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A string of mass 4.25 Kg is under a tension of  200 N. The length of the stretched string is  40 cm. If the  transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?

Option: 1

0.0922 sec


Option: 2

0.0822 sec


Option: 3

0.0722 sec


Option: 4

0.0622 sec


Answers (1)

Tension in the string  T =200 N

Mass of string  M=4.25 Kg

\text { length } \,l=40 \mathrm{~cm}=40 \times 10^{-2} \mathrm{~m}

\therefore Mass \,\,\,Per\,\,\, unit\,\,\, length ,\,\mu=\frac{M}{l}

=\frac{4.25}{40 \times 10^{-2}}=\frac{425}{4000 \times 10^{-2}}=\frac{425}{4 \times 10}=\frac{425}{40} \\

=10.625 \mathrm{~Kg} / \mathrm{m} \\
\therefore \text { Velocity }V=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{200}{10.625}} \\

=\sqrt{\frac{200 \times 10^3}{10625}}=\sqrt{\frac{20 \times 10^4}{10625}}=\sqrt{\frac{20}{10625}} \times 10^2 \\

=\frac{4.47}{103.07} \times 10^2=\frac{447}{10307} \times 10^2 \\

=0.0434 \times 10^2 \mathrm{~m} / \mathrm{s}
time taken by the disturbance to reach other end,  f=\frac{l}{V}

=\frac{40 \times 10^{-2}}{0.0434 \times 10^2}=\frac{40 \times 10^{-2} \times 10^{-2} \times 10^4}{434} \\

=\frac{40}{434}=0.0922 \mathrm\,{sec}

Posted by

Sumit Saini

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