Fluids 2

6 Answers

1357
Manish Shankar ·

yeah consider it less that H/2 as shown in the figure, for Q1 and Q2

106
Asish Mahapatra ·

sir, even if we consider h < H/2 and h > H/2 for (3) I am not getting 3H/4 ..

24
eureka123 ·

Ans1
v_e=\sqrt{\frac{2g[(\rho \frac{H}{2})+2\rho (\frac{H}{2}-h)]}{2 \rho}}

24
eureka123 ·

Ans2
-h=-\frac{gt^2}{2}=>t=\sqrt{\frac{2h}{g}}

and x=v_et

=> x=\sqrt{3Hh-4h^2}

24
eureka123 ·

v_e=\sqrt{\frac{2g[(\rho \frac{H}{2})+2\rho (\frac{H}{2}-y)]}{2 \rho}} ;t=\sqrt{\frac{2y}{g}};x=v_e.t

For xmax
\frac{dx}{dy}=0

1357
Manish Shankar ·

For third one I think there will be only one answer
both the answer are incorrect

the answer I m getting is h=H/2+0

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