Question

Water from a tap emerges vertically downwards
with an initial speed of 5 m/s. The cross-sectional
area of tap is 10 cm². Assume that the pressure
(due to atmosphere) is constant throughout the
stream of water and that the flow is steady, the
cross-sectional area of stream 3.75 m below the
tap is :-

Answer 1

Answer:

Given:

Initial Velocity = 5 m/s

Cross section of tap = 10 cm²

Distance travelled = 3.75 m

To find:

Area of cross-section after travelling 3.75 m

Concept:

Since external atmospheric pressure remains constant whole throughout the displacement , we can use Equation of Continuity to solve this problem.

Let final Velocity be v

Calculation:

${v}^{2} = {u}^{2} + 2gh$

$= > {v}^{2} = {5}^{2} + (2 \times 10 \times 3.75)$

$= > {v}^{2} = 25 + 75$

$= > {v}^{2} = 100$

$= > v = 10 \: m {s}^{ - 1}$

Now applying Equation Of Continuity :

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

$= > 10 \times 5 = A_{2} \times 10$

$= > A_{2} = 5 \: {cm}^{2}$

So final answer :

$\boxed{ \huge{ \green{ \sf{ \bold{A_{2} = 5 \: {cm}^{2}}}}}}$

Answer 2

$\underline{ \huge \boxed{ \bold{ \mathfrak{ \purple{Answer}}}}}$

Given :

initial speed of water = 5 m/s

cross-sectional area of tap = 10 cm^2

distance travelled by water = 3.75 m

To Find :

Area of cross-section after travelling 3.75 m below

Formula :

Here pressure of atmosphere is constant throughout streaming of water so, we can use continuity equation...

$\: \: \: \: \: \: \: \: \: \: \: \dag \: \underline{ \boxed{ \bold{ \rm{ \pink{Av = constant}}}}} \: \dag$

Kinematic equation is also helpful here...

$\: \: \: \: \: \: \: \: \: \: \: \: \dag \: \underline{ \boxed{ \bold{ \rm{ \blue{ {v}^{2} = {u}^{2} + 2gS}}}}} \: \dag$

Calculation :

$\implies \rm \: {v}^{2} = {(5)}^{2} + 2(10)(3.75) \\ \\ \therefore \: \boxed{ \rm\red{v = 10 \: \frac{m}{s} }} \\ \\ \implies \rm \: we \: have \begin{cases} \star \rm \: \green{A_1 = 10 \: {cm}^{2} } \\ \star \rm \: \blue{v_1 = 5 \: \frac{m}{s}} \\ \star \rm \: \red{v_2 = 10 \: \frac{m}{s} } \\ \star \rm \: \purple{A_2 = ?} \end{cases} \\ \\ \implies \rm \: A_1v_1 = A_2v_2 \\ \\ \therefore \rm \: A_2 = \frac{A_1 \times v_1}{v_2} = \frac{10 \times 5}{10} \\ \\ \therefore \: \underline{ \boxed{ \bold{ \rm{ \orange{A_2 = 5 \: {cm}^{2} }}}}} \: \purple{ \clubsuit}$