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Eshan sir... TensionT(h)= m(h/l)g How sir h/l is coming. I did not understand please explain sir.

Eshan sir...
TensionT(h)= m(h/l)g
How sir h/l is coming.
I did not understand please explain sir.

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1 Answers

Eshan
askIITians Faculty 2095 Points
5 years ago
Dear Raju,

Tension at any point is balancing the weight of the rope below it.
The mass per unit length of the rope can be written as\lambda=\dfrac{m}{l}
Hence the mass of part of rope hanging uptil lengthhwill be\lambda h=\dfrac{m}{l}h

Hence the weight of that part will be\dfrac{m}{l}hg.

Or you could see it this way- The fraction of length the given part is of the whole rope, in the same fraction would be its mass.

Hence\dfrac{m'}{m}=\dfrac{h}{l}
\implies m'=m(\dfrac{h}{l})

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