A uniform chain of mass m and length l hangs from a hook in ceiling as shown in fig I. The bottom of link is now raised vertically with the help of upward external force $F (y)$ very slowly as shown in fig. II. Choose CORRECT option(s)

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$F (y) = \frac{m g y}{2 l}$

Work done by $F (y)$ in raising end B to the position of end A is $\frac{m g l}{4}$

$F (y) = \frac{m g y}{4 l}$

Work done by $F (y)$ in raising end B to the position of end A is $\frac{m g l}{8}$

answer is B, C.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Uniform Chain Raised Slowly: Tension and Force Analysis

For a uniform chain of mass m and length l hanging from a hook, where the bottom is raised upward very slowly:

The tension at a distance y from the bottom is equal to the weight of the portion of the chain below that point:
Tension at y = (m/l) g y
The required upward force F(y) to just lift the chain at point y increases linearly with y.
The tension at the bottom of the chain is zero, because there is no chain below that point to support.
The maximum tension is at the topmost point, equal to the total weight of the chain: T_top = mg.