A steady flow of an ideal fluid occurs through a horizontal pipe of varying cross-section. The kinetic energy per unit volume of the fluid at section A is 4 times the kinetic energy per unit volume at section B. The

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Quantrex Academy · Accepted Answer

Correct answer: C. 1 : 2

Kinetic energy per unit volume is given by

K = \dfrac{1}{2} \rho v^2

.

Given that

K_A = 4 K_B

, we have:

\dfrac{1}{2} \rho v_A^2 = 4 \left( \dfrac{1}{2} \rho v_B^2 \right)

v_A^2 = 4 v_B^2 \Rightarrow v_A = 2 v_B

According to the equation of continuity,

A_A v_A = A_B v_B

.
Since the cross-sectional area

A = \dfrac{\pi d^2}{4}

, we can write:

d_A^2 v_A = d_B^2 v_B

\left( \dfrac{d_A}{d_B} \right)^2 = \dfrac{v_B}{v_A} = \dfrac{1}{2}

<br><br>Taking the square root on both sides, we get:<br>

\dfrac{d_A}{d_B} = \dfrac{1}{\sqrt{2}}

<br><br>Thus, the ratio

d_A : d_B

is

1 : \sqrt{2}

.<br><br>Answer:

1 : \sqrt{2}

A steady flow of an ideal fluid occurs through a horizontal pipe of...

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