Question

A portion of a tube shown in the figure. Fluid is flowing from cross-section area $A_1$ to $A_2$. The two cross-sections are at a distance ${}^{\prime }{l}^{\prime }$ from each other. The velocity of the fluid at section $A_2$ is $\sqrt{\frac{gl}{2}}$. If the pressure at $A_1$ & $A_2$ is the same, then the angle made by the tube with the horizontal will be.

• A. ${37}^{\circ }$
• B. $si{n}^{-1}\frac{3}{4}$
• C. ${53}^{\circ }$
• D. None of these

Answer 1

The correct option is B $si{n}^{-1}\frac{3}{4}$

equation of continuity

if $P_1=P_2$

the ${\rho }_1{A}_1{v}_1={\rho }_{2}A{v}_2[{\rho }_1={\rho }_2=\rho ]$

$A_1{v}_1=A_2{v}_2$

$A{v}_1=2A{v}_2$

$v_1=2v_2$

& given $v_2=\sqrt{\frac{9l}{2}}$

So $v_1=\sqrt{2gl}$

by energy balance

$△KE=mgl\sin \theta$

$\frac{1}{2}m(v_1^2-v_2^2)=mgl\sin \theta$

$(2gl-\frac{gl}{2})=2gl\sin \theta$

$\frac{3}{2}=2\sin \theta$

$\theta ={\sin }^{-1}\frac{3}{4}$