Question

A rope of length L and mass M is hanging from a rigid support. The tension in the rope at a distance l from the rigid support is:
$\left( A \right)Mg$
$\left( B \right)\left( {\dfrac{{L - l}}{L}} \right)Mg$
$\left( C \right)\left( {\dfrac{L}{{L - l}}} \right)Mg$
$\left( D \right)\dfrac{1}{L}Mg$

Answer 1

Hint:First of all, draw a rough diagram. We need to find the tension at a distance l from the rigid support so first take a small portion of small mass at a distance of l from the top of the rope. After that, balance the small tension and gravitational force at a distance of l then integrate that small tension and gravitational force to get the total tension at a distance of l.

Complete step by step solution:
As per the problem A rope of length L and mass M is hanging from a rigid support.
We need to calculate the tension in the rope at a distance l from the rigid support is.
Given that,
Mass of the rope = M
Length of the rope = L
Tension on the rope = T
Take a small portion of mass dm where dT is the small tension acting at a distance of l from the rigid support,
Hence small tension dT is equals to,
$dT = dmg$
Integrating both the side we will get,
$\int {dT} = \int {dmg} $
Here we can calculate the small mass as,
$\dfrac{{dm}}{{dx}} = \dfrac{M}{L}$
Rearranging the above equation we will get,
$dm = \dfrac{M}{L}dx$
Where dx is the small portion length.
Now putting this small mass in the above tension formula we will get,
$\int {dT} = \int {\dfrac{M}{L}dx\,g} $
The small portion length varies from $0\,to\,L - l$ and the tension varies from $0\,to\,T$.
Now the integration the above equation we will get,
$\int\limits_0^T {dT} = \int\limits_0^{L - l} {\dfrac{M}{L}dx\,g} $
After integration we will get,
$\left[ T \right]_0^T = \dfrac{M}{L}g\left[ x \right]_0^{L - l}$
After putting the limit we will get,
$T = \dfrac{M}{L}g\left( {L - l} \right)$
Taking L as coonm terms we will get,
$T = \dfrac{M}{L}gL\left( {1 - \dfrac{l}{L}} \right)$
Cancelling the common terms we will get,
$T = Mg\left( {1 - \dfrac{l}{L}} \right)$
On further solving we will get,
$T = Mg\left( {\dfrac{{L - l}}{L}} \right)$
Therefore the correct option is $\left( B \right)$.

Note:
Be careful while taking the limit in this type of problem. Here first take the base of the rope as reference so that you can get the limit easily and get the answer in less time. The rope will experience different tension at different positions of the length and the mass of the room also differs.