A car is moving on a plane inclined at 30 ° to the horizontal with an… | JEE Main 2021 (31 Aug, Shift 1)

Q17JEE Main 2021 · 31 AUG · SHIFT 1Difficult

Equilibrium, Third Law & Frames

A car is moving on a plane inclined at 30 $^\circ$ to the horizontal with an acceleration of 10 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 g = 10 ms ${}^{-2}$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 30

In the car frame, effective gravity $\vec{g}_{eff} = \vec{g} - \vec{a}$ ; the string lies along $\vec{g}_{eff}$ .
$\vec{a}$ : horizontal $10\cos 30^\circ = 5\sqrt{3}$ , vertical $10\sin 30^\circ = 5$ (upward).
$g_{eff}$ : horizontal $5\sqrt{3}$ , vertical $10 + 5 = 15$
$\tan\phi = \frac{5\sqrt{3}}{15} = \frac{1}{\sqrt{3}} \Rightarrow \phi = 30^\circ$
So the angle with the vertical is $30^\circ$ .

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