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Standard XII

Physics

Coefficient of Viscosity

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Question

The space between two large horizontal metal plates 6cm apart is filled with a liquid of viscosity $0.8 N s / m^{2}$ . A thin planck of surface area $1 m^{2}$ is moved parallel to the length of the plates such that the planck is at a distance of 2cm from one plate and 4cm from the other.The force F required to drag it with a constant speed of is $1 m s^{-} 1$ is

20N

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40N

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60N

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80N

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Solution

The correct option is C
60N

$F = η A {(\frac{d v}{d x})}_{1} + η {(\frac{d v}{d x})}_{2} = (0.8) (1) (\frac{10^{2}}{4}) + (0.8) (1) (\frac{10^{2}}{2}) = 0.8 \times 10^{2} [\frac{1}{4} + \frac{1}{2}] = 0.8 \times 10^{2} (\frac{3}{4}) = 60 N$

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The correct option is C 60N F=ηA(dvdx)1+η(dvdx)2=(0.8)(1)(1024)+(0.8)(1)(1022)=0.8×102[14+12]=0.8×102(34)=60N

The correct option is C
60N

$F = η A {(\frac{d v}{d x})}_{1} + η {(\frac{d v}{d x})}_{2} = (0.8) (1) (\frac{10^{2}}{4}) + (0.8) (1) (\frac{10^{2}}{2}) = 0.8 \times 10^{2} [\frac{1}{4} + \frac{1}{2}] = 0.8 \times 10^{2} (\frac{3}{4}) = 60 N$