A sand dropper drops sand of mass m(t) on a conveyer belt at a rate proportional to the square root of speed (v) of the belt, i.e. fracdmdt propto sqrtv . If P is the power delivered to run the belt at constant speed then which of the following relationship is true?

Solution & Explanation

### Related Formula F_textthrust = left(fracdmdtright)v P = F cdot v ### Core Logic Given the sand dropping rate rule: fracdmdt = C sqrtv quad (textwhere C text is a constant) To maintain a constant velocity v, the continuous force applied by the conveyor system must balance the rate of gain of momentum of the dropped sand: F = left(fracdmdtright) v = (C sqrtv) cdot v = C v^3/2 Power delivered is the product of force and speed: P = F cdot v = (C v^3/2) cdot v = C v^5/2 Squaring both sides of the expression: P^2 propto v^5 ### Pattern Recognition In variable mass problems involving dropping dust/sand at rest onto a moving frame, the thrust force always simplifies to v cdot fracdmdt, which makes power scale as v^2 cdot fracdmdt. ### 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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