Answer
a) $\omega=\sqrt{\dfrac{g}{L\sin\theta}}$
b) $71.6\;\rm rpm$
Work Step by Step
a) First, we need to draw the force diagram of the object.
The net force exerted on the object in the $x$-direction, which is the radial direction of rotation, is given by
$$\sum F_r=T\cos\theta =ma_r=m\dfrac{v^2}{R}$$
We know that $v=\omega R$;
$$T\cos\theta =m\dfrac{(\omega R)^2}{R}=mR\omega^2$$
Thus,
$$T=\dfrac{mR\omega^2 }{\cos\theta}\tag 1$$
Noting that $R$ here is given by
$$\cos\theta =\dfrac{R}{L}$$
where $L$ is the length of the wire.
Thus, $$R=L\cos\theta$$
Plugging into (1);
$$T=\dfrac{mL\cos\theta\omega^2 }{\cos\theta}=mL\omega^2$$
$$T= mL\omega^2\tag 2 $$
Applying Newton's second law in the vertical direction.
$$\sum F_y=T\sin\theta-mg=ma_y=m(0)=0$$
$$T\sin\theta=mg$$
Thus,
$$T=\dfrac{mg}{\sin\theta}$$
Plugging into (2);
$$\dfrac{ \color{red}{\bf\not}mg}{\sin\theta}= \color{red}{\bf\not}mL\omega^2 $$
$$\dfrac{ g}{\sin\theta}= L\omega^2 $$
Therefore,
$$\boxed{\omega=\sqrt{\dfrac{g}{L\sin\theta}}}$$
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b) Plugging the given data into the boxed formula above to find the angular velocity of the rock.
$$\omega=\sqrt{\dfrac{g}{L\sin\theta}}=\sqrt{\dfrac{9.8}{1\times \sin10^\circ}}=\bf 7.5\;\rm rad/s$$
$$\omega=\rm \dfrac{7.5\;\rm rad}{\rm 1\;s}\left(\dfrac{1\;rev}{2\pi\;rad}\right) \left(\dfrac{60\;s}{1\;min}\right) =\bf 71.6\;rev/min$$
Therefore,
$$\omega=\rm \color{red}{\bf 71.6}\;rpm$$
