A car is moving on a plane inclined at $30^{\circ}$ to the horizontal with an acceleration of $10 \mathrm{~ms}^{-2}$ parallel to the plane upward. A bob is suspended by a string from the roof of the car.The angle in degrees which the string makes with the vertical is (Take $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )
$\tan (30+\theta)=\frac{m g \sin 30^{\circ}+m a}{m g \cos 30^{\circ}}$
$\tan (30+\theta)=\frac{5+10}{5 \sqrt{3}}=\frac{1+2}{\sqrt{3}}$
$\frac{\tan \theta+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}} \tan \theta}=\sqrt{3}$
$\sqrt{3} \tan \theta+1=3-\sqrt{3} \tan \theta$
$2 \sqrt{3} \tan \theta=2$
$\tan \theta=\frac{1}{\sqrt{3}}$
$\theta=30^{\circ}$
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