MCQ
A body of mass $m$ is rotating in a vertical circle of radius $'r\ '$ with critical speed. The difference in its $K.E.$ at the top and the bottom is $.............$
- ✓$\text{2mgr}$
- B$\text{4mgr}$
- C$\text{6mgr}$
- D$\text{3mgr}$
✓
Answer
Correct option: A.
$\text{2mgr}$
The change of kinetic energy will be equet to the change in potential energy.
$\triangle\text{KE}=\triangle\text{PE}=\text{mg}(\text{h}_1-\text{h}_2)$
Since radious $r,$
$h_1- h_2= r$
$\triangle\text{KE}=\triangle\text{PE}=\text{mg}(\text{h}_1-\text{h}_2)$
Since radious $r,$
$h_1- h_2= r$
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