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QJEEM22Phfcba: JEE Main 2022 Physics - Class 11 JEEM22Phfcba Solution: A ball is released from rest from point P of a smooth…

A ball is released from rest from point P of a smooth semi-spherical vessel as shown in figure. The ratio of the centripetal force and normal reaction on the ball at point Q is A while angular position of point Q is α with respect to point P. Which of the following graphs represent the correct relation between A and α when ball goes from Q to R?

JEE Main 2022 (Online) 26th June Morning Shift Physics - Circular Motion Question 41 English

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/cadd17e52ed97af69e9f14748ea326aa039f47be36f0fac2233a35276e914669.webp
  • B https://ik.imagekit.io/novatra/photos/ba8c886defcf7a203f6959dab111fcfb0e9ed337f52a406bd146a139b6c8e79c.webp
  • C https://ik.imagekit.io/novatra/photos/a24867a51e860c52fff8670d39043f984faf60906e81585fb05e229d9fc90b2a.webp
  • D https://ik.imagekit.io/novatra/photos/17c76e9f5177ec6c35f7e9183553bc17d00d1cc75c22f9a8ed190c0406ec2589.webp

Step-by-Step Verified Solution

JEE Main 2022 (Online) 26th June Morning Shift Physics - Circular Motion Question 41 English Explanation

At Q ;

The equation of motion of the ball is

mv2R=N−mgsin⁡αN=mv2R+mgsin⁡α.......(i)

From figure, h=Rsin⁡α

From law of conservation of energy,

(mg)Rsin⁡α=12mv2

∴mv2/R=2mgsin⁡α\therefore \quad m v^{2} / R=2 m g \sin \alpha

So, Eq. (i) becomes,

N=2mgsin⁡α+mgsin⁡α=3mgsin⁡αN=2 m g \sin \alpha+m g \sin \alpha=3 m g \sin \alpha

∴\therefore Ratio, mv2RN=23=\frac{m v^{2}}{R N}=\frac{2}{3}= constant

A=2/3A=2 / 3 , which is independent of α\alpha .

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