Q. A heavy uniform rope changes vertically from the ceiling, with its lower end free. $A$ disturbance on the rope travelling upwards from the lower end has a velocity $v$ at a distance $x$ from the lower end such that
A) $v \propto x$
B) $v \propto \sqrt{x}$
C) $v \propto \frac{1}{x}$
D) $v \propto \frac{1}{\sqrt{x}}$
Explanation
(b) Let m=mass per unit length of rope
T=tension in the rope at a distance x from the lower end
$\therefore$ T=$(mg)x = weight\ of\ x\ meter\ of\ rope$
As $v = \sqrt{\frac{T}{m}}\ \ \therefore v = \sqrt{\frac{mgx}{m}} = \sqrt{gx}$
$ie\ \ v \propto \sqrt{x}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.