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Two mercury drops with radii in the ratio of 3 : 4 fall from a great height through the viscous liquid. The ratio of their terminal velocities is
- a)3 : 4
- b)4 : 3
- c)9 : 16
- d)16 : 9
Correct answer is option 'C'. Can you explain this answer?
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Two mercury drops with radii in the ratio of 3 : 4 fall from a great h...
Given: Two mercury drops with radii in the ratio of 3 : 4 fall from a great height through the viscous liquid.
To find: The ratio of their terminal velocities.
Solution:
Terminal Velocity: The constant speed attained by a freely falling object when the resistance of the medium through which it is falling prevents further acceleration.
When a body falls through a viscous medium like a liquid, it experiences a viscous force in the opposite direction of motion. This force is proportional to the velocity of the object and the radius of the object.
The terminal velocity of a falling object can be given as:
V ∝ r²/η
Where,
V is the terminal velocity
r is the radius of the object
η is the coefficient of viscosity of the medium
Given that the radii of the two mercury drops are in the ratio of 3:4. Let the radii of the drops be 3r and 4r.
The ratio of their terminal velocities can be given as:
V1/V2 = (3r)²/(4r)²
V1/V2 = 9/16
Therefore, the ratio of their terminal velocities is 9:16.
Hence, the correct option is (c) 9:16.
To find: The ratio of their terminal velocities.
Solution:
Terminal Velocity: The constant speed attained by a freely falling object when the resistance of the medium through which it is falling prevents further acceleration.
When a body falls through a viscous medium like a liquid, it experiences a viscous force in the opposite direction of motion. This force is proportional to the velocity of the object and the radius of the object.
The terminal velocity of a falling object can be given as:
V ∝ r²/η
Where,
V is the terminal velocity
r is the radius of the object
η is the coefficient of viscosity of the medium
Given that the radii of the two mercury drops are in the ratio of 3:4. Let the radii of the drops be 3r and 4r.
The ratio of their terminal velocities can be given as:
V1/V2 = (3r)²/(4r)²
V1/V2 = 9/16
Therefore, the ratio of their terminal velocities is 9:16.
Hence, the correct option is (c) 9:16.
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Community Answer
Two mercury drops with radii in the ratio of 3 : 4 fall from a great h...
Answer is option c.
terminal velocity directly proportional to square of radius .
terminal velocity directly proportional to square of radius .
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Two mercury drops with radii in the ratio of 3 : 4 fall from a great height through the viscous liquid. The ratio of their terminal velocities isa)3 : 4b)4 : 3c)9 : 16d)16 : 9Correct answer is option 'C'. Can you explain this answer?
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