A narrow tube of radius 1 mm made out of soda lime glass is dipped in a tank containing mercury. The angle of contact of mercury with the surface of the glass is 1400 and surface tension of mercury at the temperature of the experiment is 0.465 N m- 1 . Mercury dips down in the tube through a depth of: (Density of mercury =13.6× 103 kg m- 3 )

Question

A narrow tube of radius 1 mm made out of soda lime glass is dipped in a tank containing mercury. The angle of contact of mercury with the surface of the glass is 1400 and surface tension of mercury at the temperature of the experiment is 0.465 N m- 1 . Mercury dips down in the tube through a depth of: (Density of mercury =13.6× 103 kg m- 3 ) (A) 2.34 mm (B) 5.34 mm (C) 6.25 mm (D) 1.44 mm

Tardigrade Admin · Accepted Answer

Here, θ =140° ,r=1× 10- 3m S=0.465 N m- 1,ρ =13.6× 103kg m- 3 As h=(2 S cos θ /r ρ g)=(2 × 0.465 × (. - cos â¡ 40 ° .)/(10)- 3 × 13.6 × (10)3 × 9.8 ) (∵ cos 140 ° = - cos â¡ 40 ° ) =-5.34× 10- 3m=-5.34 mm Note: Negative value of h shows that the mercury level is depressed in the tube

$2.34 mm$

$5.34 mm$

$6.25 mm$

$1.44 mm$

2.34mm

5.34mm

6.25mm

1.44mm

Here, θ=140∘,r=1×10−3m S=0.465Nm−1,ρ=13.6×103kgm−3 As h=rρg2Scosθ​=(10)−3×13.6×(10)3×9.82×0.465×(−cos⁡40∘)​ (∵cos140∘=−cos⁡40∘) =−5.34×10−3m=−5.34mm Note: Negative value of h shows that the mercury level is depressed in the tube

$2.34 mm$

$5.34 mm$

$6.25 mm$

$1.44 mm$