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Electrostatics

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

Question image for Eshan sir... TensionT(h)= m(h/l)g How sir h/l is

raju nagula

8 Years agoGrade

1 Answer

Eshan

Approved Tutor Answer8 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 length $h$ will 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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