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Correct option is D)

Force when it reaches bottom is $F_{b}=rmv_{2} +mg$

Now using energy balance between topmost point and bottom most point ,

$21 mv_{2}=21 m(rg )_{2}+mgh$ (h is the height difference between the top and bottom point ; $gr $ is critical velocity at topmost point )

$v_{2}=rg+2gh$

or, $v_{2}=rg+4gr=5gr$

So, $F_{b}=6mg$

Using this in: $e_{1}e_{2} =F_{1}F_{2} $, we get:

$e_{2}=1.5×6=9.0cm$

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