A pendulum of  length 1cm hangs from an inclined wall. Suppose that this pendulum is released at an initial angle of  ${10}^{\circ }$ and it bounces off the wall elastically when it reaches an angle of  $-5^{\circ }$ as shown in the figure. (Take $g={\text{π}}^2\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}$) The period of this pendulum is (in second)

In  right half of oscillation of bob it oscillates ${10}^{\circ }$ so amplitude is  ${\theta }_0=\left(10°\right)$

In left half it undergoes only $5°$ due to the obstruction of wall. It travels only $\left(5°\right)$ that is $\frac{{\text{θ}}_0}{2}$.
Using the equation for angular SHM  $\text{θ}={\text{θ}}_0\sin \omega t$ ……( 1 )
Time (t) to travel from 0 to  $\frac{{\text{θ}}_{\text{0}}}{\text{2}}$,substituting $\text{θ}=\frac{{\text{θ}}_{\text{0}}}{\text{2}}$ in equation ( 1 )
$\frac{{\text{θ}}_{\text{0}}}{\text{2}}={\text{θ}}_0\sin \omega t$
$\Rightarrow \omega t={30}^{\circ }$
$\left(\because \omega =\frac{2\pi }{T}\right)$
$\Rightarrow t=\frac{T}{12}$
Simple harmonic oscillator takes  $\frac{T}{12}$ seconds. So total time period of the oscillation is
$T\text{'}=\frac{T}{4}+\frac{T}{12}+\frac{T}{4}+\frac{T}{12}=\frac{2}{3}T$
where $T=2\text{π}\sqrt{\frac{l}{g}}=2s$
$\therefore T\text{'}=\frac{2}{3}.2=\frac{4}{3}=1.33s$
Therefore, the correct answer is 1.33