
A body floats with one- third of its volume of outside water and $\dfrac{3}{4}$of its volume outside another liquid. The density of another liquid is:
$(A)\dfrac{9}{4}\dfrac{g}{{cc}}$
$(B)\dfrac{4}{9}\dfrac{g}{{cc}}$
$(C)\dfrac{1}{3}\dfrac{g}{{cc}}$
$(D)\dfrac{3}{8}\dfrac{g}{{cc}}$
Answer
458.1k+ views
Hint: Buoyancy is the main reason that an object floats in water or in any other fluid. The upward force which is exerted by an object when an object is partly immersed or fully immersed in a fluid is known as the Buoyant Force.
Complete answer:
Let us consider the mass of the body to be $M$ and volume of the body to be $v$
We also know that,
Weight of the body $ = mg$ which is basically bounced by the buoyant force
Now, let the density of the two given liquids be ${\rho _1}$ and ${\rho _2}$respectively.
The first liquid given to us is only $\dfrac{{2v}}{3}$ immersed,
So, $Mg = \dfrac{{2v}}{3}{\rho _1}g.......(1)$
(as the buoyant force exerted is always equal to the weight of the liquid which is displaced)
Similarly, for the second liquid which is given to us,
$Mg = \dfrac{{8v}}{4}{\rho _2}g......(2)$
On comparing equation (1) and equation (2), we get,
$\dfrac{{2v}}{3}{\rho _1}g = \dfrac{{8v}}{4}{\rho _2}g$
$\dfrac{{2v}}{3}{\rho _1} = \dfrac{{8v}}{4}{\rho _2}$
On further solving, we get,
${\rho _1} = 3{\rho _2}$
For water, ${\rho _1} = 1\dfrac{g}{{cc}}$
So, ${\rho _2} = \dfrac{1}{3}{\rho _1}$
${\rho _2} = \dfrac{1}{3}\left( {1\dfrac{g}{{cc}}} \right)$
${\rho _2} = \dfrac{1}{3}\dfrac{g}{{cc}}$
So, the final answer is $(C)\dfrac{1}{3}\dfrac{g}{{cc}}$
Additional information:
The density of water is close to 1 gram per milliliter but the important point to note is that this value changes with the change in temperature. Even if there are substances dissolved in it, then also the value of it changes. Ice is less dense than liquid water and this is the reason why ice cubes float in your glass.
Note:
Density is basically defined as the mass per unit volume. Density is a scalar quantity which is dependent on only the magnitude of the quantity and not on the direction of the body. The unit of density is given as kilogram per cubic metre.
Complete answer:
Let us consider the mass of the body to be $M$ and volume of the body to be $v$
We also know that,
Weight of the body $ = mg$ which is basically bounced by the buoyant force
Now, let the density of the two given liquids be ${\rho _1}$ and ${\rho _2}$respectively.
The first liquid given to us is only $\dfrac{{2v}}{3}$ immersed,
So, $Mg = \dfrac{{2v}}{3}{\rho _1}g.......(1)$
(as the buoyant force exerted is always equal to the weight of the liquid which is displaced)
Similarly, for the second liquid which is given to us,
$Mg = \dfrac{{8v}}{4}{\rho _2}g......(2)$
On comparing equation (1) and equation (2), we get,
$\dfrac{{2v}}{3}{\rho _1}g = \dfrac{{8v}}{4}{\rho _2}g$
$\dfrac{{2v}}{3}{\rho _1} = \dfrac{{8v}}{4}{\rho _2}$
On further solving, we get,
${\rho _1} = 3{\rho _2}$
For water, ${\rho _1} = 1\dfrac{g}{{cc}}$
So, ${\rho _2} = \dfrac{1}{3}{\rho _1}$
${\rho _2} = \dfrac{1}{3}\left( {1\dfrac{g}{{cc}}} \right)$
${\rho _2} = \dfrac{1}{3}\dfrac{g}{{cc}}$
So, the final answer is $(C)\dfrac{1}{3}\dfrac{g}{{cc}}$
Additional information:
The density of water is close to 1 gram per milliliter but the important point to note is that this value changes with the change in temperature. Even if there are substances dissolved in it, then also the value of it changes. Ice is less dense than liquid water and this is the reason why ice cubes float in your glass.
Note:
Density is basically defined as the mass per unit volume. Density is a scalar quantity which is dependent on only the magnitude of the quantity and not on the direction of the body. The unit of density is given as kilogram per cubic metre.
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