Tag: upthrust in fluids, archimedes' principle and floatation

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

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 } } $


Correct Option: B

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


Correct Option: A

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. $24\, cm$


Correct Option: C

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. $0.55$


Correct Option: A

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 $225\, gf$


Correct Option: B

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. $1:3$


Correct Option: C

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


Correct Option: A

A wooden cube of size $ 1 m\times 1 m\times 1 m$ of relative density 0.5 floats in water with its four faces vertical. The work done by in just submerging the tube by pushing it downward is

  1. 1250 J

  2. 2550 J

  3. 850 J

  4. 1570 J


Correct Option: C

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


Correct Option: C
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.

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. $0.8\%$


Correct Option: B
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} $
$\therefore  \dfrac vV \times 100 = 11 \times \dfrac {100}{103} \cong  11\%$