Question

A car is moving on a plane inclined at $30^{\circ}$ to the horizontal with an acceleration of $10 \,m/s^{2}$ parallel to the plane upward. A bob is suspended by a string from the roof. The angle in degrees which the string makes with the vertical is: (Assume that the bob does not move relative to car) $[g= 10 \,m/s^{2}]$

• A. $20^{\circ}$
• B. $30^{\circ}$
• C. $45^{\circ}$
• D. $60^{\circ}$

Answer 1

Answer: (B)

In $F.B.D$. of the bob resolving the components of pseudo force and tension in horizontal and vertical direction

$T\,sin\,\alpha=ma\,cos\,\theta \ldots(i)$
$T\,cos\,\alpha=ma\,sin\,\alpha + mg \ldots(ii)$
From (i) and (ii), $tan\, \alpha=\frac{a\,sin\,\theta}{a\,sin\,\theta+g}$
Here $a = 10 \,m/s^{2}$,
$\theta=30^{\circ}$
After solving we get $tan \,\alpha=\frac{1}{\sqrt{3}}$ or $\alpha=30^{\circ}$