
What is the ratio of the volume of ice immersed in water to that in oil if a cube of ice floats partly in water and partly in oil? It is given that specific gravity of oil is $0.8$ and that of ice is $0.9$.
Answer
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HintCalculate the specific gravity of water from the expression –
$S = \dfrac{{{\rho _O}}}{{{\rho _W}}}$
where, ${\rho _O}$ is the density of an object and ${\rho _W}$ is the density of water.
The ice floats partly in water and partly in oil therefore, volume of ice is equal to the volume of water and volume of oil.
Now, use the buoyancy floatation to find the ratio between volume of oil and volume of water.
Complete step-by-step solution:
Let the specific gravity of water be ${S_w}$, specific gravity of oil be ${S_o}$ and specific gravity of ice be ${S_{ice}}$.
Now, we have to find out the specific gravity of water to find the specific gravity, the formula used is –
$S = \dfrac{{{\rho _O}}}{{{\rho _W}}} \cdots (1)$
Now, we know that, density of water is $1000kg/{m^3}$ so, the density of object is also the density of water –
$
\therefore {S_w} = \dfrac{{1000}}{{1000}} \\
\Rightarrow {S_w} = 1 \\
$
According to the question, it is given that –
${S_{ice}} = 0.9$
${S_o} = 0.8$
Now, let the volume of ice, oil and water be ${V_{ice}},{V_o}$ and ${V_w}$ respectively
Because the ice is partly in water and partly in oil
$\therefore {V_{ice}} = {V_o} + {V_w} \cdots (1)$
Using the buoyancy floatation, we get –
${V_{ice}} \times {S_{ice}} = {V_o} \times {S_o} + {V_w} \times {S_w}$
Using the equation $(1)$
$({V_o} + {V_w}){S_{ice}} = {V_o} \times {S_o} + {V_w} \times {S_w}$
Putting the values in the above equation –
$
0.9{V_o} + 0.9{V_w} = 0.8{V_o} + {V_w} \\
0.1{V_o} = 0.1{V_w} \\
\dfrac{{{V_o}}}{{{V_w}}} = \dfrac{1}{1} \\
$
So, the volume of an ice cube immersed in water is equal to the volume immersed in oil. Hence, their ratio is $1:1$.
Note:- When a body is immersed in fluid, an upward force is exerted by the fluid on the body. This upward force is equal to the weight of fluid displaced by the body and is called the force of buoyancy.
$S = \dfrac{{{\rho _O}}}{{{\rho _W}}}$
where, ${\rho _O}$ is the density of an object and ${\rho _W}$ is the density of water.
The ice floats partly in water and partly in oil therefore, volume of ice is equal to the volume of water and volume of oil.
Now, use the buoyancy floatation to find the ratio between volume of oil and volume of water.
Complete step-by-step solution:
Let the specific gravity of water be ${S_w}$, specific gravity of oil be ${S_o}$ and specific gravity of ice be ${S_{ice}}$.
Now, we have to find out the specific gravity of water to find the specific gravity, the formula used is –
$S = \dfrac{{{\rho _O}}}{{{\rho _W}}} \cdots (1)$
Now, we know that, density of water is $1000kg/{m^3}$ so, the density of object is also the density of water –
$
\therefore {S_w} = \dfrac{{1000}}{{1000}} \\
\Rightarrow {S_w} = 1 \\
$
According to the question, it is given that –
${S_{ice}} = 0.9$
${S_o} = 0.8$
Now, let the volume of ice, oil and water be ${V_{ice}},{V_o}$ and ${V_w}$ respectively
Because the ice is partly in water and partly in oil
$\therefore {V_{ice}} = {V_o} + {V_w} \cdots (1)$
Using the buoyancy floatation, we get –
${V_{ice}} \times {S_{ice}} = {V_o} \times {S_o} + {V_w} \times {S_w}$
Using the equation $(1)$
$({V_o} + {V_w}){S_{ice}} = {V_o} \times {S_o} + {V_w} \times {S_w}$
Putting the values in the above equation –
$
0.9{V_o} + 0.9{V_w} = 0.8{V_o} + {V_w} \\
0.1{V_o} = 0.1{V_w} \\
\dfrac{{{V_o}}}{{{V_w}}} = \dfrac{1}{1} \\
$
So, the volume of an ice cube immersed in water is equal to the volume immersed in oil. Hence, their ratio is $1:1$.
Note:- When a body is immersed in fluid, an upward force is exerted by the fluid on the body. This upward force is equal to the weight of fluid displaced by the body and is called the force of buoyancy.
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