JEE MAIN - Physics (2020 - 2nd September Evening Slot - No. 10)
A capillary tube made of glass of radius 0.15
mm is dipped vertically in a beaker filled with
methylene iodide (surface tension = 0.05 Nm–1,
density = 667 kg m–3) which rises to height h in
the tube. It is observed that the two tangents
drawn from liquid-glass interfaces (from opp.
sides of the capillary) make an angle of 60o
with one another. Then h is close to (g = 10 ms–2)
0.049 m
0.087 m
0.137 m
0.172 m
Explanation
_2nd_September_Evening_Slot_en_10_1.png)
h = $${{2T\cos \theta } \over {\rho gr}}$$
= $${{2 \times 0.05 \times {\sqrt 3 \over 2}} \over {667 \times 10 \times 0.15 \times {{10}^{ - 3}}}}$$
= 0.087 m
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