Practice question
Question
A capillary tube of radius 0.65 mm is dipped in water ( S = 0.0727 N/m, rho = 1000 kg/m³, cos θ = 1 ). What is the capillary rise? (Take g = 9.8 m/s² )
Explanation
Given:
A capillary tube of radius 0.65 mm is dipped in water ( S = 0.0727 N/m, rho = 1000 kg/m³, cos θ = 1 ). What is the capillary rise? (Take g = 9.8 m/s² )
These values define the system as per NCERT data.
Formula:
h = 2 S cos θ/rho g a.
This is the standard NCERT relation for this phenomenon.
Substitution & Calculation:
S = 0.0727 N/m, rho = 1000 kg/m³, g = 9.8 m/s², a = 0.65 × 10â»Â³ m . h = frac2 × 0.0727 × 11000 × 9.8 × 0.65 × 10â»Â³= 0.02285 m = 2.285 cm .
Result:
The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kgâ»Â¹ Kâ»Â¹, m/s², 10â»âµ are properly used as per NCERT.
Discussion
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