JEE Main & Advanced Physics NLM, Friction, Circular Motion JEE PYQ - NLM Friction Circular Motion

Sand is being dropped on a conveyer belt at the rate of 2 kg per second. The force necessary to keep the belt moving with a constant speed of 3 \[\text{m}{{\text{s}}^{\text{-1}}}\]will be [JEE ONLINE 19-05-2012]

A) 12N

B) 6N

C) Zero

D) 18N

Correct Answer: B

Solution :

[b] Here, rate of fall of sand on conveyer belt,\[\frac{dm}{dt}=2kg\,{{s}^{-1}}\]

Conveyer belt speed \[v=3\,m{{s}^{-1}}\]

Force F = ?

\[F=v\frac{dm}{dt}=3\times 2=6N\]