Question

a thin walled glass tube of radius 0.5 cm is dipped vertically in the water of surface tension 70 dyne /cm the force required to take it out from water is

Answer 1

Answer:

r=0.5mm=0.5×10

3

=5×10

−4

m

T=0.04N/m

ρ=0.8gm/cc

=

10

−6

0.8×10

−3

θ=10⁰

g=9.8m/s²

T=rhρg/2cosθ

∴h=2Tcosθ/rgρ

∴h=2×0.04cos

5

10

×800×9.8

h=2×0.04×

5

0.9848

×800×9.8×10

−4

m

h=2.01×10

−2

m

∴ The height of the capillary rise is 2.01×10

−2

m

Hence,

option (A) is correct answer.

Answer 2

Answer:

2.01 multiples 10

Explanation:

r=0.5mm=0.5×10

3

=5×10

−4

m

T=0.04N/m

ρ=0.8gm/cc

=

10

−6

0.8×10

−3

θ=10⁰

g=9.8m/s²

T=rhρg/2cosθ

∴h=2Tcosθ/rgρ

∴h=2×0.04cos

5

10

×800×9.8

h=2×0.04×

5

0.9848

×800×9.8×10

−4

m

h=2.01×10

−2

m

∴ The height of the capillary rise is 2.01×10

−2

m

hope this helps you

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