Question

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) $T+\frac{mv^{2}}{l}$(b) $T-\frac{mv^{2}}{l}$(c) Zero(d) T

Answer 1

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Question 5.4 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 circle with speed v the net force on the particle (directed towards the centre) is:

(i) T,
(ii) T-((mv^2) / l),
(iii) T+((mv^2) / l),
(iv) 0

T is the tension in the string. [Choose the correct alternative].


We know,
When , a particle tied with sting and moves in circular , the required centripital force = tension in string

Here length of string 'l ' is connected to a particle of mass m . after moving of string in a circle with speed v.
Centripital force = mv²/l directed towards the centre . and tension in string is T
Hence, net Force of string
directed towards centre = centripital force = Tension in string
e.g T = mv²/l
hence , correct alternative is T