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A string of mass $2.5 \mathrm{~kg}$ is under tension of $200 \mathrm{~N}$. The length of the stretched string is $20.0 \mathrm{~m}$. If the transverse jerk is struck at one end of the string, the disturbance will reach the other end in
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The correct answer is:
$0.5 \mathrm{~s}$
$0.5 \mathrm{~s}$
As given that, Mass $M=2.5 \mathrm{~kg}$
(Mass per unit length)
$$
m=\frac{M}{l}=\frac{2.5 \mathrm{~kg}}{20}=\frac{125}{10}=0.125 \mathrm{~kg} / \mathrm{m}
$$
Speed $v=\sqrt{\frac{T}{m}}=\sqrt{\frac{200}{0.125}}$
[speed of transverse waves in any string]
Distance $l=v \times t$
$$
\begin{aligned}
&\Rightarrow 20=\sqrt{\frac{200}{0.125}} \times t \\
&t=20 \times \sqrt{\frac{125}{2 \times 10^5}}=20 \times \sqrt{\frac{25 \times 5}{2 \times 10^5}} \\
&=20 \times \sqrt{25 \times \frac{1}{0.4 \times 10^5}} \\
&=20 \times 5 \sqrt{\frac{1}{4 \times 10^4}}=\frac{20 \times 5}{2 \times 10^2} \\
&t=\frac{1}{2}=0.5 \mathrm{sec} . \\
&
\end{aligned}
$$
(Mass per unit length)
$$
m=\frac{M}{l}=\frac{2.5 \mathrm{~kg}}{20}=\frac{125}{10}=0.125 \mathrm{~kg} / \mathrm{m}
$$
Speed $v=\sqrt{\frac{T}{m}}=\sqrt{\frac{200}{0.125}}$
[speed of transverse waves in any string]
Distance $l=v \times t$
$$
\begin{aligned}
&\Rightarrow 20=\sqrt{\frac{200}{0.125}} \times t \\
&t=20 \times \sqrt{\frac{125}{2 \times 10^5}}=20 \times \sqrt{\frac{25 \times 5}{2 \times 10^5}} \\
&=20 \times \sqrt{25 \times \frac{1}{0.4 \times 10^5}} \\
&=20 \times 5 \sqrt{\frac{1}{4 \times 10^4}}=\frac{20 \times 5}{2 \times 10^2} \\
&t=\frac{1}{2}=0.5 \mathrm{sec} . \\
&
\end{aligned}
$$
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