An iron cube floats in a vessel containing mercury at $ 20^\circ C $ . If the temperature is increased by $ 100^\circ C $ , then the cube will float.
A) Lower
B) Higher
C) At the same level
D) Lower or higher depending on the mass of the cube
Answer
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Hint : When an object is placed in liquid, it will float or sink depending on the density of the object as compared to the density of the liquid. The dimensions of the iron cube will change when it is heated. The density of the iron cube will change inversely with respect to the volume of the iron cube.
Formula used: In this solution, we will use the following formula
Buoyant force: $ {F_B} = \rho Vg $ where $ \rho $ is the density of the liquid, $ V $ is the volume of the object immersed in the liquid, and $ g $ is the gravitational acceleration.
Complete step by step answer:
We’ve been given that an iron cube floats in a vessel containing mercury at $ 20^\circ C $ and then its temperature is increased by $ 100^\circ C $ .
When the iron cube if partly immersed in mercury, the buoyant force that is acting on it will be
$ {F_B} = \rho Vg $ where $ V $ is the volume of the iron cube immersed in the liquid.
Now, when we increase the temperature of the iron cube, its length increases due to the increase in temperature. As a result, the volume of the iron cube will increase. Since we have a constant mass of the iron cube, from the relation of density with mass and volume for the iron cube
$ {\rho _i} = \dfrac{{{m_i}}}{{{V_i}}} $
We can say that the density will decrease as the volume increases.
Since the density of the iron block will decrease, to produce the same amount of buoyant force $ \left( {{F_B} = \rho Vg} \right) $ , the volume of the iron block immersed in the liquid will have to increase for the cube to not sink in the mercury.
This will be done by immersing a higher volume of the iron cube in the mercury by immersing more part of the mercury that is the cube will float lower in the liquid which corresponds to choice (A).
Note:
Here we have assumed that the temperature of the liquid does not change despite the change in the temperature of the iron cube. While calculating the buoyant force, the density that we use is of the liquid mercury but the volume that we use is of the iron cube that is immersed in water so we must be careful to use the appropriate values.
Formula used: In this solution, we will use the following formula
Buoyant force: $ {F_B} = \rho Vg $ where $ \rho $ is the density of the liquid, $ V $ is the volume of the object immersed in the liquid, and $ g $ is the gravitational acceleration.
Complete step by step answer:
We’ve been given that an iron cube floats in a vessel containing mercury at $ 20^\circ C $ and then its temperature is increased by $ 100^\circ C $ .
When the iron cube if partly immersed in mercury, the buoyant force that is acting on it will be
$ {F_B} = \rho Vg $ where $ V $ is the volume of the iron cube immersed in the liquid.
Now, when we increase the temperature of the iron cube, its length increases due to the increase in temperature. As a result, the volume of the iron cube will increase. Since we have a constant mass of the iron cube, from the relation of density with mass and volume for the iron cube
$ {\rho _i} = \dfrac{{{m_i}}}{{{V_i}}} $
We can say that the density will decrease as the volume increases.
Since the density of the iron block will decrease, to produce the same amount of buoyant force $ \left( {{F_B} = \rho Vg} \right) $ , the volume of the iron block immersed in the liquid will have to increase for the cube to not sink in the mercury.
This will be done by immersing a higher volume of the iron cube in the mercury by immersing more part of the mercury that is the cube will float lower in the liquid which corresponds to choice (A).
Note:
Here we have assumed that the temperature of the liquid does not change despite the change in the temperature of the iron cube. While calculating the buoyant force, the density that we use is of the liquid mercury but the volume that we use is of the iron cube that is immersed in water so we must be careful to use the appropriate values.
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