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  • Please help! - Properties of Solids and Liquids - JEE Main-3

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 \tfrac{1}{4}, then the radius r2
will be equal to :

 

  • Option 1)

    r_{1}

  • Option 2)

    2r_{1}

  • Option 3)

    4r_{1}

  • Option 4)

    \frac{r_{1}}{2}

 

Answers (1)

best_answer

Rate of liquid flow through a tube =\frac{\Delta P.\pi r^{4}}{8\eta l}

Since rate of flow is same across the two tubes.

\therefore\ \; \frac{\pi P_{1}r_{1}^{4}}{8\eta l_{1}}=\frac{\pi P_{2}r_{2}^{4}}{8\eta l_{2}}

    \frac{P_{1}r_{1}^{4}}{l_{1}}=\frac{P_{2}r_{2}^{4}}{l_{2}}

Since P_{1}=4P_{1}\ \; \; l_{2}l_{2}=l_{1}4

\therefore\ \; r_{2}^{4}=\frac{r_{1}^{2}}{16}

\therefore\ \; r_{2}=\frac{r_{1}}{2}

Correct option is 4.


Option 1)

r_{1}

This is an incorrect option.

Option 2)

2r_{1}

This is an incorrect option.

Option 3)

4r_{1}

This is an incorrect option.

Option 4)

\frac{r_{1}}{2}

This is an incorrect option.

Posted by

SudhirSol

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