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Physics-

General

Easy

Question

In figure, a particle is placed at the highest point $A$ of a smooth sphere of radius $r$ . It is given slight push, and it leaves the sphere at $B$ , at a depth $h blank$ vertically below $A$ such that $h$ is equal to

$\frac{r}{6}$
$\frac{1}{4} r$
$\frac{1}{3} r$
$\frac{1}{2} r$

The correct answer is: $fraction numerator 1 over denominator 3 end fraction r$

If $v$ velocity acquired at $B$ , then
$v to the power of 2 end exponent equals 2 g h$
The particle will leave the sphere at $B$ , when
$fraction numerator m v to the power of 2 end exponent over denominator r end fraction greater or equal than m g cos invisible function application theta$
$fraction numerator 2 g h over denominator r end fraction equals fraction numerator g left parenthesis r minus h right parenthesis over denominator r end fraction comma$

Which gives $h equals fraction numerator r over denominator 3 end fraction$

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