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When the capillary tube is lowered into water, the mass of water raised in the tube, above the outside water level is 5gm. If the radius of the tube is doubled, the mass of water that rises in the capillary tube above the outside water level is
- a)1.25gm
- b)5gm
- c)10gm
- d)20gm
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
When the capillary tube is lowered into water, the mass of water raise...
When a capillary tube is lowered, the water outside it is raised due to more volume of tube inserted in it. If the radius is doubled , then the periphery of tube is also doubled, doubling its volume.
∴ water rising outside, hence, is 10 gm.
Most Upvoted Answer
When the capillary tube is lowered into water, the mass of water raise...
Given:
- The mass of water raised in the capillary tube above the outside water level is 5g.
- The radius of the tube is doubled.
To find:
The mass of water that rises in the capillary tube above the outside water level when the radius of the tube is doubled.
Solution:
When a capillary tube is lowered into water, the water rises in the tube due to capillary action. The height to which the water rises in the tube depends on the radius of the tube.
Let's assume the original radius of the capillary tube is 'r' and the new radius after doubling is '2r'.
Step 1: Calculate the original height of the water column in the capillary tube.
The height 'h' of the water column in the capillary tube can be calculated using the formula:
h = (2Tcosθ)/(ρgr)
Where:
T = Surface tension of water
θ = Contact angle between water and the tube material
ρ = Density of water
g = Acceleration due to gravity
r = Radius of the capillary tube
Since the contact angle and surface tension remain the same, the height of the water column is directly proportional to the radius of the tube.
Step 2: Calculate the new height of the water column in the capillary tube.
When the radius of the tube is doubled, the new height 'H' of the water column can be calculated using the formula:
H = (2Tcosθ)/(ρg(2r))
Step 3: Calculate the mass of water raised in the capillary tube when the radius is doubled.
The mass of water raised in the capillary tube is directly proportional to the height of the water column. Therefore,
mass = (ρπr^2h)
The new mass of water raised in the capillary tube when the radius is doubled is given by:
new_mass = (ρπ(2r)^2H)
Step 4: Substitute the values and calculate the new mass.
Given that the original mass of water raised in the capillary tube is 5g, we can write:
mass = 5g
ρπr^2h = 5g
Substituting the value of 'h' from Step 1:
ρπr^2((2Tcosθ)/(ρgr)) = 5g
Cancelling out common terms:
2Tcosθ/r = 5g
Now, substitute the value of 'H' from Step 2:
ρπ(2r)^2((2Tcosθ)/(ρg(2r))) = new_mass
Cancelling out common terms:
16Tcosθ/g = new_mass
From the given options, we can see that the correct answer is option 'C', which is 10g.
- The mass of water raised in the capillary tube above the outside water level is 5g.
- The radius of the tube is doubled.
To find:
The mass of water that rises in the capillary tube above the outside water level when the radius of the tube is doubled.
Solution:
When a capillary tube is lowered into water, the water rises in the tube due to capillary action. The height to which the water rises in the tube depends on the radius of the tube.
Let's assume the original radius of the capillary tube is 'r' and the new radius after doubling is '2r'.
Step 1: Calculate the original height of the water column in the capillary tube.
The height 'h' of the water column in the capillary tube can be calculated using the formula:
h = (2Tcosθ)/(ρgr)
Where:
T = Surface tension of water
θ = Contact angle between water and the tube material
ρ = Density of water
g = Acceleration due to gravity
r = Radius of the capillary tube
Since the contact angle and surface tension remain the same, the height of the water column is directly proportional to the radius of the tube.
Step 2: Calculate the new height of the water column in the capillary tube.
When the radius of the tube is doubled, the new height 'H' of the water column can be calculated using the formula:
H = (2Tcosθ)/(ρg(2r))
Step 3: Calculate the mass of water raised in the capillary tube when the radius is doubled.
The mass of water raised in the capillary tube is directly proportional to the height of the water column. Therefore,
mass = (ρπr^2h)
The new mass of water raised in the capillary tube when the radius is doubled is given by:
new_mass = (ρπ(2r)^2H)
Step 4: Substitute the values and calculate the new mass.
Given that the original mass of water raised in the capillary tube is 5g, we can write:
mass = 5g
ρπr^2h = 5g
Substituting the value of 'h' from Step 1:
ρπr^2((2Tcosθ)/(ρgr)) = 5g
Cancelling out common terms:
2Tcosθ/r = 5g
Now, substitute the value of 'H' from Step 2:
ρπ(2r)^2((2Tcosθ)/(ρg(2r))) = new_mass
Cancelling out common terms:
16Tcosθ/g = new_mass
From the given options, we can see that the correct answer is option 'C', which is 10g.
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When the capillary tube is lowered into water, the mass of water raised in the tube, above the outside water level is 5gm. If the radius of the tube is doubled, the mass of water that rises in the capillary tube above the outside water level isa)1.25gmb)5gmc)10gmd)20gmCorrect answer is option 'C'. Can you explain this answer?
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