A) \[F\]
B) \[\frac{l}{L}F\]
C) \[\frac{L}{l}F\]
D) \[\left( 1-\frac{l}{L} \right)F\]
Correct Answer: D
Solution :
Mass of length\[l=mL\], where m is mass per unit length of the rope. From Newtons 2nd law, \[Force=mass\times acceleration\] \[F=mL\,\,a\] \[\Rightarrow \] \[a=\frac{F}{mL}\] Tension in rope at distance I from the given end \[T=m(L-l)a\] \[=m(L-l)\frac{F}{mL}\] \[T=\left( 1-\frac{l}{L} \right)F\]You need to login to perform this action.
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