Coefficient of viscosity - is 2-76-10 -4 Pa-S at -40 oC- If E for viscosity is 5984 J mol -1- calculate - at -50oC-

The correct option is D 3.17×10 ^ 4 Pa.S logη=log A +E/2.303RT logη_2/η_1=E/2.303R[1/T_2 1/T_1] logη_2/2.76× 10^ 4=5984/2.303× 8 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Coefficient of Viscosity

Coefficient o...

Question

Coefficient of viscosity $(η)$ is $2.76 \times 10$ $^{- 4}$ Pa.S at $- 40$ $^{o}$ C. If E (for viscosity) is $5984$ J mol $^{- 1}$ , calculate $η$ at $- 50^{o}$ C:

6.17 \times 10

^{- 5}

Pa.S

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3.17 \times 10

^{- 4}

Pa.S

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

5.19 \times 10

^{- 6}

Pa.S

No worries! We‘ve got your back. Try BYJU‘S free classes today!

7.10 \times 10

^{- 6}

Pa.S

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $3.17 \times 10$ $^{- 4}$ Pa.S
$log η = log$ A $+ \frac{E}{2.303 R T}$
$log \frac{η_{2}}{η_{1}} = \frac{E}{2.303 R} [\frac{1}{T_{2}} - \frac{1}{T_{1}}]$
$log \frac{η_{2}}{2.76 \times 10^{- 4}} = \frac{5984}{2.303 \times 8.314} [\frac{1}{223} - \frac{1}{233}]$
$η_{2} = 3.17 \times 10^{- 4}$ Pa.S

Suggest Corrections

Similar questions

Q. If ball of steel (density

ρ = 7.8 {c m}^{- 3}

) attains a terminal velocity of

10 c m s^{- 1}

when falling in a tank of water (coefficient of viscosity

η_{w a t e r} = 8.5 \times 10^{- 4} P a . s

) then its terminal velocity in glycerin (

ρ = 1.2 g {c m}^{- 3} η = 13.2 P a . s

) would be nearly:

Q. If a ball of steel (density

ρ = 7.8 g c m^{- 3}

) attains a terminal velocity of

10 c m s^{- 1}

when falling in a tank of water (coefficient of viscosity

η_{w a t e r} = 8.5 \times 10^{- 4} P a . s

) then its terminal velocity in glycerine

(ρ = 12 g c m^{- 3}, η = 13.2 p a . s)

would be nearly:-

Q. The velocity of water in a river is

18 {kmh}^{- 1}

near the surface. If the river is

5 m

deep, then the shearing stress between the surface layer and the bottom layer is:
(Given - coefficient of viscosity of water,

η = 10^{- 3} Pa.s

)

Q. The velocity of water in a river is

18 {kmh}^{- 1}

near the surface. If the river is

5 m

deep, then the shearing stress between the surface layer and the bottom layer is:
(Given - coefficient of viscosity of water,

η = 10^{- 3} Pa.s

)

Q. A metal ball of radius

1 m m

and density

10^{4}

kg m

^{- 3}

. falls freely in air through a height '

h

' before falling in a tank full of water. If on falling in water its velocity remains unchanged, then the value of h will be (Coefficient of viscosity of water

= 10^{- 3}

Pa.s and

g = 10 m s^{- 2}

)

Join BYJU'S Learning Program

Coefficient of viscosity (η) is 2.76×10−4 Pa.S at −40oC. If E (for viscosity) is 5984 J mol−1, calculate η at −50oC:

The correct option is D 3.17×10−4 Pa.Slogη=log A +E2.303RTlogη2η1=E2.303R[1T2−1T1]logη22.76×10−4=59842.303×8.314[1223−1233]η2=3.17×10−4 Pa.S

Coefficient of viscosity $(η)$ is $2.76 \times 10$ $^{- 4}$ Pa.S at $- 40$ $^{o}$ C. If E (for viscosity) is $5984$ J mol $^{- 1}$ , calculate $η$ at $- 50^{o}$ C: