The surface tension of water was determined in a laboratory by using the drop weight method. 100 drops were released from a bure
1 answer:
Answer:
γ = 0.06563 N / m
9.78% difference
Explanation:
Solution:-
- Surface tension is the ability of any fluid to resist any external force which causes a decreases in surface area of the impact area due to inward compressive forces. These compressive forces occur due to cohesive nature of the fluid molecules.
- Mathematically, surface tension ( γ ) is defined as the force felt per unit length by the fluid.
γ = F / L
Where,
F: Force imparted
L: The length over which force is felt
- We are given the mass ( M ) of ( n = 100 ) water droplets to e 3.78 g. The mass of a single droplet ( m ) can be evaluated as follows:
m = M / n
m = 3.78 / 100
m = 0.0378 g
- The force ( F ) imparted by a single drop of water from the burette can be determined from the force balance on a single droplet. Assuming the distance over which the drop falls is negligible and resistive forces are negligible. Then the only force acting on the droplet is due to gravity:
F = m*g
F = 0.0378*9.81*10^-3
F = 0.000370818 N
- The length over which the force is felt can be magnified into a circular area with diameter equal to that of a single droplet ( d ). The circumferential length ( L ) of the droplet would be as follows:
L = π*d
L = π*( 0.0018 )
L = 0.00565 m
- Then the surface tension would be:
γ = F / L
γ = 0.000370818 / 0.00565
γ = 0.06563 N / m
- The tabulated value of water's surface tension is given as follows:
γa = 0.07275 N/m
- We will determine the percentage difference between the value evaluated and tabulated value as follows:
- The %difference between is within the allowable practical limits of 10%. Hence, the evaluated value ( γ = 0.06563 N / m ) can be accepted with 9.78% error.
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Incomplete part of the question
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 in place by a stiff rod of negligible volume and weight. Throughout the problem, assume the acceleration due to gravity is g = 9.81 m/s2 . What is the weight Wb of the ball? Express your answer numerically in newtons.
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.
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.
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?
Answer:
(a) 2.94 N
(b) 1 N
(c) 2.42 N
(d) 3.42 N
Explanation:
(a)
From the definition of density, it's mass per unit volume hence mass is a product of density and volume. To get weight, we multiply mass by acceleration due to gravity
The weight of the ball is
Where is the density, V is volume and g is acceleration due to gravity
Substituting density for 5000 Kg/m3 and g for 9.8 m/s2 and v for 0.00006 m3 then
(b)
Because the ball is being held up mostly by the rod, the fluid pressure on the bottom of the cylinder is just the same as before.
The scale does not "know" the ball is there at all.
That's why it still reads 1 N.
Therefore, the reading is 1 N
(c)
The buoyant force of the fluid on the ball is equal to the weight of the displaced fluid, namely,
so the force needed for the rod to hold up the ball is 2.94 N - 0.52 N = 2.42 N.
(d)
Now the scale "feels" the weight of the ball,
so the scale reads the weight of the ball
PLUS the weight of the original fluid
MINUS the weight of the fluid that was displaced
= 2.94 N + 1.00 N - 0.52 N = 3.42 N