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A liquid is taken in a long vertical cylindrical vessel and the cylinder is rotated about its vertical axis as shown in the figure. During rotation, the liquid rises along its sides. If the radius of vessel is $0.05 \mathrm{~m}$ and speed of rotation is $10 \mathrm{rad} \mathrm{s}^{-1}$, then the height difference between the liquid at the centre of the vessel and its sides is $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$
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Verified Answer
The correct answer is:
$125 \times 10^{-4} \mathrm{~m}$
From the Bernoulli's theorem
$$
P_1+\frac{1}{2} \rho v_1^2+\rho g h_1=P_2+\frac{1}{2} \rho v_1^2+\rho g h_2
$$
Hire $h_1=h_2$
$$
\begin{aligned}
& P_1-P_2=\rho\left(v_1^2-v_1^2\right) \\
& v_1=0 v_2=r \omega \\
& \rho g h=\frac{1}{2} \rho r^2 \omega^2 \Rightarrow h=\frac{r^2 \omega^2}{2 g} \\
& h=\frac{(0.05)^2 \times 100}{2 \times 10}=125 \times 10^{-4} \mathrm{~m}
\end{aligned}
$$
$$
P_1+\frac{1}{2} \rho v_1^2+\rho g h_1=P_2+\frac{1}{2} \rho v_1^2+\rho g h_2
$$
Hire $h_1=h_2$
$$
\begin{aligned}
& P_1-P_2=\rho\left(v_1^2-v_1^2\right) \\
& v_1=0 v_2=r \omega \\
& \rho g h=\frac{1}{2} \rho r^2 \omega^2 \Rightarrow h=\frac{r^2 \omega^2}{2 g} \\
& h=\frac{(0.05)^2 \times 100}{2 \times 10}=125 \times 10^{-4} \mathrm{~m}
\end{aligned}
$$
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