Water flows through a vertical tube of variable cross section. The cross-sectional areas at A and B are 4 mm^2 and 2 mm^2 respectively. If 1 cc of water enters through A per second and A is vertically above B by

Correct answer: B. 25 cm/s,\ 50 cm/s,\ 0 Using continuity, Q = 1 cc/s = 10^ -6 m^3/s . At A , A_A = 4× 10^ -6 m^2 , so v_A = (Q)/(A_A) = 10^ -6 4× 10^ -6 = 0.25 m/s = 25 cm/s . At B , A_B = 2× 10^ -6 m^2 , so v_B = (Q)/(A_B) = 10^ -6 2× 10^ -6 = 0.5 m/s = 50 cm/s . By Bernoulli's equation, P_A + (1)/(2) v_A^2 + g z_A = P_B + (1)/(2) v_B^2 + g z_B . Since A is above B by (15)/(16) cm = 0.009375 m , P_A - P_B = (1)/(2) (v_B^2 - v_A^2) + g(z_B - z_A) . Substituting = 1000 kg/m^3 , g = 10 m/s^2 , v_A = 0.25 m/s and v_B = 0.5 m/s gives P_A - P_B = 93.75 - 93.75 = 0 . Hence the answers are 25 cm/s , 50 cm/s , and 0 .

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Fluid Mechanics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Fluid Mechanics

Water flows through a vertical tube of variable cross section. The...

Water flows through a vertical tube of variable cross section. The cross-sectional areas at

A

and

B

are

4\text{ mm}^2

and

2\text{ mm}^2

respectively. If

1\text{ cc}

of water enters through

A

per second and

A

is vertically above

B

\dfrac{15}{16}\text{ cm}

, find the speed of flow at

A

, the speed of flow at

B

, and the pressure difference

P_A - P_B

. Take

g = 10\text{ m s}^{-2}

Options

A $25\text{ cm/s},\ 50\text{ cm/s},\ 94\text{ N/m}^2$
B $25\text{ cm/s},\ 50\text{ cm/s},\ 0$
C $50\text{ cm/s},\ 25\text{ cm/s},\ 0$
D $25\text{ cm/s},\ 25\text{ cm/s},\ 0$

Correct answer

B. $25\text{ cm/s},\ 50\text{ cm/s},\ 0$

Step-by-step solution

Using continuity,

Q = 1\text{ cc/s} = 10^{-6}\text{ m}^3\text{/s}

.

At

A

A_A = 4\times 10^{-6}\text{ m}^2

, so

v_A = \dfrac{Q}{A_A} = \dfrac{10^{-6}}{4\times 10^{-6}} = 0.25\text{ m/s} = 25\text{ cm/s}

.

At

B

A_B = 2\times 10^{-6}\text{ m}^2

, so

v_B = \dfrac{Q}{A_B} = \dfrac{10^{-6}}{2\times 10^{-6}} = 0.5\text{ m/s} = 50\text{ cm/s}

.

By Bernoulli's equation,

P_A + \dfrac{1}{2}\rho v_A^2 + \rho g z_A = P_B + \dfrac{1}{2}\rho v_B^2 + \rho g z_B

.

Since

A

is above

B

\dfrac{15}{16}\text{ cm} = 0.009375\text{ m}

P_A - P_B = \dfrac{1}{2}\rho(v_B^2 - v_A^2) + \rho g(z_B - z_A)

.

Substituting

\rho = 1000\text{ kg/m}^3

g = 10\text{ m/s}^2

v_A = 0.25\text{ m/s}

and

v_B = 0.5\text{ m/s}

gives

P_A - P_B = 93.75 - 93.75 = 0

.

Hence the answers are

25\text{ cm/s}

50\text{ cm/s}

, and

0

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