Question

A thin ring of diameter of 8cm is pulled out of water. Surface tension is 0.07N/m. The force required to break free the ring from water is A) 0.0088N B) 0.0176N C 0.0352N D) 3.52
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Answer 1

Explanation:

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Answer 2

Given: Diameter of ring 8cm, Surface tension = 0.07N/m

To find: the force required to break free the ring from water

Solution: Surface tension is a property of a liquid with the help of which the surface of the liquid can resist any external force. It is because of the cohesive nature of the liquid molecules.

Here, in the question, we are given the diameter of the ring

R= d/2= 4 cm = 0.04m

T is surface tension= 0.07N/m

Circumference of the ring= 2πR= 2× 3.14× 0.04= 0.2512m

Upward force= 2TL = 2× 0.07× 0.2512= 0.0352N

Therefore, force required to break free the ring from the water will be c) 0.0352N.