Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

One end of the 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 a circle with speed $v$ , the net force on the particle (directed towards the centre) will be: ( $T$ represents the tension in the string)

$T$

$T + \frac{mv^2}{l}$

$T - \frac{mv^2}{l}$

Zero

Step-by-Step Solution

Identify Forces: Consider the forces acting on the particle. In the vertical direction, the weight ( $mg$ ) is balanced by the normal reaction ( $N$ ) from the table. In the horizontal plane, the only force acting on the particle is the tension ( $T$ ) exerted by the string directed towards the peg (centre).
Net Force: According to Newton's Second Law, the net force ( $F_{net}$ ) acting on an object is the vector sum of all individual physical forces acting on it. Here, $F_{net} = T$ .
Centripetal Force Concept: The term $\frac{mv^2}{l}$ represents the mass times the centripetal acceleration ( $ma_c$ ). It is not an independent force of nature but the result of the net force. The equation of motion is $T = \frac{mv^2}{l}$ .
Conclusion: The question asks for the net force, which is provided solely by the tension. Thus, the net force is $T$ .

(Reference: NCERT Class 11, Physics Part I, Chapter 4: Laws of Motion, Exercise 4.4).

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