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PhysicsFluidBernoulli's theorem and applicationsMedium2 minPYQ_2023

PhysicsMediumnumerical

The surface of water in a water tank of cross section area $750 {cm}^{2}$ on the top of a house is $h m$ . above the tap level. The speed of water coming out through the tap of cross section area $500 {mm}^{2}$ is $30 cm s^{- 1}$ . At that instant, $\frac{d h}{d t}$ is $x \times 10^{- 3} m s^{- 1}$ . The value of $x$ will be ______.

Answer:

2.00

Solution:

Using continuity equation

$A_{1} v_{1} = A_{2} v_{2}$

We have,

$750 \times 10^{- 4} v_{1} = 500 \times 10^{- 6} \times 0.3$

$v_{1} = \frac{500 \times 3 \times 10^{- 3}}{750}$

$= 2 \times 10^{- 3} m s^{- 1}$

Thus, $\frac{d h}{d t} = - 2 \times 10^{- 3} m s^{- 1}$

Hence, the value of $x = 2$ .

Stream:JEESubject:PhysicsTopic:FluidSubtopic:Bernoulli's theorem and applications

⏱ 2mℹ️ Source: PYQ_2023

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