A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter\[\text{AB=D}\]. The height h is equal to [NEET - 2018] |
A) \[\frac{7}{5}D\]
B) \[D\]
C) \[\frac{3}{2}D\]
D) \[\frac{5}{4}D\]
Correct Answer: D
Solution :
[d] |
As track is frictionless, so total mechanical energy will remain constant |
\[\text{T}\text{.M}\text{.}{{\text{E}}_{\text{l}}}\text{=T}\text{.M}\text{.}{{\text{E}}_{\text{F}}}\] |
\[\text{0+mgh=}\frac{1}{2}m{{v}_{L}}^{2}+0\] |
\[h=\frac{v_{L}^{2}}{2g}\] |
For completing the vertical circle, \[{{v}_{L}}\ge \sqrt{5gR}\] |
\[h=\frac{5gR}{2g}=\frac{5}{2}R=\frac{5}{4}D\] |
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