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Question
In the arrangement shown in figure , the string has a mass of 4⋅5 g. How much time will it take for a transverse disturbance produced at the floor to reach the pulley? Take g = 10 m s−2.
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Solution
Given,
Mass of the block = 2 kg
Total length of the string = 2 + 0.25 = 2.25 m
Mass per unit length of the string:
\[m = \frac{4 . 5 \times {10}^{- 3}}{2 . 25}\]
\[ = 2 \times {10}^{- 3} kg/m\]
\[T = 2g = 20 N\]
\[\text{ Wave speed,} \nu = \sqrt{\left( \frac{T}{m} \right)}\]
\[ = \sqrt{\frac{20}{\left( 2 \times {10}^{- 3} \right)}}\]
\[ = \sqrt{{10}^4} \]
\[ = {10}^2 m/s = 100 \text{ m/s }\]
Time taken by the disturbance to reach the pulley:
\[t = \left( \frac{s}{\nu} \right)\]
\[ = \frac{2}{100} = 0 . 02 s\]
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