An incompressible fluid flows between two section as shown in the figure- The flow across the upper surface is kV0b- where V0 is inlet velocity and - is the height of fluid region- The value of k correct up two decimal place is -

Flow across the upper surface Q=∫_0^δV_0bdy ∫_0^δV_Dsin ( π yδ )bdy =V_0b[y]_0^δ V_0b ( δπ ) [ cos ( π yδ ) ]_0^δ =V_0bδ V_0b ( δπ )(+1+1) =V_0bδ (1 2π ) =0 ...

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Standard XII

Physics

Rotational Work and Energy

An incompress...

Question

An incompressible fluid flows between two section as shown in the figure. The flow across the upper surface is $k V_{0} b δ$ , where $V_{0}$ is inlet velocity and $δ$ is the height of fluid region. The value of $k$ correct up two decimal place is _______.
0.35

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Solution

The correct option is A 0.35
Flow across the upper surface
$Q = \int_{0}^{δ} V_{0} b d y - \int_{0}^{δ} V_{D} s i n (\frac{π y}{δ}) b d y$
$= V_{0} b [y]_{0}^{δ} - V_{0} b (\frac{δ}{π}) {[- c o s (\frac{π y}{δ})]}_{0}^{δ}$
$= V_{0} b δ - V_{0} b (\frac{δ}{π}) (+ 1 + 1)$
$= V_{0} b δ (1 - \frac{2}{π})$
$= 0.363 V_{0} b δ$

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An incompressible fluid flows between two section as shown in the figure. The flow across the upper surface is kV0bδ, where V0 is inlet velocity and δ is the height of fluid region. The value of k correct up two decimal place is _______.0.35

The correct option is A 0.35Flow across the upper surface Q=∫δ0V0bdy−∫δ0VDsin(πyδ)bdy =V0b[y]δ0−V0b(δπ)[−cos(πyδ)]δ0 =V0bδ−V0b(δπ)(+1+1) =V0bδ(1−2π) =0.363V0bδ

An incompressible fluid flows between two section as shown in the figure. The flow across the upper surface is $k V_{0} b δ$ , where $V_{0}$ is inlet velocity and $δ$ is the height of fluid region. The value of $k$ correct up two decimal place is _______.
0.35