Practice question
Question
What is the excess pressure inside a mercury drop of radius 4mm at 20∘C? (Surface tension = 0.4355N/m)
Explanation
Excess pressure: ΔP = 2Sr. S = 0.4355N/m, r = 4×10−3m. ΔP = 2×0.43554×10−3 = 217.75Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 218 Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Discussion
Comments
Share your thoughts. New comments appear after admin approval.
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.