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Physics-

General

Easy

Question

A particle of mass $m$ is initially situated at the point $P$ inside a hemispherical surface of radius $r$ as shown in figure. A horizontal acceleration of magnitude $a subscript 0 end subscript$ is suddenly produced on the particle in the horizontal direction. If gravitational acceleration is neglected, the time taken by particle to touch the sphere again is

$\sqrt{\frac{4 r s i n α}{a_{0}}}$
$\sqrt{\frac{4 r t a n α}{a_{0}}}$
$\sqrt{\frac{4 r c o s α}{a_{0}}}$
None of these

The correct answer is: $square root of fraction numerator 4 r blank c o s alpha over denominator a subscript 0 end subscript end fraction end root$

Let the particle touches the sphere t the point $A.$
Let $P A equals 1$
$therefore P B equals fraction numerator l over denominator 2 end fraction$
In $increment O P B comma cos invisible function application alpha equals fraction numerator P B over denominator r end fraction$

$therefore P B equals r cos invisible function application a$
or $fraction numerator l over denominator 2 end fraction equals r cos invisible function application a$
$therefore l equals 2 r cos invisible function application alpha$
$B u t blank l equals fraction numerator 1 over denominator 2 end fraction a subscript 0 end subscript t to the power of 2 end exponent$
$therefore blank t equals square root of open parentheses fraction numerator 2 l over denominator a subscript 0 end subscript end fraction close parentheses end root equals square root of open parentheses fraction numerator 2 cross times 2 r cos invisible function application a over denominator a subscript 0 end subscript end fraction close parentheses end root equals square root of open parentheses fraction numerator 4 blank r cos invisible function application a over denominator a subscript 0 end subscript end fraction close parentheses end root$

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