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# The block is placed on a frictionless surface in gravity free space. A heavy string of a mass m is connected and force F is applied on the string, then the tension at the middle of rope is:A.$\dfrac{\left( \dfrac{m}{2}+M \right)F}{m+M}$B.$\dfrac{\left( \dfrac{M}{2}+m \right)F}{m+M}$C. zeroD.$\dfrac{MF}{m+M}$

Last updated date: 13th Jul 2024
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Hint We know that tension is nothing but the drawing force acting on the body when it is hung from objects like chain, cable, string etc. It is represented by T. The direction of tension is the pull which is given the name tension. Thus, the tension will point away from the mass in the direction of the string or rope. In case of the hanging mass, the string pulls it upwards, so the string or rope exerts an upper force on the mass and the tension will be on the upper side. The tension force is the force that is transmitted through a string, rope, cable or wire when it is pulled tight by forces acting from opposite ends. The tension force is directed along the length of the wire and pulls equally on the objects on the opposite ends of the wire. Based on this concept we have to solve this question.

Now we can say that due to force $F$, the whole system would move with an acceleration of $\mathrm{a}=\dfrac{\mathrm{F}}{\mathrm{M}+\mathrm{m}}$
$\therefore \mathrm{T}=\left(\mathrm{M}+\dfrac{\mathrm{m}}{2}\right) \dfrac{\mathrm{F}}{\mathrm{M}+\mathrm{m}}$