QJEEM21Ph56fc: JEE Main 2021 Physics - Class 11 JEEM21Ph56fc Solution: A car is moving on a plane inclined at 30 to the horizontal…

Given,

Angle of inclination, $\theta$ = 30 $^\circ$

Acceleration, a = 10 ms ^{$-$ 2}

Acceleration due to gravity, g = 10 ms ^{$-$ 2}

According to the question the car and bob is as shown below,

Here, F' is the pseudo force acting on the bob when we considered it from car's frame and T is the tension on the string.

In equilibrium, $\Sigma$ F_x = 0 and $\Sigma$ F_y = 0

$\Rightarrow$ F' cos30 $^\circ$ = T sin $\alpha$

${{ma\cos 30^\circ } \over {\sin \alpha }} = T$ ...... (i)

where, m is the mass of the bob.

$F'\sin 30^\circ + mg = T\cos \alpha$

$\Rightarrow ma\sin 30^\circ + mg = {{ma\cos 30^\circ } \over {\sin \alpha }}(\cos \alpha )$ [ $\because$ using Eq. (i)]

$\Rightarrow a\sin 30^\circ + g = {{a\cos 30^\circ } \over {\sin \alpha }}(\cos \alpha )$

$10 \times {1 \over 2} + 10 = {{10 \times \sqrt 3 } \over 2}\cot \alpha$

$\Rightarrow {1 \over 2} + 1 = {{\sqrt 3 } \over 2}\cot \alpha \Rightarrow {3 \over 2} = {{\sqrt 3 } \over 2}\cot \alpha$

or, $\cot \alpha = \sqrt 3$

$\Rightarrow \alpha = 30^\circ$