1. Choose the non-inertial frame of the car

Since the car is accelerating up the incline with acceleration $a = 10\,\text{m s}^{-2}$, in the car's frame the bob experiences a pseudo force opposite to the car's acceleration, i.e. down the incline.

2. Forces on the bob in the car frame

The bob is in equilibrium relative to the car, so the string aligns along the resultant of:

• Weight: $mg$ vertically downward
• Pseudo force: $ma$ down the incline

Given: $g = 10\,\text{m s}^{-2}, \qquad a = 10\,\text{m s}^{-2}$

3. Resolve the pseudo acceleration into horizontal and vertical components

The incline makes $30^\circ$ with the horizontal. So the direction down the incline is $30^\circ$ below the horizontal.

Hence the pseudo acceleration $a$ has components:

• Horizontal component: $a\cos 30^\circ$
• Vertical downward component: $a\sin 30^\circ$

So the effective gravity components are:

• Horizontal: $g_x = a\cos 30^\circ = 10\cdot \frac{\sqrt{3}}{2} = 5\sqrt{3}$

• Vertical downward: $g_y = g + a\sin 30^\circ = 10 + 10\cdot \frac{1}{2} = 15$

4. Find the angle made by the string with the vertical

If the string makes angle $\theta$ with the vertical, then $\tan\theta = \frac{\text{horizontal component}}{\text{vertical component}}$

Therefore, $\tan\theta = \frac{5\sqrt{3}}{15} = \frac{\sqrt{3}}{3}$

So, $\theta = 30^\circ$

5. Final answer

The string makes an angle $\boxed{30^\circ}$ with the vertical.