Problem 63 - A Complete Resource Book in Physics for JEE Main

Question

One end of a string of length \(l\) is connected to a particle of mass \(m\) and the other to a small peg on a smooth horizontal table. If the particle moves in a circular motion with speed \(v\), the net force on the particle (directed towards the centre) is: (A) \(T\) (B) \(T-\frac{m v^{2}}{l}\) (C) \(T+\frac{m v^{2}}{l}\) (D) 0

Step-by-Step Solution

Answer

(A) \(T\)

Step 1: Write down the given information and the formula for centripetal force

We are given: - Mass of the particle, \(m\) - Length of the string, \(l\) - Speed of the particle, \(v\) The formula for centripetal force (\(F_c\)) acting on a particle moving in a circle of radius \(r\) with speed \(v\) is given by: \(F_c = \frac{m v^2}{r}\) In this case, the radius of the circle is equal to the length of the string, \(l\). So, the centripetal force (directed towards the center) acting on the particle is: \(F_c = \frac{m v^2}{l}\)

Step 2: Compare the centripetal force with the given options

Now let's compare the centripetal force, \(\frac{m v^2}{l}\), with the given options: (A) \(T\) (B) \(T - \frac{m v^2}{l}\) (C) \(T + \frac{m v^2}{l}\) (D) 0 We can see that none of the options match the given centripetal force directly. However, considering that the net force directed towards the center would be equal to the centripetal force, we can safely assume that the tension in the string (\(T\)) must be equal to the centripetal force acting on the particle. In other words, the tension in the string provides the centripetal force required to keep the particle moving in a circle. Therefore, the net force acting on the particle towards the center is equal to the centripetal force, which is equal to the tension in the string (\(T\)). The correct answer is: (A) \(T\).