A) 1.8 N
B) zero
C) 9 N
D) 0.9N
Correct Answer: B
Solution :
The force is given by \[F=600-2\times {{10}^{5}}t=0\] So, time for which force acts is \[t=\frac{600}{2\times {{10}^{5}}}\] \[=3\times {{10}^{-3}}\sec \] \[\therefore \] Average impulse \[\text{= force }\!\!\times\!\!\text{ time}\] \[=(600-2\times {{10}^{5}}\times 3\times {{10}^{-3}})\] \[=0\]You need to login to perform this action.
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