Answer :

Diameter of the first tube = 3 mm


Diameter of the second tube = 6 mm


Thus, Radius of the first tube, r1 = 1.5 mm = 1.5 × 10-3 m


Radius of the first tube, r2 = 3 mm = 3 × 10-3 m


Surface tension of water, S = 7.3 × 10-2 N/m


Density of water, ρ = 1000 kg/m3


Angle of contact between bore surface and water, θ = 00


Acceleration due to gravity, g = 9.8 m/s


Height of water rise,


Where,


S = Surface tension


θ = Contact angle


ρ = Density


g = acceleration due to gravity


r = radius


Let h1 and h2 be the height rise of water in tube1 and tube2 respectively. Then the height are given by,


;


Thus, the difference between heights is given by,


Δh = h1 – h2






= 4.996 × 10-3 m


≈ 4.97 mm


The difference between the water levels in two bores = 4.97 mm


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

The surface tensiPhysics - Exemplar

Surface tension iPhysics - Exemplar

A hot air balloonPhysics - Exemplar

If a drop of liquPhysics - Exemplar

A wire forming a HC Verma - Concepts of Physics Part 1

For a surface molPhysics - Exemplar

The capillaries sHC Verma - Concepts of Physics Part 1

The free surface Physics - Exemplar

Consider a small HC Verma - Concepts of Physics Part 1

Find the excess pHC Verma - Concepts of Physics Part 1