JEE Main 2021 — Laws Of Motion Question with Solution

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$$\alpha$$

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

where, m is the mass of the bob.

$$F^{\prime }\sin 30^{\circ }+mg=T\cos \alpha$$

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

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

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

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

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

$$\Rightarrow \alpha =30^{\circ }$$