The rate of flow of a liquid Q through a capillary tube depends upon the pressure gradient Pl- radius of the capillary r- coefficient viscosity - and constant of proportionality is -8- The equation for the rate of flow of the liquid Q is given by -

The correct option is B Q=π pr ^ 4 8η l Q=π8(M^1L^0T^ 2)^α #160; #160; (L^1)^β #160; #160; (M^1L^ 1T^ 1)^γ Q=M^0L^3T^ 1 α =1 , # ...

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Byju's Answer

Standard VIII

Physics

Viscosity

The rate of f...

Question

The rate of flow of a liquid Q through a capillary tube depends upon the pressure gradient ( $\frac{P}{l}$ ), radius of the capillary ( $r$ ), coefficient viscosity $η$ and constant of proportionality is $\frac{π}{8}$ . The equation for the rate of flow of the liquid Q is given by :

Q = \frac{{π p r}^{2}}{8 η l}

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Q = \frac{{π p r}^{4}}{8 η l}

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Q = \frac{π}{8} {p r}^{2} η^{2} l

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Q = \frac{8 η l}{{π p r}^{4}}

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Solution

The correct option is B $Q = \frac{{π p r}^{4}}{8 η l}$
$Q = \frac{π}{8} (M^{1} L^{0} T^{- 2})^{α} (L^{1})^{β} (M^{1} L^{- 1} T^{- 1})^{γ}$
$Q = M^{0} L^{3} T^{- 1}$
$α = 1, β = 4, γ = - 1$
$Q = \frac{π}{8} \frac{p r^{4}}{η l}$

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Similar questions

Q. The rate of flow of liquid in a tube of radius

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, length

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,whose ends are maintained at a pressure difference

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, where

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Q. The rate of flow of a liquid through a capillary tube of length

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V = π p r^{4} / 8 η L

. Determine the dimensions of the coefficient of viscosity

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Q. A capillary tube of radius

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and viscosity

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\frac{V}{t} = \frac{π p r^{4}}{8 μ L}

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pressure difference across the tube. Velocity with which level of liquid sinks in vessel

Q. The volume of a liquid (v) flowing per second through a cylindrical tube depends upon the pressure gradient (p/l) radius of the tube (r) coefficient of viscosity (

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Q. Two capillary of length

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The rate of flow of a liquid Q through a capillary tube depends upon the pressure gradient (Pl), radius of the capillary (r), coefficient viscosity η and constant of proportionality is π8. The equation for the rate of flow of the liquid Q is given by :

The correct option is B Q=πpr48ηlQ=π8(M1L0T−2)α(L1)β(M1L−1T−1)γQ=M0L3T−1α=1,β=4,γ=−1Q=π8pr4ηl

The rate of flow of a liquid Q through a capillary tube depends upon the pressure gradient ( $\frac{P}{l}$ ), radius of the capillary ( $r$ ), coefficient viscosity $η$ and constant of proportionality is $\frac{π}{8}$ . The equation for the rate of flow of the liquid Q is given by :

The correct option is B $Q = \frac{{π p r}^{4}}{8 η l}$
$Q = \frac{π}{8} (M^{1} L^{0} T^{- 2})^{α} (L^{1})^{β} (M^{1} L^{- 1} T^{- 1})^{γ}$
$Q = M^{0} L^{3} T^{- 1}$
$α = 1, β = 4, γ = - 1$
$Q = \frac{π}{8} \frac{p r^{4}}{η l}$