A string of length L is fixed at one end and carries a mass of M at the other end. The mass makes left(frac3pi ight) rotations per second about the vertical axis passing through end of the string as shown. The tension in the string is ____ ML.

Numerical Answer Type:
Enter a numerical value Answer: 36 to 36 +4 marks

Solution & Explanation

### Related Formula For a conical pendulum: T cos theta = Mg T sin theta = Momega^2 R where R = L sin theta. ### Core Logic Substituting R = L sin theta into the centripetal equation:
Conical pendulum string force vectors diagram Q22
Conical pendulum string force vectors diagram Q22
T sin theta = M omega^2 (L sin theta) T = M omega^2 L Given rotational frequency: f = frac3pi\ mathrmrev/s Angular velocity: omega = 2pi f = 2pi left(frac3pi ight) = 6\ mathrmrad/s Substituting omega back into the simplified tension equation: T = M (6)^2 L = 36\ ML ### Pattern Recognition In a conical pendulum, the horizontal projection of tension provides the exact centripetal force. The sintheta terms cancel out, making string tension independent of the semi-vertical angle. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

Reference Study Guides

More Laws of Motion Previous-Year Questions — Page 6

Q47 jee_main_2024_31_jan_morning Circular Motion Friction
A coin is placed on a disc. The coefficient of friction between the coin and the disc is mu. If the distance of the coin from the center of the disc is r, the maximum angular velocity which can be given to the disc, so that the coin does not slip away, is:
  • A. fracmu gr
  • B. sqrtfracrmu g
  • C. sqrtfracmu gr
  • D. fracmusqrtrg

Solution

### Related Formula f_s leq mu_s N F_c = mromega^2 ### Core Logic
Circular Motion Friction diagram for Q47 - JEE Main 2024 Morning
Circular Motion Friction diagram for Q47 - JEE Main 2024 Morning
Circular Motion Friction diagram for Q47 - JEE Main 2024 Morning
Circular Motion Friction diagram for Q47 - JEE Main 2024 Morning
To prevent the coin from slipping, the static friction must provide the necessary centripetal force for circular motion. f = momega^2 r The normal force on the flat disc is N = mg. The maximum static friction is f_textmax = mu N = mu mg. For no slipping: m r omega^2 leq mu mg omega^2 leq fracmu gr omega_textmax = sqrtfracmu gr ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws Of Motion

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