For a particle in uniform circular motion the acceleration $\vec { a }$ at a point $P ( R , \theta )$ on the circle of radius R is (here $\theta$ is measured from the $x$-axis)
(1) $- \frac { v ^ { 2 } } { R } \cos \theta \hat { i } + \frac { v ^ { 2 } } { R } \sin \theta \hat { j }$
(2) $- \frac { v ^ { 2 } } { R } \sin \theta \hat { i } + \frac { v ^ { 2 } } { R } \cos \theta \hat { j }$
(3) $- \frac { v ^ { 2 } } { R } \cos \theta \hat { i } - \frac { v ^ { 2 } } { R } \sin \theta \hat { j }$
(4) $\frac { v ^ { 2 } } { R } \hat { i } + \frac { v ^ { 2 } } { R } \hat { j }$

A particle is moving in a circular path of radius $a$, with a constant velocity $v$ as shown in the figure. The centre of circle is marked by 'C'. The angular momentum from the origin O can be written as:
(1) $va(1+\cos 2\theta)$
(2) $va(1+\cos\theta)$
(3) $va\cos 2\theta$
(4) $va$

A particle is moving along a circular path with a constant speed of $10 \mathrm {~ms} ^ { - 1 }$. What is the magnitude of the change in velocity of the particle, when it moves through an angle of $60 ^ { \circ }$ around the centre of the circle?
(1) $10 \sqrt { 3 } \mathrm {~m} / \mathrm { s }$
(2) zero
(3) $10 \sqrt { 2 } \mathrm {~m} / \mathrm { s }$
(4) $10 \mathrm {~m} / \mathrm { s }$

The centre of a wheel rolling on a plane surface moves with a speed $v _ { 0 }$. A particle on the rim of the wheel at the same level as the centre will be moving at a speed $\sqrt { x } v _ { 0 }$. Then the value of $x$ is $\_\_\_\_$ .