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The velocity of upper layer of water in a river is $18 k m / h r$ . Shearing stress between horizontal layers of water is $10^{- 3} N / m^{2} .$ Depth of the river is _______m. (Coefficient of viscosity of water is $10^{- 2} P a - s$ )

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answer is 50.

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Detailed Solution

$V_{1} = 18 k m / h r = 18 \times \frac{5}{18} = 5 m / s$

$Δ V = V_{1} - V_{2} = 5 - 0 = 5 m / s$

$F = η A \frac{Δ V}{Δ Y}$

$\Rightarrow \frac{F}{A} = η . \frac{Δ V}{Δ Y}$

$10^{- 3} = 10^{- 2} \times \frac{5}{Δ Y}$

$Δ Y = 50 m$

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