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.

Answer 1

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: 

\[t = \left( \frac{s}{\nu} \right)\] 

\[   = \frac{2}{100} = 0 . 02  s\]