Water (ideal fluid) is flowing through a horizontal tube having cross sectional areas of its two ends being A and $A^{'}$ such that the ratio $A / A^{'}$ is 5.If the pressure difference of water between the two ends is $3 \times 10^{5} N m^{- 2}$ , the velocity of water with which it enters the tube will be (neglect gravity effects)

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$5 m s^{- 1}$

$10 m s^{- 1}$

$25 m s^{- 1}$

$50 \sqrt{10} m s^{- 1}$

answer is A.

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Detailed Solution

According to Bernouli’s theorem
$P_{1} + \frac{1}{2} ρ {v_{1}}^{2} = P_{2} + \frac{1}{2} ρ {v_{2}}^{2}$
From equation ,
$P_{1} - P_{2} = 3 \times 10^{5}, \frac{A_{1}}{A_{2}} = 5$
According to equation of continuity
$A_{1} v_{1} = A_{2} v_{2}$
Or, $\frac{A_{1}}{A_{2}} = \frac{v_{2}}{v_{1}} = 5 \Rightarrow v_{2} = 5 v_{1}$
From equation (i)
$P_{1} - P_{2} = \frac{1}{2} ρ ({v_{2}}^{2} - {v_{1}}^{2})$
Or $3 \times 10^{5} = \frac{1}{2} \times 1000 (25 {v_{1}}^{2} - {v_{1}}^{2})$
$\Rightarrow 600 = 6 v_{1} \times 4 v_{1} \Rightarrow {v_{1}}^{2} = 25 ∴ v_{1} = 5 m / s$