A) \[\frac{1}{2}F\]
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 :
Force F is applied \[\therefore \] Acceleration in string\[=\frac{F}{M}\] \[m=\]mass of string mass per unit length of string \[=\frac{M}{L}\] Tension at a distance \[(l)\] \[=m\times a=mass\text{ }of\text{ }AB\times a\] \[\frac{m}{L}(L-l)\times \frac{F}{M}=\left( 1-\frac{1}{L} \right)F\]You need to login to perform this action.
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