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# A uniform rope of mass M and length L is fixed to its upper end vertically from a rigid support. Thus the tension in the rope at the distance I from the rigid support is ${\text{A}}{\text{. Mg}}\dfrac{{\text{L}}}{{{\text{L + I}}}} \\ {\text{B}}{\text{. }}\dfrac{{{\text{Mg}}}}{{\text{L}}}({\text{L - I)}} \\ {\text{C}}{\text{. Mg}} \\ {\text{D}}{\text{. }}\dfrac{{\text{I}}}{{\text{L}}}{\text{Mg}} \\$

Last updated date: 20th Jun 2024
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Hint : Tension is described as the pulling force transmitted axially by the means of a string, a cable, chain, or by each end of a rod.

Complete step by step solution:
Mass = M
Length = L
Tension = mass × acceleration due to gravity
T = mg
Now we have to do the differentiating we get
dT = dmg
${\text{dm = }}\dfrac{{{\text{Mdx}}}}{{\text{L}}}$
Integrating from I to L
${\text{T = }}\dfrac{{{\text{Mdxg}}}}{{\text{I}}} = \dfrac{{{\text{Mg(L - I)}}}}{{\text{L}}}$ 
The ends of a string or the other object transmitting tension will exert forces on the object to which the string or the rod is connected in the direction of the string at the point of attachment.

Thus option B is the correct answer.