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Ail object of mass m = 100 kg and volume V= 0.18 m3 is dropped from a height h = 100m above the surface of the water. Upon entering water (β= 1000 kg/m3 ), the object experiences a resistance force given by F = -kv. where k = 10N sec in. The maximum depth to which the object will sink before starting to rise again is _____ m. (Neglect air resistance and assume the object looses negligible energy when going through the surface of the water). (Nearest integer)
Correct answer is between '125,129'. Can you explain this answer?
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Ail object of mass m = 100 kg and volume V= 0.18 m3 is dropped from a ...
Let the velocity with which the object hits the water be 
The equation of motion after it enters the water.
The equation of motion after it enters the water.
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Ail object of mass m = 100 kg and volume V= 0.18 m3 is dropped from a ...
Assuming the water is at rest), the object will experience a buoyant force and undergo a deceleration due to the resistance of water.
To determine the buoyant force, we can use Archimedes' principle, which states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
The weight of the fluid displaced can be calculated using the density of water (ρ = 1000 kg/m^3) and the volume of the object (V = 0.18 m^3):
Weight of fluid displaced = density of water * volume of object
= 1000 kg/m^3 * 0.18 m^3
= 180 kg
Therefore, the buoyant force acting on the object is 180 kg.
Now, let's consider the deceleration of the object. The deceleration depends on the drag force acting on the object. The drag force can be calculated using the equation:
Drag force = 0.5 * density of water * velocity^2 * cross-sectional area * drag coefficient
Since the object is initially dropped from a height and enters the water vertically, its initial velocity can be calculated using the equation:
Initial velocity = sqrt(2 * acceleration due to gravity * height)
= sqrt(2 * 9.8 m/s^2 * 100 m)
= 44.27 m/s
The cross-sectional area of the object can be calculated using its volume and height:
Cross-sectional area = volume / height
= 0.18 m^3 / 100 m
= 0.0018 m^2
Assuming a drag coefficient of 0.5 (typical value for a submerged object), we can calculate the drag force:
Drag force = 0.5 * 1000 kg/m^3 * (44.27 m/s)^2 * 0.0018 m^2 * 0.5
= 110.5 N
Now, we can calculate the net force acting on the object:
Net force = buoyant force - drag force
= 180 kg - 110.5 N
= 69.5 N
Finally, we can calculate the deceleration of the object using Newton's second law:
Net force = mass * acceleration
69.5 N = 100 kg * acceleration
acceleration = 0.695 m/s^2
Therefore, upon entering the water, the object will experience a deceleration of 0.695 m/s^2.
To determine the buoyant force, we can use Archimedes' principle, which states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
The weight of the fluid displaced can be calculated using the density of water (ρ = 1000 kg/m^3) and the volume of the object (V = 0.18 m^3):
Weight of fluid displaced = density of water * volume of object
= 1000 kg/m^3 * 0.18 m^3
= 180 kg
Therefore, the buoyant force acting on the object is 180 kg.
Now, let's consider the deceleration of the object. The deceleration depends on the drag force acting on the object. The drag force can be calculated using the equation:
Drag force = 0.5 * density of water * velocity^2 * cross-sectional area * drag coefficient
Since the object is initially dropped from a height and enters the water vertically, its initial velocity can be calculated using the equation:
Initial velocity = sqrt(2 * acceleration due to gravity * height)
= sqrt(2 * 9.8 m/s^2 * 100 m)
= 44.27 m/s
The cross-sectional area of the object can be calculated using its volume and height:
Cross-sectional area = volume / height
= 0.18 m^3 / 100 m
= 0.0018 m^2
Assuming a drag coefficient of 0.5 (typical value for a submerged object), we can calculate the drag force:
Drag force = 0.5 * 1000 kg/m^3 * (44.27 m/s)^2 * 0.0018 m^2 * 0.5
= 110.5 N
Now, we can calculate the net force acting on the object:
Net force = buoyant force - drag force
= 180 kg - 110.5 N
= 69.5 N
Finally, we can calculate the deceleration of the object using Newton's second law:
Net force = mass * acceleration
69.5 N = 100 kg * acceleration
acceleration = 0.695 m/s^2
Therefore, upon entering the water, the object will experience a deceleration of 0.695 m/s^2.
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Ail object of mass m = 100 kg and volume V= 0.18 m3 is dropped from a height h = 100m above the surface of the water. Upon entering water (β= 1000 kg/m3 ),the object experiences a resistance force given by F = -kv. where k = 10N sec in. The maximum depth to which the object will sink before starting to rise again is _____ m. (Neglect air resistance and assume the object looses negligible energy when going through the surface of the water). (Nearest integer)Correct answer is between '125,129'. Can you explain this answer?
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Ail object of mass m = 100 kg and volume V= 0.18 m3 is dropped from a height h = 100m above the surface of the water. Upon entering water (β= 1000 kg/m3 ),the object experiences a resistance force given by F = -kv. where k = 10N sec in. The maximum depth to which the object will sink before starting to rise again is _____ m. (Neglect air resistance and assume the object looses negligible energy when going through the surface of the water). (Nearest integer)Correct answer is between '125,129'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Ail object of mass m = 100 kg and volume V= 0.18 m3 is dropped from a height h = 100m above the surface of the water. Upon entering water (β= 1000 kg/m3 ),the object experiences a resistance force given by F = -kv. where k = 10N sec in. The maximum depth to which the object will sink before starting to rise again is _____ m. (Neglect air resistance and assume the object looses negligible energy when going through the surface of the water). (Nearest integer)Correct answer is between '125,129'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Ail object of mass m = 100 kg and volume V= 0.18 m3 is dropped from a height h = 100m above the surface of the water. Upon entering water (β= 1000 kg/m3 ),the object experiences a resistance force given by F = -kv. where k = 10N sec in. The maximum depth to which the object will sink before starting to rise again is _____ m. (Neglect air resistance and assume the object looses negligible energy when going through the surface of the water). (Nearest integer)Correct answer is between '125,129'. Can you explain this answer?.
Ail object of mass m = 100 kg and volume V= 0.18 m3 is dropped from a height h = 100m above the surface of the water. Upon entering water (β= 1000 kg/m3 ),the object experiences a resistance force given by F = -kv. where k = 10N sec in. The maximum depth to which the object will sink before starting to rise again is _____ m. (Neglect air resistance and assume the object looses negligible energy when going through the surface of the water). (Nearest integer)Correct answer is between '125,129'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Ail object of mass m = 100 kg and volume V= 0.18 m3 is dropped from a height h = 100m above the surface of the water. Upon entering water (β= 1000 kg/m3 ),the object experiences a resistance force given by F = -kv. where k = 10N sec in. The maximum depth to which the object will sink before starting to rise again is _____ m. (Neglect air resistance and assume the object looses negligible energy when going through the surface of the water). (Nearest integer)Correct answer is between '125,129'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Ail object of mass m = 100 kg and volume V= 0.18 m3 is dropped from a height h = 100m above the surface of the water. Upon entering water (β= 1000 kg/m3 ),the object experiences a resistance force given by F = -kv. where k = 10N sec in. The maximum depth to which the object will sink before starting to rise again is _____ m. (Neglect air resistance and assume the object looses negligible energy when going through the surface of the water). (Nearest integer)Correct answer is between '125,129'. Can you explain this answer?.
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