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In the figure shown all pulleys are light and there is no friction between pulley and strings. Masses $m_{1} = m_{2} = 2 kg$ and all strings have a linear mass density $μ = 3 \times 10^{- 2} kg / m$ which makes masses of strings negligibly small compared to m₁ and m₂. Initially separation between pulley P₂ and mass m₂is 20 cm. At this instant system is released from rest and a transverse pulse is created at the location of m₂

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Pulse will reach P₂ in time 0.01s

Pulse will reach P₂ in time 0.1s

Pulse with reach P₂ in time 1s

Pulse will reach P₂ in time 10s

answer is A.

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Detailed Solution

We have
2T – 20 = 2b ….. (i)
20 – T = 2a ….. (ii)

a – 2b = 0 $\Rightarrow$ a = 2b ….. (iii)
2T – 20 = 2b
40 – 2T = 8b

$20 = 10 b \Rightarrow b = 2 m / s^{2} ∴ a = 4 m / s^{2}$

$\begin{array}{l} Also T = 12 N ∴ V_{wave} = \sqrt{\frac{12}{3 \times 10^{- 2}}} = 20 m / s \\ a_{pulley} = 2 m / s^{2} ∴ 0.2 = 20 t - \frac{1}{2} 2 t^{2} \\ \Rightarrow t^{2} - 20 t + 0.2 = 0 t = 0.01 s \end{array}$

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