A) \[\frac{l}{2}F\]
B) \[\frac{L}{l}F\]
C) \[\left( 1-\frac{l}{L} \right)F\]
D) \[\left( 1+\frac{l}{L} \right)F\]
Correct Answer: C
Solution :
Mass per unit length of string \[=\frac{M}{L}\]acceleration in string due to force F \[a=\frac{F}{M}\] Tension at a distance ( from end (Q) is T = force on part PO = mass of part \[PO\times \]acceleration or \[T=\frac{M}{L}(L-l)\times \frac{F}{M}\] \[\Rightarrow \] \[T=\left( 1-\frac{l}{L} \right)F\]You need to login to perform this action.
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