A) \[\sqrt{\frac{5L}{m}(qE+mg)}\]
B) \[\sqrt{\frac{5L}{m}(2qE+mg)}\]
C) \[\sqrt{\frac{5L}{m}(qE+2mg)}\]
D) \[\sqrt{\frac{5L}{m}(2qE+3mg)}\]
Correct Answer: C
Solution :
[c] \[{{T}_{P}}-mg+qE=\frac{mv_{P}^{2}}{L}\] \[{{T}_{P}}=15mg\] \[mv_{P}^{2}=L\,\,(14mg+qE)\] ...(i) From work-energy theorem \[(qE-mg)2L=\frac{1}{2}mv_{Q}^{2}-\frac{1}{2}mv_{P}^{2}\] \[{{V}_{Q}}=\sqrt{(5qE+10mg)\frac{L}{m}}=\sqrt{\frac{5L}{m}(qE+2mg)}\]You need to login to perform this action.
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