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A car is moving on a plane inclined at 30^{a} 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} )
 

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best_answer



\tan \theta = \frac{a+g\sin 30^{\circ}}{g\cos 30^{\circ}}
a= g= 10
\tan \theta = \frac{10+10\times \frac{1}{2}}{10\times \frac{\sqrt{3}}{2}}= \frac{15\times 2}{10\sqrt{3}}= \frac{\sqrt{3}}{1}
\theta = 60^{\circ}
Angle made by the steing with no of 60^{\circ} & with vertical it is 30^{\circ}
 

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vishal kumar

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