
A body floats in water with \[40\% \] of its volume outside water. When the same body floats in oil \[60\% \] of its volume remains outside oil. The relative density of oil is:
A. \[0.9\]
B. \[1.0\]
C. \[1.2\]
D. \[1.5\]
Answer
477k+ views
Hint:Some things when we put them into water will float and the others will sink. Water with more salt content makes things float on them easily. There will be a force that acts on an object when the object is surrounded by fluids. Archimedes' Principle deals with these forces. This principle states that when an object is immersed in a fluid there will be an upward buoyant force acting on the body. This buoyant force will be equal to the fluid’s weight that the body displaces and acts in the upward direction at the center of the mass of the displaced fluid.
Complete step by step answer:
Let the volume and the density of the human body are equal to \[V\] and \[d\] respectively. By Archimedes' principle, the weight of the body submerged in the fluid will be equal to the weight of the fluid displaced by the body. Therefore the volume and the density of the displaced fluid are \[{V_1}\]and \[{d_w}\] respectively.
Given that the body floats in water with \[40\% \] of its volume outside water , so the volume of the body outside the water is \[0.4V\]. Therefore the volume of the body inside the water is \[{V_1} = 0.6V\]. Using Archimedes principle,
\[Vdg = {V_1}{d_w}g\]
Substituting for\[{V_1}\],
\[Vdg = 0.6V{d_w}g\] ……….. (1)
Now coming to the body immersed in the oil. Let the volume and the density of the oil is \[{V_2}\] and \[{d_0}\] respectively.
Given that the same body floats in oil \[60\% \] of its volume remains outside oil, so the volume of the oil outside the body is \[0.6V\]. Therefore the volume of the body outside the volume is \[{V_2} = 0.4V\]. Using Archimedes principle,
\[Vdg = {V_2}{d_0}g\]
Substituting for \[{V_2}\] we get,
\[Vdg = 0.4V{d_0}g\] ……… (2)
Therefore from (1) and (2), we get,
\[\dfrac{{{d_0}}}{{{d_w}}} = \dfrac{{0.6}}{{0.4}}\]
\[\therefore \dfrac{{{d_0}}}{{{d_w}}} = 1.5\]
Therefore the relative density is \[1.5\].
Note:There are many applications for this Archimedes principle, one of which is submarines. The reason behind why submarines are consistently submerged is that they have a part called a tank for the load which permits the water to enter causing the submarine to be in its position submerged as the heaviness of the submarine is more noteworthy than the buoyant force.
Complete step by step answer:
Let the volume and the density of the human body are equal to \[V\] and \[d\] respectively. By Archimedes' principle, the weight of the body submerged in the fluid will be equal to the weight of the fluid displaced by the body. Therefore the volume and the density of the displaced fluid are \[{V_1}\]and \[{d_w}\] respectively.
Given that the body floats in water with \[40\% \] of its volume outside water , so the volume of the body outside the water is \[0.4V\]. Therefore the volume of the body inside the water is \[{V_1} = 0.6V\]. Using Archimedes principle,
\[Vdg = {V_1}{d_w}g\]
Substituting for\[{V_1}\],
\[Vdg = 0.6V{d_w}g\] ……….. (1)
Now coming to the body immersed in the oil. Let the volume and the density of the oil is \[{V_2}\] and \[{d_0}\] respectively.
Given that the same body floats in oil \[60\% \] of its volume remains outside oil, so the volume of the oil outside the body is \[0.6V\]. Therefore the volume of the body outside the volume is \[{V_2} = 0.4V\]. Using Archimedes principle,
\[Vdg = {V_2}{d_0}g\]
Substituting for \[{V_2}\] we get,
\[Vdg = 0.4V{d_0}g\] ……… (2)
Therefore from (1) and (2), we get,
\[\dfrac{{{d_0}}}{{{d_w}}} = \dfrac{{0.6}}{{0.4}}\]
\[\therefore \dfrac{{{d_0}}}{{{d_w}}} = 1.5\]
Therefore the relative density is \[1.5\].
Note:There are many applications for this Archimedes principle, one of which is submarines. The reason behind why submarines are consistently submerged is that they have a part called a tank for the load which permits the water to enter causing the submarine to be in its position submerged as the heaviness of the submarine is more noteworthy than the buoyant force.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

