A car is moving on a plane inclined at $30^{o}$ 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. The angle in degrees which the string makes with the vertical , for equilibrium of the bob, is: (Assume that the bob does not move relative to car) [ g= l0 $m / s^{2}$ ]

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$20^{0}$

$30^{0}$

$45^{0}$

$60^{0}$

answer is D.

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Detailed Solution

In F.B.D of the bob resolving the components of pseudo force and tension in horizontal and vertical direction.

$T \sin α = ma \cos θ$ -------(i)

$T \cos α = ma \sin α + mg$ ---------(ii)

From (i) and (ii), $\tan α = \frac{a \sin θ}{a \sin θ + g}$

Here a = 10 $m / s^{2}$ , $θ = 30^{0}$

After solving we get $\tan α = \frac{1}{\sqrt{3}} or α = 30^{0}$

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