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A particle is given an initial speed u inside a smooth spherical shell of radius $R=1m$ that it is just able to complete the circle. Acceleration of the particle when its velocity is vertical is

$(a)\;g \sqrt {10} \quad (b)\;g \quad (c)\;g \sqrt 2 \quad (d)\;g \sqrt 6$
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Since the body is given just enough speed to complete the vertical circle
$u^2=5gR$
$v^2=u^2-2gR$
$\quad=5gR-2gR$
$\quad=3gR$
When the velocity is vertical, the body is acted by two acceleration $a_T='g'$ tangentially and centripetal acceleration $a_c=\large\frac{v^2}{R}$ radially.
$a_c=\large\frac{v^2}{R}=\frac{3gR}{R}$
$\quad=3g$
Total acceleration $=\sqrt {a_T^4+a_c^2}$
$\quad=\sqrt {g^2+3g^2}$
$\quad=g\sqrt {10}$
Hence a is the correct answer.

answered Jul 19, 2013 by
edited Feb 10, 2014 by meena.p
The V can also be found out using theorem of conservation of machanical energy.

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