$A$ cylindrical vessel partially filled with water is rotated about its vertical central axis. Its surface will

  • A
    Rise equally
  • B
    Rise from the sides
  • C
    Rise from the middle
  • D
    Lowered equally

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The diagram shows a cup of tea seen from above. The tea has been stirred and is now rotating without turbulence. $A$ graph showing the speed $v$ with which the liquid is crossing points at a distance $X$ from $O$ along a radius $XO$ would look like:

Two blocks of ice when pressed together join to form one block because

$A$ table tennis ball has radius $(3 / 2) \times 10^{-2} \text{ m}$ and mass $(22 / 7) \times 10^{-3} \text{ kg}$. It is slowly pushed down into a swimming pool to a depth of $d = 0.7 \text{ m}$ below the water surface and then released from rest. It emerges from the water surface at speed $v$,without getting wet,and rises up to a height $H$. Which of the following option$(s)$ is (are) correct?
[Given: $\pi = 22 / 7, g = 10 \text{ ms}^{-2}$,density of water $= 1 \times 10^3 \text{ kg m}^{-3}$,viscosity of water $= 1 \times 10^{-3} \text{ Pa-s}$.]
$(A)$ The work done in pushing the ball to the depth $d$ is $0.077 \text{ J}$.
$(B)$ If we neglect the viscous force in water,then the speed $v = 7 \text{ m/s}$.
$(C)$ If we neglect the viscous force in water,then the height $H = 1.4 \text{ m}$.
$(D)$ The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is $500 / 9$.

$A$ cylinder of radius $4 \ cm$ and height $10 \ cm$ is immersed in two liquids as shown. Specific gravity of oil is $0.5$. $2 \ cm$ of the cylinder is in the air. Select the $INCORRECT$ statement. Neglect atmospheric pressure.

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