Question

A right angled tube is fixed horizontally on a horizontal surface and an ideal liquid of density p is flowing into the tube at the rate of $Q=4m^3/s$ Cross sectional area of the tube at intake and outlet positions are $A=2m^2$ and $S=1m^2$ respectively. The magnitude of net force exerted by liquid on the tube is (Given that the value of atmospheric pressure $P_o=4pN/m^2$)

• A. $3\sqrt{12}p$
• B. $4\sqrt{41}p$
• C. $5\sqrt{21}p$
• D. None of there

Answer 1

The correct option is B $4\sqrt{41}p$

From momentum conservation principle,

$F_{net}=\rho Q(v_f-v_i)$

in X-direction,

$P_1{A}_1-F_X=\rho Q(0-\frac{Q}{A_1})$

$\Rightarrow P_{0}A-F_X=\rho Q(0-\frac{Q}{A})$

$\Rightarrow 4p\times 2-F_X=p\times 4(0-\frac{4}{2})$

$\Rightarrow F_X=16p$

in Y-direction,

$F_Y-P_2{A}_2=\rho Q(\frac{Q}{A_2}-0)$

$\Rightarrow F_Y-P_{0}S=\rho Q(\frac{Q}{S}-0)$

$\Rightarrow F_Y-4p\times 1=p\times 4(\frac{4}{1}-0)$

$\Rightarrow F_Y=20p$

So, net force=$\sqrt{{\left(F_X\right)}^2+{\left(F_Y\right)}^2}$

$\Rightarrow F=\sqrt{{\left(16p\right)}^2+{\left(20p\right)}^2}$

$\Rightarrow F=4\sqrt{41}p$