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A circular ring of thin wire of 3cm radius is suspended with its plane horizontal by a thread passing through a 10cm mark of a meter scale. pivoted at the centre and is balanced by 8gm-wt suspended from 80cm mark. when is just brought into the contact with the surface of liquid 3gm-wt has to be moved to 90cm mark just to detach the ring from the liquid. find surface tension of the liquid. (theeta= 15°, g=9.8)
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A circular ring of thin wire of 3cm radius is suspended with its plane...
Problem:
Find the surface tension of the liquid when a circular ring of thin wire of 3cm radius is suspended with its plane horizontal by a thread passing through a 10cm mark of a meter scale. It is pivoted at the center and is balanced by 8gm-wt suspended from 80cm mark. When just brought into contact with the surface of liquid 3gm-wt has to be moved to 90cm mark just to detach the ring from the liquid. Given, theeta= 15°, g=9.8.
Solution:
Step 1: Calculate the force acting on the ring due to its weight and tension in the thread.
The weight of the ring = mg = (8/1000) x 9.8 = 0.0784 N
The tension in the thread = weight of the ring = 0.0784 N
The horizontal component of the tension = T cosθ = 0.0784 cos15° = 0.0760 N
The vertical component of the tension = T sinθ = 0.0784 sin15° = 0.0201 N
Step 2: Calculate the force acting on the ring due to surface tension.
Let F be the force acting on the ring due to surface tension.
When the 3gm-wt is moved from the 80cm mark to the 90cm mark, the force acting on the ring due to surface tension is just equal to the weight of the 3gm-wt i.e., 0.003 x 9.8 = 0.0294 N
At this point, the ring is about to detach from the liquid.
Step 3: Calculate the radius of the meniscus formed by the liquid.
The weight of the liquid in contact with the ring = 0.003 x 9.8 = 0.0294 N
Let r be the radius of the meniscus formed by the liquid.
The force due to surface tension acting on the circumference of the ring = 2πrTcosθ
At equilibrium, this force is balanced by the weight of the liquid in contact with the ring.
Therefore, 2πrTcosθ = 0.0294 N
r = 0.0294/2πTcosθ = 0.0294/2π x 0.0784 x cos15° = 0.0038 m
Step 4: Calculate the surface tension of the liquid.
The force due to surface tension acting on the circumference of the ring = 2πrTcosθ
Substituting the values of r, T and θ, we get
F = 2π x 0.0038 x 0.0784 x cos15° = 0.0009 N
Therefore, the surface tension of the liquid = F/2πr = 0.0009/2π x 0.0038 = 0.037 N/m
Find the surface tension of the liquid when a circular ring of thin wire of 3cm radius is suspended with its plane horizontal by a thread passing through a 10cm mark of a meter scale. It is pivoted at the center and is balanced by 8gm-wt suspended from 80cm mark. When just brought into contact with the surface of liquid 3gm-wt has to be moved to 90cm mark just to detach the ring from the liquid. Given, theeta= 15°, g=9.8.
Solution:
Step 1: Calculate the force acting on the ring due to its weight and tension in the thread.
The weight of the ring = mg = (8/1000) x 9.8 = 0.0784 N
The tension in the thread = weight of the ring = 0.0784 N
The horizontal component of the tension = T cosθ = 0.0784 cos15° = 0.0760 N
The vertical component of the tension = T sinθ = 0.0784 sin15° = 0.0201 N
Step 2: Calculate the force acting on the ring due to surface tension.
Let F be the force acting on the ring due to surface tension.
When the 3gm-wt is moved from the 80cm mark to the 90cm mark, the force acting on the ring due to surface tension is just equal to the weight of the 3gm-wt i.e., 0.003 x 9.8 = 0.0294 N
At this point, the ring is about to detach from the liquid.
Step 3: Calculate the radius of the meniscus formed by the liquid.
The weight of the liquid in contact with the ring = 0.003 x 9.8 = 0.0294 N
Let r be the radius of the meniscus formed by the liquid.
The force due to surface tension acting on the circumference of the ring = 2πrTcosθ
At equilibrium, this force is balanced by the weight of the liquid in contact with the ring.
Therefore, 2πrTcosθ = 0.0294 N
r = 0.0294/2πTcosθ = 0.0294/2π x 0.0784 x cos15° = 0.0038 m
Step 4: Calculate the surface tension of the liquid.
The force due to surface tension acting on the circumference of the ring = 2πrTcosθ
Substituting the values of r, T and θ, we get
F = 2π x 0.0038 x 0.0784 x cos15° = 0.0009 N
Therefore, the surface tension of the liquid = F/2πr = 0.0009/2π x 0.0038 = 0.037 N/m
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A circular ring of thin wire of 3cm radius is suspended with its plane horizontal by a thread passing through a 10cm mark of a meter scale. pivoted at the centre and is balanced by 8gm-wt suspended from 80cm mark. when is just brought into the contact with the surface of liquid 3gm-wt has to be moved to 90cm mark just to detach the ring from the liquid. find surface tension of the liquid. (theeta= 15°, g=9.8)
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A circular ring of thin wire of 3cm radius is suspended with its plane horizontal by a thread passing through a 10cm mark of a meter scale. pivoted at the centre and is balanced by 8gm-wt suspended from 80cm mark. when is just brought into the contact with the surface of liquid 3gm-wt has to be moved to 90cm mark just to detach the ring from the liquid. find surface tension of the liquid. (theeta= 15°, g=9.8) for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A circular ring of thin wire of 3cm radius is suspended with its plane horizontal by a thread passing through a 10cm mark of a meter scale. pivoted at the centre and is balanced by 8gm-wt suspended from 80cm mark. when is just brought into the contact with the surface of liquid 3gm-wt has to be moved to 90cm mark just to detach the ring from the liquid. find surface tension of the liquid. (theeta= 15°, g=9.8) covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A circular ring of thin wire of 3cm radius is suspended with its plane horizontal by a thread passing through a 10cm mark of a meter scale. pivoted at the centre and is balanced by 8gm-wt suspended from 80cm mark. when is just brought into the contact with the surface of liquid 3gm-wt has to be moved to 90cm mark just to detach the ring from the liquid. find surface tension of the liquid. (theeta= 15°, g=9.8) .
A circular ring of thin wire of 3cm radius is suspended with its plane horizontal by a thread passing through a 10cm mark of a meter scale. pivoted at the centre and is balanced by 8gm-wt suspended from 80cm mark. when is just brought into the contact with the surface of liquid 3gm-wt has to be moved to 90cm mark just to detach the ring from the liquid. find surface tension of the liquid. (theeta= 15°, g=9.8) for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A circular ring of thin wire of 3cm radius is suspended with its plane horizontal by a thread passing through a 10cm mark of a meter scale. pivoted at the centre and is balanced by 8gm-wt suspended from 80cm mark. when is just brought into the contact with the surface of liquid 3gm-wt has to be moved to 90cm mark just to detach the ring from the liquid. find surface tension of the liquid. (theeta= 15°, g=9.8) covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A circular ring of thin wire of 3cm radius is suspended with its plane horizontal by a thread passing through a 10cm mark of a meter scale. pivoted at the centre and is balanced by 8gm-wt suspended from 80cm mark. when is just brought into the contact with the surface of liquid 3gm-wt has to be moved to 90cm mark just to detach the ring from the liquid. find surface tension of the liquid. (theeta= 15°, g=9.8) .
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