Question

Difference in height between the liquid columns in the vertical arms in situation (ii) will be

• A. $\frac{{\omega }^2{L}^2}{2g}$
• B. $\frac{L{h}^2{\omega }^2}{Dg}$
• C. $\frac{h^2{(L+2h)}^2{\omega }^{2}d}{HgM}$
• D. $\sqrt{\frac{L{h}^2{\omega }^2}{g}}$

Answer 1

The correct option is A $\frac{{\omega }^2{L}^2}{2g}$

$dF=\left(dm\right){\omega }^{2}x\Rightarrow \left(dP\right)A=\left(\rho Adx\right){\omega }^{2}x$

$\Rightarrow {\int }_{P1}^{P2}dP=\rho {\omega }^2{\int }_0^{L}xdx$

$\Rightarrow P_2-P_1=\rho g{h}_1=\rho {\omega }^2{L}^2/2$

$\Rightarrow h_1=\frac{{\omega }^2{L}^2}{2g}$