9. For light incident from air on a meta-material, the appropriate ray diagram is

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{(A)}
  \includegraphics[alt={},max width=\textwidth]{4e050d28-8dd5-4115-bb8a-1c7b16f3381f-05_521_518_1518_262}
\end{center}
\end{figure}

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{(B)}
  \includegraphics[alt={},max width=\textwidth]{4e050d28-8dd5-4115-bb8a-1c7b16f3381f-05_521_509_1518_1088}
\end{center}
\end{figure}

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{(C)}
  \includegraphics[alt={},max width=\textwidth]{4e050d28-8dd5-4115-bb8a-1c7b16f3381f-05_508_533_2056_266}
\end{center}
\end{figure}

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{(D)}
  \includegraphics[alt={},max width=\textwidth]{4e050d28-8dd5-4115-bb8a-1c7b16f3381f-05_505_514_2052_1095}
\end{center}
\end{figure}

\begin{enumerate}
  \setcounter{enumi}{9}
  \item Choose the correct statement.\\
(A) The speed of light in the meta-material is $v = c | n |$\\
(B) The speed of light in the meta-material is $v = \frac { c } { | n | }$\\
(C) The speed of light in the meta-material is $v = \mathrm { c }$.\\
(D) The wavelength of the light in the meta-material $\left( \lambda _ { m } \right)$ is given by $\lambda _ { m } = \lambda _ { \text {air } } | n |$, where $\lambda _ { \text {air } }$ is the wavelength of the light in air.
\end{enumerate}

ANSWER : B

\section*{Paragraph for Questions 11 and 12}
The $\beta$-decay process, discovered around 1900, is basically the decay of a neutron ( $n$ ). In the laboratory, a proton ( $p$ ) and an electron ( $e ^ { - }$) are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process, i.e. $n \rightarrow p + e ^ { - } + \bar { v } _ { e }$, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino $\left( \bar { v } _ { e } \right)$ to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is $0.8 \times 10 ^ { 6 } \mathrm { eV }$. The kinetic energy carried by the proton is only the recoil energy.\\