Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1 mm made of this glass is dipped in a trough containing mercury. The surface tension of mercury at the temperature of the experiment is 0.465 N m-1. The distance by which the mercury dip down in the tube relative to the mercury surface outside is (Density of mercury = 13.6 × 103 kg m-3)

Question

Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1 mm made of this glass is dipped in a trough containing mercury. The surface tension of mercury at the temperature of the experiment is 0.465 N m-1. The distance by which the mercury dip down in the tube relative to the mercury surface outside is (Density of mercury = 13.6 × 103 kg m-3) (A) 5.34 mm (B) 2.35 mm (C) 6.25 mm (D) 1.44 mm

Tardigrade Admin · Accepted Answer

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

$5.34 mm$

$2.35 mm$

$6.25 mm$

$1.44 mm$

5.34mm

2.35mm

6.25mm

1.44mm

Here, θ=140∘,r=1×10−3m S=0.465Nm−1,ρ=13.6×103kgm−3 As h=rρg2Scosθ​=10−3×13.6×103×9.82×0.465×(−cos40∘)​ (∵cos140∘=−cos40∘) =−5.34×10−3m=−5.34mm Note: Negative value of h shows that the mercury level is depressed in the tube

$5.34 mm$

$2.35 mm$

$6.25 mm$

$1.44 mm$