A car is moving on a plane inclined at 30^{ci | vedclass

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(C) In the frame of the car,the bob experiences a pseudo force $ma$ acting down the incline. The effective gravitational force components are $mg \sin

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$30$

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(C) In the frame of the car,the bob experiences a pseudo force $ma$ acting down the incline. The effective gravitational force components are $mg \sin 30^{\circ}$ (down the incline) and $mg \cos 30^{\circ}$ (perpendicular to the incline).Let $\alpha$ be the angle the string makes with the normal to the inclined plane. The forces acting on the bob in the frame of the car are:$1$. Pseudo force $ma$ (down the incline).$2$. Component of gravity $mg \sin 30^{\circ}$ (down the incline).$3$. Component of gravity $mg \cos 30^{\circ}$ (perpendicular to the incline).The angle $\alpha$ with the normal is given by $	an \alpha = \frac{ma + mg \sin 30^{\circ}}{mg \cos 30^{\circ}}$.Substituting the values: $	an \alpha = \frac{10m + 10m \sin 30^{\circ}}{10m \cos 30^{\circ}} = \frac{10 + 5}{5\sqrt{3}} = \frac{15}{5\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$.So,$\alpha = 60^{\circ}$.The angle the string makes with the vertical is $	heta = \alpha - 30^{\circ} = 60^{\circ} - 30^{\circ} = 30^{\circ}$.

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