A uniform rope of length L and mass M is hanging from a rigid support. The tension in the rope at a distance x from the rigid support is
A
Mg
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B
Mg(L−xL)
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C
Mg(LL−x)
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D
Mg(xL)
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Solution
The correct option is BMg(L−xL)
Since the rope is uniform, mass per unit length for the rope remain same at each section.
Therefore ML=mL−x(L−x) ⇒mL−x=M(L−xL)
From the FBD, T=(mL−x)g=M(L−xL)g
Hence, tension in the rope at a distance x from the rigid support T=M(L−xL)g