A U tube is rotated about one of it's limbs with an angular velocity `omega`. Find the difference in height H of the liquid (density `rho`) level, where diameter of the tube `dlt lt L`.

Answer

`H=(L^(2)omega^(2))/(2g)`

Explanation

For circular motion of small element dx `dF=(dm)xomega^(2)`
`therefore (dp)xxA=(pAdx)xxxomega^(2)`
or `dp=rho omega^(2)xdx`
or `underset(P_(1))overset(P_(2))(int)dp=rho omega^(2)underset(0)overset(L)(int)xdx`
or `(p_(2)-p_(1))=rho omega^(2)[(x^(2))/(2)]_(0)^(L)`
or `p_(2)-p_(1)=(rho omega^(2)L^(2))/(2)`
or `rho gH=(rho omega^(2)L^(2))/(2)` or `H=(omega^(2)L^(2))/(2g)`