A string of length L and mass M is lying on a horizontal table. A force F is applied at one of its ends. Tension in the string at a distance x from the end at which force is applied is

Options

A. Zero
B. F
C. `(F(L-y))/(L)`
D. `(F(l-y))/(M )`

Answer

C

Explanation

Acceleration of the rope ` a= (F )/(M)`

Now , considering the motion of the part AB of the rope [ which has mas `(M//L)y` and acceleration given by eq(I) ] assuming that tension at B is T
` F-T =((M)/(L ) y) xx a or F- T =(M )/(L ) y xx (F) /(M )=(Fy)/(L )`
` or t-F -F (y)/(L ) =F (1- (y)/(L ) ) = (F(L-y))/( L)`