A uniform chain of length and total mass hangs from a support

PHYSICS · Single correct · CBSE · Physics · Laws of Motion · Laws of Motion

Problem

A uniform chain of length LL and total mass MM hangs from a support. The entire chain is pulled upward with a constant acceleration aa (where 0<a<g0 < a < g). Let xx be the distance measured downward from the support. Which expression gives the tension T(x)T(x) at position xx while the chain is accelerating? (a) T(x)=ML(g+a)x\displaystyle T(x)=\frac{M}{L}(g+a)x (b) T(x)=MLg (L−x)\displaystyle T(x)=\frac{M}{L}g\,(L-x) (c) T(x)=ML(g+a)(L−x)\displaystyle T(x)=\frac{M}{L}(g+a)(L-x) (d) T(x)=ML(g−a)(L−x)\displaystyle T(x)=\frac{M}{L}(g-a)(L-x)

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