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Instantaneous Velocity

Three rivers ...

Question

Three rivers $R_{1}, R_{2}$ and $R_{3}$ merge together to form a river, $R_{4}$ . The cross sectional areas of the three rivers $R_{1}, R_{2}$ and $R_{3}$ are in the ratio $1 : 2 : 3$ and speed of water flowing in these are in the ratio $1 : \frac{1}{2} : \frac{1}{3}$ . Assuming streamline flow, if $R_{4}$ has cross sectional area equal to that of $R_{1}$ , the ratio of speed of water in $R_{4}$ to that in $R_{1}$ is

4 : 1

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3 : 1

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2 : 1

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1 : 1

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Solution

The correct option is B $3 : 1$
$D i s c h a r g e = v e l o c i t y \times c r o s s s e c t i o n a r e a$
$R 1 : R 2 : R 3 = a r e a = 1 : 2 : 3$
$R 1 = 1 x, R 2 = 2 x, R 3 = 3 x$ are the cross section areas
and their respective velocities are $1 : 1 / 2 : 1 / 3 = 6 : 3 : 2$
$R 1 = 6 y, R 2 = 3 y, R 3 = 2 y$
$R 1$ discharge $= 1 x \times 6 y = 6 x y$
$R 2. = 6 x y$
$R 3. = 6 x y$
$R 4$ discharge $= R 1 + R 2 + R 3 = 18 x y$
$R 4$ has he same cross section as $R 1 = 1 x$
so, $R 4$ has velocity $= 18 y$
$R 4 : R 3$ (velocity ratio) $= 18 y : 6 y$
which is $3 : 1$

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Similar questions

R_{1} : R_{2} : R_{3} : R_{4} = 1 : 2 : 3 : 4

R_{1}

R_{2}

R_{3}

R_{4} = A \times A

R_{1}, R_{2}, R_{3}, R_{4},

f (x) = 4 x^{4} - a x^{3} + b x^{2} - c x + 5

(a, b, c \in R)

r_{1}, r_{2}, r_{3}, r_{4}

\frac{r_{1}}{2} + \frac{r_{2}}{4} + \frac{r_{3}}{5} + \frac{r_{4}}{8} = 1

a

f (x) = 4 x^{4} - a x^{3} + b x^{2} - c x + 5

(a, b, c ϵ R)

r_{1}, r_{2}, r_{3}, r_{4}

\frac{r_{1}}{2} + \frac{r_{2}}{4} + \frac{r_{3}}{5} + \frac{r_{4}}{8} = 1

1 : 1.5

1 : 2 : 3

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