A block is lying on the horizontal frictionless surface. One end of a uniform rope is fixed to the block which is pulled in the horizontal direction by applying a force F at the other end. If the mass of the rope is half the mass of the block, the tension in the middle of the rope will be nF/6. Find the value of n (Neglect sagging of rope under its own weight)
Physics · Mixed Chapter
JEE Mains

Question

A block is lying on the horizontal frictionless surface. One end of a uniform rope is fixed to the block which is pulled in the horizontal direction by applying a force F at the other end. If the mass of the rope is half the mass of the block, the tension in the middle of the rope will be nF/6. Find the value of n (Neglect sagging of rope under its own weight)

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Let block mass be M and rope mass M/2. With external pull F and no horizontal friction, total acceleration a=F/(M+M/2)=2F/(3M). Step 2: Tension at the rope midpoint must accelerate the block plus the half-rope behind that point, total mass M+(M/4)=5M/4. Step 3: midpoint tension=(5M/4)a=(5M/4)(2F/3M)=5F/6. Step 4: Comparing with nF/6 gives n=5. Answer: 5.