Question

A weight Mg ig suspended from the middle of a rope whose ends are at the same level. The rope is no longer horizontal. The minimum tension required to completely straighten the rope is

• A. $\frac{Mg}{2}$
• B. $Mg\cos \theta$
• C. 2$Mg\cos \theta$
• D. Infinitely large

Answer 1

The correct option is D Infinitely large

$\begin{array}{c}Mg=2T\sin \theta \\ T=\frac{Mg}{2\sin \theta }\\ Ropehorizontal\theta =0\\ T=Mg/2\sin \theta \\ \frac{Mg}{2\times 0}=\infty \end{array}$