
An aluminium sphere is dipped into water. Which of the following is true?
A. Buoyancy will be less in water at ${0^o}C$ than that in water at ${4^o}C$.
B. Buoyancy will be more in water at ${0^o}C$ than that in water at ${4^o}C$.
C. Buoyancy in water at ${0^o}C$ will be the same as that in water at ${4^o}C$.
D. Buoyancy may be more or less in water at ${4^o}C$ depending upon the radius of the sphere.
Answer
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Hint:In order to solve this question, we should know about Buoyancy. Buoyancy is the upward force exerted on all the objects by the liquid in which they are dipped and here we will use the general formula of buoyancy force and also use the concept that density of water at higher temperature is more than the density of water at lower temperature.
Complete step by step answer:
According to the question, we have given that a spherical aluminium object is dipped in water so, let us suppose the volume of sphere is denoted by V. and acceleration due to gravity is g, At ${0^o}C$ let, density of water is denoted by ${\rho _o}$ so according to general formula of buoyant force we know that,
${F_B} = V{\rho _o}g$
where ${F_B}$ denotes the buoyant force.
Now, considering second case when temperature of water is at ${4^o}C$ and let density of water at this temperature be denoted by ${\rho _4}$ and since volume of sphere will still be V and acceleration due to gravity will also be g, so new buoyant force can be written as
$F{'_B} = V{\rho _4}g$
Now, divide the both equations of buoyant force we get,
$\dfrac{{{F_B}}}{{F{'_B}}} = \dfrac{{Vg{\rho _o}}}{{Vg{\rho _4}}}$
$\Rightarrow \dfrac{{{F_B}}}{{F{'_B}}} = \dfrac{{{\rho _o}}}{{{\rho _4}}}$
Now as we know that, density of water at higher temperature is more than the density of water at lower temperature so,
${\rho _4} > {\rho _0}$ so we get,
$\Rightarrow \dfrac{{{F_B}}}{{F{'_B}}} = \dfrac{{{\rho _o}}}{{{\rho _4}}} < 1$
$\therefore {F_B} < F{'_B}$
So, Buoyancy will be less in water at ${0^o}C$ than that in water at ${4^o}C$.
Hence, the correct option is A.
Note: It should be remembered that, Density of water is also maximum at a temperature of ${4^o}C$ and it’s because as temperature increases the molecules of water came close to each other and they turns slow down and hence in result the water molecules start to form clusters which results in increase in density of water.
Complete step by step answer:
According to the question, we have given that a spherical aluminium object is dipped in water so, let us suppose the volume of sphere is denoted by V. and acceleration due to gravity is g, At ${0^o}C$ let, density of water is denoted by ${\rho _o}$ so according to general formula of buoyant force we know that,
${F_B} = V{\rho _o}g$
where ${F_B}$ denotes the buoyant force.
Now, considering second case when temperature of water is at ${4^o}C$ and let density of water at this temperature be denoted by ${\rho _4}$ and since volume of sphere will still be V and acceleration due to gravity will also be g, so new buoyant force can be written as
$F{'_B} = V{\rho _4}g$
Now, divide the both equations of buoyant force we get,
$\dfrac{{{F_B}}}{{F{'_B}}} = \dfrac{{Vg{\rho _o}}}{{Vg{\rho _4}}}$
$\Rightarrow \dfrac{{{F_B}}}{{F{'_B}}} = \dfrac{{{\rho _o}}}{{{\rho _4}}}$
Now as we know that, density of water at higher temperature is more than the density of water at lower temperature so,
${\rho _4} > {\rho _0}$ so we get,
$\Rightarrow \dfrac{{{F_B}}}{{F{'_B}}} = \dfrac{{{\rho _o}}}{{{\rho _4}}} < 1$
$\therefore {F_B} < F{'_B}$
So, Buoyancy will be less in water at ${0^o}C$ than that in water at ${4^o}C$.
Hence, the correct option is A.
Note: It should be remembered that, Density of water is also maximum at a temperature of ${4^o}C$ and it’s because as temperature increases the molecules of water came close to each other and they turns slow down and hence in result the water molecules start to form clusters which results in increase in density of water.
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