A simple pendulum of length 1 m and having a bob of mass 100 g is suspended in a car, moving on a circular track radius 100 m with uniform speed 10 m/s. If the pendulum makes small oscillation in a radial direction about its equilibrium position, then its time period can be given by $T=2\pi /{\alpha }^{1/4}$ . The value of $\alpha$ is [Take $g=10 m/s^2$ ]

The bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car. 
Acceleration due to gravity = g
Centripetal force acceleration =  $V^2/R$
 Effective acceleration will be 
$g\text{'}=\sqrt{g^2+\frac{v^4}{R^2}}$

Time period, T = $2\pi \sqrt{\frac{L}{g\text{'}}}$

$\Rightarrow \frac{2\pi }{2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}=2\pi \sqrt{\frac{1}{\sqrt{g^2+\frac{v^4}{R^2}}}}$

$\Rightarrow \alpha =g^2+\frac{v^4}{R^2}={\left(10\right)}^2+\frac{{\left(10\right)}^4}{{\left(100\right)}^2}=101$