Capillary action is influenced by the forces between the fluid and the glass, with option B being favored due to the greater weight of fluid outside the tube compared to the inside. This weight creates a stronger upward pull on the fluid within the tube. The discussion emphasizes that the surface area in contact with the fluid remains constant, making it easier for the fluid to rise inside. The argument is presented as a matter of common sense rather than complex theory. Understanding these dynamics is crucial for grasping the principles of capillary action.
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lovelyrwwr
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How is B more correct than D? Both show that forces between glass and fluid are stronger than those between fluid molecules. Thank you for any insight!
B. because on the outside. there is more fluid. therefore. more weight that is the fluid putting on the glass. but on the inside. less fluid and the same surface are touching the tube! so it would be easier for the fluid to rise inside the tube. this question is really common sense! i never did this stuff. but its obvious!
Kindly see the attached pdf. My attempt to solve it, is in it.
I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction.
I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook.
Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water.
I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Does the direction my second finger points in indicate the literal direction of magnetic field at that point in space, or do I need to further use right hand rule or something for a circular path around the finger