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 (l1/4), then the radius r2 will be equal to :

Question

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 (l1/4), then the radius r2 will be equal to : (A) r1 (B) 2 r1 (C) 4 r1 (D) (r1/2)

Tardigrade Admin · Accepted Answer

(φ v/dt)=(π/8) (pr4/nL) (p1r41/L1)=(p2r42/L2) (p1r41/l2)=(4p1r42/l1/4)=r24=(r14/16) r2=(r1/2)

$r_{1}$

$2 r_{1}$

$4 r_{1}$

$\frac{r _{1}}{2}$

$\frac{ϕ v}{d t} = \frac{π}{8} \frac{p r ^{4}}{n L}$
$\frac{p _{1} r _{1}^{4}}{L _{1}} = \frac{p _{2} r _{2}^{4}}{L _{2}}$
$\frac{p _{1} r _{1}^{4}}{l _{2}} = \frac{4 p _{1} r _{2}^{4}}{l _{1/4}} = r_{2}^{4} = \frac{r _{1}^{4}}{16}$
$r_{2} = \frac{r _{1}}{2}$

r1​

2r1​

4r1​

2r1​​

dtϕv​=8π​nLpr4​ L1​p1​r14​​=L2​p2​r24​​ l2​p1​r14​​=l1/4​4p1​r24​​=r24​=16r14​​ r2​=2r1​​

$r_{1}$

$2 r_{1}$

$4 r_{1}$

$\frac{r _{1}}{2}$

$\frac{ϕ v}{d t} = \frac{π}{8} \frac{p r ^{4}}{n L}$
$\frac{p _{1} r _{1}^{4}}{L _{1}} = \frac{p _{2} r _{2}^{4}}{L _{2}}$
$\frac{p _{1} r _{1}^{4}}{l _{2}} = \frac{4 p _{1} r _{2}^{4}}{l _{1/4}} = r_{2}^{4} = \frac{r _{1}^{4}}{16}$
$r_{2} = \frac{r _{1}}{2}$