Laws of Motion - Result Question 21
21. 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) :
[2017]
(a) $T+\frac{mv^{2}}{l}$
(b) $T-\frac{mv^{2}}{l}$
(c) Zero
(d) $T$
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Answer:
Correct Answer: 21. (d)
Solution:
- (d) Net force on particle in uniform circular motion is centripetal force $(\frac{m v^{2}}{\ell})$ which is provided by tension in string so the net force will be equal to tension i.e., $T$.
