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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

At a certain confluence, river Yamuna mixes with the river Chambal...

At a certain confluence, river Yamuna mixes with the river Chambal and becomes river Kaveri. Suppose Yamuna has a width of 12 m12 \text{ m}, Chambal has a width of 8 m8 \text{ m} and Kaveri has a width of 16 m16 \text{ m}. Assume that the depth of water is identical in all three rivers. If the average speed of water in Yamuna is 20 km h−120 \text{ km h}^{-1} and in Chambal is 16 km h−116 \text{ km h}^{-1}, determine the average speed of water in the river Kaveri.

Options

  1. A23 km h−123 \text{ km h}^{-1}
  2. B7 km h−17 \text{ km h}^{-1}
  3. C36 km h−136 \text{ km h}^{-1}
  4. D18 km h−118 \text{ km h}^{-1}

Correct answer

A. 23 km h−123 \text{ km h}^{-1}

Step-by-step solution

According to the equation of continuity, the total volume flow rate of water before the confluence is equal to the volume flow rate after the confluence.

QKaveri=QYamuna+QChambalQ_{Kaveri} = Q_{Yamuna} + Q_{Chambal}

AKvK=AYvY+ACvCA_K v_K = A_Y v_Y + A_C v_C

Since the depth dd is identical in all three rivers, the cross-sectional area is given by A=w×dA = w \times d, where ww is the width.

wKdvK=wYdvY+wCdvCw_K d v_K = w_Y d v_Y + w_C d v_C

Dividing by dd gives:

wKvK=wYvY+wCvCw_K v_K = w_Y v_Y + w_C v_C

Substituting the given values:

16×vK=12×20+8×1616 \times v_K = 12 \times 20 + 8 \times 16

16vK=240+12816 v_K = 240 + 128

16vK=36816 v_K = 368

vK=36816=23 km h−1v_K = \dfrac{368}{16} = 23 \text{ km h}^{-1}
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