jee-main 2025 Q11

jee-main · India · session2_04apr_shift2 Not Maths
Q11. The number of electrons flowing per second in the filament of a 110 W bulb operating at 220 V is : ( Given $\mathrm { e } = 1.6 \times 10 ^ { - 19 } \mathrm { C }$ )
(1) $6.25 \times 10 ^ { 17 }$
(2) $1.25 \times 10 ^ { 19 }$
(3) $6.25 \times 10 ^ { 18 }$
(4) $31.25 \times 10 ^ { 17 }$
Q12. Match List-I with List-II :
(A) $\mathrm { Y } =$ magnetic susceptibility $\mathrm { X } =$ magnetising field
(B) $\mathrm { Y } =$ magnetic field
(C) $\mathrm { Y } =$ magnetic field $\mathrm { X } =$ distance from centre of a current carrying wire for $x < a$ (where $a =$ radius of wire)
List-1
Y vs X
List-II
Shape of Graph (II) (I) [Figure]
(II) [Figure]
(III) [Figure]
\begin{verbatim} X = distance from centre of a current carrying wire for x>a (where $a =$ radius of wire) \end{verbatim}
(D) $\mathrm { Y } =$ magnetic field inside solenoid
(IV) [Figure]
$\mathrm { X } =$ distance from centre Choose the correct answer from the options given below :
(1) (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
(2) (A)-(I), (B)-(III), (C)-(II), (D)-(IV)
(3) (A)-(III), (B)-(IV), (C)- (I), (D)-(II)
(4) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
Q11. The number of electrons flowing per second in the filament of a 110 W bulb operating at 220 V is : ( Given $\mathrm { e } = 1.6 \times 10 ^ { - 19 } \mathrm { C }$ )\\
(1) $6.25 \times 10 ^ { 17 }$\\
(2) $1.25 \times 10 ^ { 19 }$\\
(3) $6.25 \times 10 ^ { 18 }$\\
(4) $31.25 \times 10 ^ { 17 }$

Q12.\\
Match List-I with List-II :\\
(A) $\mathrm { Y } =$ magnetic susceptibility\\
$\mathrm { X } =$ magnetising field\\
(B) $\mathrm { Y } =$ magnetic field\\
(C) $\mathrm { Y } =$ magnetic field\\
$\mathrm { X } =$ distance from centre of a current carrying wire for $x < a$ (where $a =$ radius of wire)

\section*{List-1}
Y vs X

\section*{List-II}
Shape of Graph\\
(II)\\
(I)\\
\includegraphics[max width=\textwidth, alt={}, center]{0b8a1bc0-351f-46d1-9f54-c9f2aa997d2e-03_271_260_376_938}

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{(II)}
  \includegraphics[alt={},max width=\textwidth]{0b8a1bc0-351f-46d1-9f54-c9f2aa997d2e-03_300_266_671_934}
\end{center}
\end{figure}

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{(III)}
  \includegraphics[alt={},max width=\textwidth]{0b8a1bc0-351f-46d1-9f54-c9f2aa997d2e-03_278_275_1114_927}
\end{center}
\end{figure}

\begin{verbatim}
X = distance from centre of a current carrying
wire for x>a
(where $a =$ radius of wire)
\end{verbatim}

(D) $\mathrm { Y } =$ magnetic field inside solenoid

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{(IV)}
  \includegraphics[alt={},max width=\textwidth]{0b8a1bc0-351f-46d1-9f54-c9f2aa997d2e-03_259_250_1535_936}
\end{center}
\end{figure}

$\mathrm { X } =$ distance from centre\\
Choose the correct answer from the options given below :\\
(1) (A)-(IV), (B)-(I), (C)-(III), (D)-(II)\\
(2) (A)-(I), (B)-(III), (C)-(II), (D)-(IV)\\
(3) (A)-(III), (B)-(IV), (C)- (I), (D)-(II)\\
(4) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)