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Standard XII

Chemistry

Vapour Pressure

Two tubes of ...

Question

Two tubes of radii $r_{1}$ and $r_{2}$ , and lengths $l_{1}$ and $l_{2}$ , respectively, are connected in series and a liquid flows through each of them in stream line conditions. $P_{1}$ and $P_{2}$ are pressure differences across the two tubes. If $P_{2}$ is $4 P_{1}$ and $l_{2}$ is $\frac{l_{1}}{4}$ , then the radius $r_{2}$ will be equal to :

2 r_{1}

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4 r_{1}

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r_{1}

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\frac{r_{1}}{2}

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Solution

The correct option is A $2 r_{1}$
Given, $P_{2} = {4 P}_{1}$ & $l_{2} = \frac{l_{1}}{4}$
By the equation of continuity:
$A_{1} v_{1} = A_{2} v_{2}$
$\frac{A_{1}}{A_{2}}$ = $\frac{v_{1}}{v_{2}}$
Pressure is inversely proportional to velocity.
So, $\frac{P_{1}}{P_{2}}$ = $\frac{v_{2}}{v_{1}}$
So, $\frac{v_{2}}{v_{1}}$ = $\frac{1}{4}$
Thus,
$\frac{A_{1}}{A_{2}}$ = $\frac{1}{4}$
$\frac{r_{1}^{2}}{r_{2}^{2}}$ = $\frac{1}{4}$
$r_{2} = 2 r_{1}$

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Two tubes of radii r1 and r2, and lengths l1 and l2, respectively, are connected in series and a liquid flows through each of them in stream line conditions. P1 and P2 are pressure differences across the two tubes. If P2 is 4P1 and l2 is l14, then the radius r2 will be equal to :

The correct option is A 2r1Given, P2=4P1 & l2=l14By the equation of continuity:A1v1=A2v2A1A2 = v1v2Pressure is inversely proportional to velocity.So, P1P2 = v2v1So, v2v1 = 14Thus,A1A2 = 14r21r22 = 14r2=2r1