A target object is placed at the bottom of a tank of depth 20m and filled with a liquid of density 500gm/cc .Another object of density 100gm/cc is dropped onto it from a height of 45m from the upper surface of the liquid:- (take :g=10m/s^(2)) (Neglect viscous force)
Answers
(B) The second object does not collide with the target object.
(C) The least separation between the object and the target is 11.25 m
(D) The retardation of the object in the liquid is 40 m/s²
Explanation:
Note: The given options are:
(A) The second object will collide with the target
(B) The second object does not collide with the target object
(C) The least separation between the object and the target is 11.25 m
(D) The retardation of the object in the liquid is 40 m/s²
Density of the object ρ = 100 g/cc
Density of the liquid ρ' = 500 g/cc
Height from which the object is dropped h = 45 m
Depth of the target object = 20 m
Let the mass of the object be m
If the velocity just at the surface of the liquid be v then
By the conservation of energy
or,
m/s
After entering the liquid, a buoyant force will act on the object
Buoyant force in upward direction
Thus, the retardation due to this force, acting on the object of mass m
m/s²
Let the distance travelled by the object in liquid be s then
Using the third equation of motion
m
Thus, the object will only travel a distance of 11.25 m in the liquid before it comes to rest and then starts moving upwards as the density of the liquid is more than that of the object
Thus,
The second object does not collide with the target object.
The least separation between the object and the target is 11.25 m
The retardation of the object in the liquid is 40 m/s²
Hope this answer is helpful.
Know More:
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