Question

A glass tube of fine bore of radius 5 x 10-4 m is dipped in large tank of
water with surface tension, 0.073 Nm-1. The angle of contact of liquid
with walls of glass tube is 0°. The height of liquid rise in glass tube is
approximately
1) 2.98 cm
2) 1.75 cm
3) 6.28 cm
4) 3.50 cm

Don't give irrelevant answers I'll report it! ​

Answer 1

Points to note before solving:-

• the angle theta between the tangent to the liquid surface and the surface is called contact angle formed by the adhesive and cohesive forces between them
• the tendency of a fluid to be raised or suppressed in a tube is called capillary action
• h= 2Tcos@/đgr( where T is surface tension,ď id density of fluid)
• the above formula is the result of capillary action and surface tension..

Given:-

radius=5×10^-4m

density =1 kg/m3(fluid is water)

T=0.073N/m

angle=0°,cos(0)=1

h=?

$h = \frac{2tcos \alpha}{ dgr } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: h = \frac{2 \times 0.073}{1 \times 10 \times {5 \times 10}^{ - 4} } \\ \: \: \: \: \: h = \frac{ 2 \times 73 \times {10}^{ - 3}}{5 \times {10}^{ - 3} } \\ h = \frac{146}{5} \\ h = 29.2metres$

h=29.2m

=2.92cm

(h will be 2.98 if you'll take g as 9.8m/s^2)

hope it helps!

Good luck ♧♧

Answer 2

OPTION 1) 2.98 IS THE CORRECT ANSWER TO YOUR QUESTION!!❤