Question

of 3 s 3) A glass tumblex in any position is perpendicularly immey sed po a trou trough btaining water. Water con ho h conto not rise into the tumble Why? Which property of air is shown here of​

Answer 1

Suppose pressures at the points A, B, C and D be P

A

,P

B

,P

C

and P

D

respectively

The pressure on the concave side of the liquid surface is greater than that on the other side by 2T/R

An angle of contact θ is given to be 0

∘

hence R cos 0

∘

= r or R = r

∴ P

A

=P

B

+2T/r

1

and P

C

=P

D

+2T/r

2

where r

1

and r

2

are the radii of the two limbs

But P

A

=P

C

∴ P

B

+

r

1

2T

=P

D

+

r

2

2T

or P

D

−P

B

=2T(

r

1

1

−

r

2

1

)

where h is the difference in water levels in the two limbs

Now h=

ρg

2T

(

r

1

1

−

r

2

1

) Given that T = 0.07Nm

−1

,ρ=1000kgm

−3

r

1

=

2

3

mm=

20

3

cm=

20×100

3

m=1.5×10

−3

m,r

2

=3×10

−3

m

∴ h=

1000×9.8

2×0.07

(

1.5×10

−3

1

−

3×10

−3

1

)m=4.76×10

−3

m=4.76mm

$Suppose pressures at the points A, B, C and D be PA,PB,PC and PD respectively</h3><h3></h3><h3>The pressure on the concave side of the liquid surface is greater than that on the other side by 2T/R</h3><h3>An angle of contact θ is given to be 0∘ hence         R cos 0∘ = r or R = r</h3><h3>∴ PA=PB+2T/r1 and                    PC=PD+2T/r2</h3><h3>where r1 and r2are the radii of the two limbs</h3><h3>But        PA=PC ∴       PB+r12T=PD+r22T</h3><h3>or            PD−PB=2T(r11−r21)</h3><h3>where h is the difference in water levels in the two limbs</h3><h3>Now                h=ρg2T(r11−r21) Given that           T = 0.07Nm−1,ρ=1000kgm−3</h3><h3>r1=23mm=203cm=20×1003m=1.5×10−3m,r2=3×10−3m</h3><h3>∴             h=1000×9.82×0.07(1.5×10−31−3×10−31)m=4.76×10−3m=4.76mm$

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