is true.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

An ice-berg floating partly immersed in sea water of density $1.03 g/cm^3$. The density of ice is $0.92 g/cm^3$. The fraction of the total volume of the iceberg above the level of sea water is

  1. $8.1\%$
  2. $11\%$
  3. $34\%$
  4. $0.8\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $v$ be the volume of the ice-berg outside the sea water and $V$ be the total volume of ice-berg. Then as per question
$0.92V = 1.03(V-v)$

or, $\dfrac vV = 1-\dfrac {0.92}{1.03}= \dfrac{11}{103} $
$\therefore  \dfrac vV \times 100 = 11 \times \dfrac {100}{103} \cong  11\%$