Q.
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 $a_{0}$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
A) $\sqrt{\frac{4r\ \sin\alpha}{a_{0}}}$
B) $\sqrt{\frac{4r\ \tan\alpha}{a_{0}}}$
C) $\sqrt{\frac{4r\ \cos\alpha}{a_{0}}}$
D) None of these
Explanation
(c) Let the particle touches the sphere t the point $A.$
Let $PA = 1$
$\therefore PB = \frac{l}{2}$
In $\mathrm{\Delta}OPB,\cos{\alpha = \frac{PB}{r}}$
[image:3_image199.emf]
$\therefore PB = r\cos a$
or $\frac{l}{2} = r\cos a$
$\therefore l = 2r\cos\alpha$
$But\ l = \frac{1}{2}a_{0}t^{2}$
$\therefore\ t = \sqrt{\left( \frac{2l}{a_{0}} \right)} = \sqrt{\left( \frac{2 \times 2r\cos a}{a_{0}} \right)} = \sqrt{\left( \frac{4\ r\cos a}{a_{0}} \right)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.