Submerged Sphere in a Beaker
A cylindrical beaker of height 0.100m
and negligible weight is filled...
Submerged Sphere in a Beaker
A cylindrical beaker of height 0.100m
and negligible weight is filled to the brim with a fluid of
density
?= 890kg/m3 . When the beaker is placed on
a scale, its weight is measured to be 1.00N
.(Figure 1)
A ball of density ?b=
5000kg/m3 and volume V=
60.0cm3 is then submerged in the fluid,
so that some of the fluid spills over the side of the beaker. The
ball is held in place by a stiff rod of negligible volume and
weight. Throughout the problem, assume the acceleration due to
gravity is g= 9.81m/s2 .
1. What is the weight Wb of the ball?
Express your answer numerically in newtons.
2. What is the reading W2 of the scale when the ball is
held in this submerged position? Assume that none of the
water that spills over stays on the scale.
Calculate your answer from the
quantities given in the problem and express it numerically in
newtons.
3.
What is the force Fr applied to the ball by the
rod? Take upward forces to be positive (e.g., if the force
on the ball is downward, your answer should be negative).
Express your answer numerically in newtons.
4. The rod is now shortened and attached to the bottom of the
beaker. The beaker is again filled with fluid, the ball is
submerged and attached to the rod, and the beaker with fluid and
submerged ball is placed on the scale. What weight W3 does the scale now
show?
Solutions
Expert Solution
Concepts and reason
The concepts required to solve the given problem are Archimedes Principle and weight.
Calculate the weight of the ball by taking the product of mass and acceleration due to gravity.
Calculate the reading on scale with the help of equilibrium condition of force.
Fundamentals
Weight is the force with which the earth attracts every object towards itself.
The weight of an object is given by,
W=mg
Here, m is the mass of object and g is the acceleration due to gravity.
Archimedes Principle states that the upward force exerted on the body immersed in a fluid is equal to the weight of the fluid that is displaced.
The buoyant force can be calculated by the formula,
Fbuoyancy=ρVg
Here, ρ is the density of water, V is the volume of water displaced by the ball and g is the acceleration due to gravity.
The equilibrium condition for force gives,
ΣF=0
Here, F is force
(1)
The weight of the ball can be found by the formula,
Wball=mballg …… (1)
Here, mball is the mass of ball and g is the acceleration due to gravity.
The density, mass and volume of an object are related as,
ρ=Vmball
Here, ρ is the density, mball is the mass of ball and V is the volume.
Rearrange the above equation.
mball=ρV
Substitute 5000kg/m3 for ρand 60.0cm3 for V in the above equation.
mball=(5000kg/m3)(60cm3)(106cm31m3)=0.3kg
Substitute 0.3kg for mball and 9.81m/s2 for g in equation (1).
Wball=(0.3kg)(9.81m/s2)=2.94N
(2)
According to Archimedes’ principle, the weight of the beaker as shown by the scale will be,
Wbeaker=W(water−spilledwater)+W(object) …… (2)
But, W(object)=W(spilledwater)
Substitute W(spilledwater) for W(object) in equation (2).
A cylindrical beaker of height 0.100 m and negligible weight is filled to the brim with a fluid of density ρ = 890 kg/m3 . When the beaker is placed on a scale, its weight is measured to be 1.00 N .(Figure 1)
A ball of density ρb = 5000 kg/m3 and volume V = 60.0 cm3 is then submerged in the fluid so that some of the fluid spills over the side of the beaker. The ball is held...
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