Difficult
$A$ cylindrical vessel containing a liquid is rotated about its axis so that the liquid rises at its sides as shown in the figure. The diameter of the vessel is $10 \, cm$ and the angular speed of rotation is $\omega \, rad \, s^{-1}$. The difference in the height,$h$ (in $cm$),of the liquid at the centre of the vessel and at the side will be
- A$\frac{25 \omega^{2}}{2 g}$
- B$\frac{2 \omega^{2}}{5 g}$
- C$\frac{5 \omega^{2}}{2 g}$
- D$\frac{2 \omega^{2}}{25 g}$
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View SolutionGiven below are two statements:
Statement $I$: When the speed of a liquid is zero everywhere,the pressure difference at any two points depends on the equation $P_1-P_2=\rho g(h_2-h_1)$.
Statement $II$: In the venturi tube shown,$2gh=v_2^2-v_1^2$.
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Statement $I$: When the speed of a liquid is zero everywhere,the pressure difference at any two points depends on the equation $P_1-P_2=\rho g(h_2-h_1)$.
Statement $II$: In the venturi tube shown,$2gh=v_2^2-v_1^2$.
In the light of the above statements,choose the most appropriate answer from the options given below.
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View Solution$A$ thin vertical uniform wooden rod is pivoted at the top and immersed in water as shown. The container is slowly raised. At a certain moment,the equilibrium becomes unstable. If the density of water is $9/5$ times the density of wood,then the ratio of the total length of the rod to the submerged length of the rod at that moment is:
$A$ large number of droplets,each of radius $r$,coalesce to form a bigger drop of radius $R$. An engineer designs a machine so that the energy released in this process is converted into the kinetic energy of the drop. The velocity of the drop is ($T=$ surface tension,$\rho =$ density)
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