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Properties of Matter question

2020 · 2 Sep · Shift 1 · Q53
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Properties of Matter question

2020 · 2 Sep · Shift 1 · Q53

JEE MainPhysicsProperties of MatterMCQ+4 / −1
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 radius of vessel is 5 cm and the angular speed of rotation is ω\omegaω rad s–1. The difference in the height, h (in cm) of liquid at the centre of vessel and at the side will be : JEE Main 2020 (Online) 2nd September Morning Slot Physics - Properties of Matter Question 212 English
  1. A
    2ω225g{{2{\omega ^2}} \over {25g}}25g2ω2​
  2. B
    5ω22g{{5{\omega ^2}} \over {2g}}2g5ω2​
  3. C
    25ω22g{{25{\omega ^2}} \over {2g}}2g25ω2​
  4. D
    2ω25g{{2{\omega ^2}} \over {5g}}5g2ω2​
View written solutionFree

Correct answer: C

  1. Shape of free surface in a rotating liquid

When a liquid rotates with angular speed ω\omegaω about a vertical axis, its free surface becomes a paraboloid. The height difference between two points at distances r1r_1r1​ and r2r_2r2​ from the axis is

Δh=ω22g(r22−r12).\Delta h = \frac{\omega^2}{2g}\left(r_2^2-r_1^2\right).Δh=2gω2​(r22​−r12​).
  1. Apply to centre and side of the vessel
  • At the centre: r1=0r_1=0r1​=0
  • At the side: r2=R=5 cmr_2=R=5\text{ cm}r2​=R=5 cm

So the difference in height between side and centre is

h=ω22g(R2).h = \frac{\omega^2}{2g}(R^2).h=2gω2​(R2).
  1. Substitute R=5R=5R=5 cm

Since the options are in cm, we use R=5R=5R=5 cm directly:

h=ω22g(52)=25ω22g.h = \frac{\omega^2}{2g}(5^2) = \frac{25\omega^2}{2g}.h=2gω2​(52)=2g25ω2​.

Thus,

h=25ω22g.\boxed{h=\frac{25\omega^2}{2g}}.h=2g25ω2​​.
  1. Check options
  • A: 2ω225g\dfrac{2\omega^2}{25g}25g2ω2​ ✗
  • B: 5ω22g\dfrac{5\omega^2}{2g}2g5ω2​ ✗
  • C: 25ω22g\dfrac{25\omega^2}{2g}2g25ω2​ ✓
  • D: 2ω25g\dfrac{2\omega^2}{5g}5g2ω2​ ✗

Therefore, the correct option is C.

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