A car is moving on a plane inclined at $30^{0}$ to the horizontal with an acceleration of $10 m s^{- 2}$ 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_____

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answer is 30.

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Detailed Solution

$\tan (30 + θ) = \frac{m g \sin 30^{0} + m a}{m g \cos 30^{0}} = \frac{5 + 10}{5 \sqrt{3}} = \frac{1 + 2}{\sqrt{3}} 2 \sqrt{3} \tan θ = 2 \Rightarrow \tan θ = \frac{1}{\sqrt{3}} H e n c e θ = 30^{0}$

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