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BITSAT 2019 Paper

Section: Physics

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Question : 19 of 150

Marks: +1, -0

Water is flowing through a horizontal tube having cross-sectional areas of its two ends beings A and A' such that the ratio A/A' is 5. If the pressure difference of water between the two ends is 3 x 10

^{5}

^{-2}

, the velocity of water with which it enters the tube will be (neglect gravity effects)

5 m s $^{-1}$
10 m s $^{-1}$
25 ms $^{-1}$
$50$ m s $^{-1}$

Solution:

(a) According to Bernoulli’s theorem

P_1 + \frac{1}{2}\rho v_1^2

=

P_2 + \frac{1}{2}\rho v_2^2

........(i)

According to the condition,

P_1 - P_2

= 3 x 10

^{5}

,

\frac{A_1}{A_2}

= 5

From equation of continuity,

A_1v_1 = A_2v_2

so,

\frac{A_1}{A_2} = \frac{v_2}{v_1}

= 5 ⇒

v_2 = 5v_1

From equation (i)

P_1 - P_2

=

\frac{1}{2}\rho (v_2^2 - v_1^2)

or 3 x 10

^{5}

=

\frac{1}{2}

× 1000

(25v_1^2 - v_1^2)

⇒ 600 = 24v

_{1}

⇒ v

_{2}

= 25

∴ v

_{1}

= 5 m/s

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