A block of mass 2m is hanging at the lower and of a ropeof mass m and length the l, the other end being fixed to the ceililng. A pulse of wavelength `lamda_(0)` is produced at the lower of the rope.

The time taken by the pulse to reach the other end of the rope is:

Options

A. `2sqrt((l)/(g))(sqrt(3 )-1)`
B. `2sqrt((l)/(g))(sqrt(3)-2)`
C. `2sqrt((l)/(g))`
D. `2sqrt((l)/(g))(sqrt(3) -sqrt(2))`

Answer

D

Explanation

Let `v_(x)` be speed of pulses x distance from bottom of rope
Then `v_(x) = sqrt((2m + (max)/(l))(g)/(m//l)) = sqrt((2l + x))g`
Then time taken to go from bottom to top will be
T = `underset(0)overset(L)(int)(dx)/(sqrt((2l +x))g) = 2sqrt((l)/(g))(sqrt(3) -sqrt(2))`.