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A string of mass 2.5 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, the disturbance will reach the other end in

(a) one second

(b) 0.5 second

(c) 2 seconds

(d) data given is insufficient.

0 2 Views | Posted 3 months ago
Asked by Shiksha User

  • 1 Answer

  • P

    Answered by

    Payal Gupta | Contributor-Level 10

    3 months ago

    This is a multiple choice answer as classified in NCERT Exemplar

    (b) m=2.5kg

    μ = mass per unit length

    =m/l =2.5/20=125/10=0.125kg/m

    V= T μ = 200 0.125

    L=v × t

    20= 125 2 × 10 5 = 20 × 25 × 5 2 × 10 5

    = 20 × 25 0.4 × 10 5 = 1 2

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