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In the figure shown, the heavy cylinder (radius R) resting on a smooth surface separates two liquids of densities 2r and 3r. The height `h' for the equilibrium of cylinder must be
- a)3R/2
- b)
- c)R√2
- d)None
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
In the figure shown, the heavy cylinder (radius R) resting on a smooth...
First, let’s concentrate on the force exerted by the liquid of density 3ρ on the cylinder in the horizontal direction.
Let the length of the cylinder be L.
Consider a small segment of length rdθ at an angle θ from the horizontal line.
Height of this segment from the topmost point of fluid 3ρ is R sinθ
Hence, the pressure exerted by the fluid will be 3ρgRsinθ
The force exerted in the horizontal direction, dF=3ρgRsinθRLcosθdθ
Similarly, proceeding for the fluid with density 2ρ
Height of any segment, above horizontal =h−R−Rsinθ
below horizontal, h−R+Rsinθ
Thus, horizontal force on the cylinder because of fluid,
For equilibrium, both the forces should be equal, hence solving the above equation,
h = R √3/2
Let the length of the cylinder be L.
Consider a small segment of length rdθ at an angle θ from the horizontal line.
Height of this segment from the topmost point of fluid 3ρ is R sinθ
Hence, the pressure exerted by the fluid will be 3ρgRsinθ
The force exerted in the horizontal direction, dF=3ρgRsinθRLcosθdθ
Similarly, proceeding for the fluid with density 2ρ
Height of any segment, above horizontal =h−R−Rsinθ
below horizontal, h−R+Rsinθ
Thus, horizontal force on the cylinder because of fluid,
For equilibrium, both the forces should be equal, hence solving the above equation,
h = R √3/2
Most Upvoted Answer
In the figure shown, the heavy cylinder (radius R) resting on a smooth...
First, let’s concentrate on the force exerted by the liquid of density 3ρ on the cylinder in the horizontal direction.
Let the length of the cylinder be L.
Consider a small segment of length rdθ at an angle θ from the horizontal line.
Height of this segment from the topmost point of fluid 3ρ is R sinθ
Hence, the pressure exerted by the fluid will be 3ρgRsinθ
The force exerted in the horizontal direction, dF=3ρgRsinθRLcosθdθ
Similarly, proceeding for the fluid with density 2ρ
Height of any segment, above horizontal =h−R−Rsinθ
below horizontal, h−R+Rsinθ
Thus, horizontal force on the cylinder because of fluid,
For equilibrium, both the forces should be equal, hence solving the above equation,
h = R √3/2
Let the length of the cylinder be L.
Consider a small segment of length rdθ at an angle θ from the horizontal line.
Height of this segment from the topmost point of fluid 3ρ is R sinθ
Hence, the pressure exerted by the fluid will be 3ρgRsinθ
The force exerted in the horizontal direction, dF=3ρgRsinθRLcosθdθ
Similarly, proceeding for the fluid with density 2ρ
Height of any segment, above horizontal =h−R−Rsinθ
below horizontal, h−R+Rsinθ
Thus, horizontal force on the cylinder because of fluid,
For equilibrium, both the forces should be equal, hence solving the above equation,
h = R √3/2
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Community Answer
In the figure shown, the heavy cylinder (radius R) resting on a smooth...
B
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In the figure shown, the heavy cylinder (radius R) resting on a smooth surface separates two liquids of densities 2r and 3r. The height `h' for the equilibrium of cylinder must be a)3R/2b)c)R√2d)NoneCorrect answer is option 'B'. Can you explain this answer?
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