One end of string of length $$l$$ is connected to a particle of mass $$'m'$$ and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed $$'v'$$ the net force on the particle (directed towards centre) will be ($$T$$ represents the tension in the string) :-
A.
$$T + \frac{{m{v^2}}}{l}$$
B.
$$T - \frac{{m{v^2}}}{l}$$
C.
Zero
D.
$$T$$
Answer :
$$T$$
Solution :
Net force on particle in uniform circular motion is centripetal force $$\left( {\frac{{m{v^2}}}{l}} \right)$$ which is provided by tension in string so the net force will be equal to tension i.e., $$T.$$
Releted MCQ Question on Basic Physics >> Uniform Circular Motion
Releted Question 1
A small block is shot into each of the four tracks as shown below. Each of the tracks rises to the same height. The speed with which the block enters the track is the same in all cases. At the highest point of the track, the normal reaction is maximum in
A block $$P$$ of mass $$m$$ is placed on a horizontal frictionless plane. A second block of same mass $$m$$ is placed on it and is connected to a spring of spring constant $$k,$$ the two blocks are pulled by distance $$A.$$ Block $$Q$$ oscillates without slipping. What is the maximum value of frictional force between the two blocks.
A ball of mass $$\left( m \right) 0.5kg$$ is attached to the end of a string having length $$\left( L \right)\,0.5m.$$ The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is $$324N.$$ The maximum possible value of angular velocity of ball (in radian/s) is
An annular ring with inner and outer radii $${R_1}$$ and $${R_2}$$ is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring, $$\frac{{{F_1}}}{{{F_2}}}$$ is
A.
$${\left( {\frac{{{R_1}}}{{{R_2}}}} \right)^2}$$