NEET Test Series from KOTA - 10 Papers In MS WORD
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Electro Magnetic Induction
154860
An inductor coil of inductance $L$ is cut into two equal parts and both the parts are connected in parallel. The net inductance is :
1 $\mathrm{L}$
2 $\mathrm{L} / 2$
3 $\mathrm{L} / 4$
4 $2 \mathrm{~L}$.
Explanation:
C If two inductors are connected in parallel combination then, $\mathrm{L}_{\mathrm{eq}}=\frac{\mathrm{L}_{1} \mathrm{~L}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}$ Since, we know that inductance of conductor directly proportional to its length $\therefore \quad \mathrm{L}_{1}=\mathrm{L}_{2}=\mathrm{L} / 2$ $\text { So, } \quad \mathrm{L}_{\text {eq }}=\frac{\frac{\mathrm{L}}{2} \times \frac{\mathrm{L}}{2}}{\frac{\mathrm{L}}{2}+\frac{\mathrm{L}}{2}}$ $\mathrm{~L}_{\mathrm{eq}}=\frac{\mathrm{L}}{4}$
AIIMS-2011
Electro Magnetic Induction
154861
With the decrease of current in the primary coil from 2 amperes to zero value in $0.01 \mathrm{~s}$ the emf generated in the secondary coil is $\mathbf{1 0 0 0}$ volts. The mutual inductance of the two coils is
1 $1.25 \mathrm{H}$
2 $2.50 \mathrm{H}$
3 $5.00 \mathrm{H}$
4 $10.00 \mathrm{H}$
Explanation:
C Given that, Primary current $\mathrm{I}_{1}=2 \mathrm{~A}$ Secondary current $\mathrm{I}_{2}=0 \mathrm{~A}$ Time $\Delta \mathrm{t}=0.01 \mathrm{sec}$ emf $(\varepsilon)=1000$ volts Mutual inductance $(\mathrm{M})=-\frac{\varepsilon}{(\mathrm{dI} / \mathrm{dt})}=-\frac{1000}{(0-2) / 0.01}$ $\mathrm{M}=5 \mathrm{H}$
AIIMS-2007
Electro Magnetic Induction
154862
The coefficient of mutual inductance, when magnetic flux changes by $2 \times 10^{-2}$ Wb and current changes by $0.01 \mathrm{~A}$ is
1 8 Henry
2 4 Henry
3 3 Henry
4 2 Henry
Explanation:
D Given, Change in Magnetic flux, $\mathrm{d} \phi=2 \times 10^{-2} \mathrm{~Wb}$ Change in current $\mathrm{dI}=0.01 \mathrm{~A}$ $\mathrm{M}=?$ $\mathrm{M}=\frac{\mathrm{d} \phi}{\mathrm{dI}}=\frac{2 \times 10^{-2}}{0.01}$ $\mathrm{M}=2 \text { Henry }$
AIIMS-2002
Electro Magnetic Induction
154864
In an inductor of self-inductance $L=\mathbf{2} \mathrm{mH}$, Current changes with time according to relation $i=t^{2} e^{-t}$. At what time emf is zero?
1 $4 \mathrm{~s}$
2 $3 \mathrm{~s}$
3 $2 \mathrm{~s}$
4 $1 \mathrm{~s}$
Explanation:
C Given, $\mathrm{L}=2 \mathrm{mH}$, and $\mathrm{i}=\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}$ We know that. $\text { emf }=\text { L. } \frac{\mathrm{di}}{\mathrm{dt}}$ Let at time $\mathrm{t}$ emf is zero. $\therefore \quad \mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}=0$ \(0=2 \times 10^{-3}\left(\frac{d}{d t}\left(t^{\frac{I I}{2}} \cdot e^{-\mathrm{t}}\right)\right)\) $0=2 \mathrm{te}^{-\mathrm{t}}+\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}(-1)$ $2 \mathrm{te}^{-\mathrm{t}}-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}=0$ $\mathrm{te}^{-\mathrm{t}}(2-\mathrm{t})=0$ $\therefore \quad 2-\mathrm{t}=0$ $\mathrm{t}=2 \mathrm{sec}$
154860
An inductor coil of inductance $L$ is cut into two equal parts and both the parts are connected in parallel. The net inductance is :
1 $\mathrm{L}$
2 $\mathrm{L} / 2$
3 $\mathrm{L} / 4$
4 $2 \mathrm{~L}$.
Explanation:
C If two inductors are connected in parallel combination then, $\mathrm{L}_{\mathrm{eq}}=\frac{\mathrm{L}_{1} \mathrm{~L}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}$ Since, we know that inductance of conductor directly proportional to its length $\therefore \quad \mathrm{L}_{1}=\mathrm{L}_{2}=\mathrm{L} / 2$ $\text { So, } \quad \mathrm{L}_{\text {eq }}=\frac{\frac{\mathrm{L}}{2} \times \frac{\mathrm{L}}{2}}{\frac{\mathrm{L}}{2}+\frac{\mathrm{L}}{2}}$ $\mathrm{~L}_{\mathrm{eq}}=\frac{\mathrm{L}}{4}$
AIIMS-2011
Electro Magnetic Induction
154861
With the decrease of current in the primary coil from 2 amperes to zero value in $0.01 \mathrm{~s}$ the emf generated in the secondary coil is $\mathbf{1 0 0 0}$ volts. The mutual inductance of the two coils is
1 $1.25 \mathrm{H}$
2 $2.50 \mathrm{H}$
3 $5.00 \mathrm{H}$
4 $10.00 \mathrm{H}$
Explanation:
C Given that, Primary current $\mathrm{I}_{1}=2 \mathrm{~A}$ Secondary current $\mathrm{I}_{2}=0 \mathrm{~A}$ Time $\Delta \mathrm{t}=0.01 \mathrm{sec}$ emf $(\varepsilon)=1000$ volts Mutual inductance $(\mathrm{M})=-\frac{\varepsilon}{(\mathrm{dI} / \mathrm{dt})}=-\frac{1000}{(0-2) / 0.01}$ $\mathrm{M}=5 \mathrm{H}$
AIIMS-2007
Electro Magnetic Induction
154862
The coefficient of mutual inductance, when magnetic flux changes by $2 \times 10^{-2}$ Wb and current changes by $0.01 \mathrm{~A}$ is
1 8 Henry
2 4 Henry
3 3 Henry
4 2 Henry
Explanation:
D Given, Change in Magnetic flux, $\mathrm{d} \phi=2 \times 10^{-2} \mathrm{~Wb}$ Change in current $\mathrm{dI}=0.01 \mathrm{~A}$ $\mathrm{M}=?$ $\mathrm{M}=\frac{\mathrm{d} \phi}{\mathrm{dI}}=\frac{2 \times 10^{-2}}{0.01}$ $\mathrm{M}=2 \text { Henry }$
AIIMS-2002
Electro Magnetic Induction
154864
In an inductor of self-inductance $L=\mathbf{2} \mathrm{mH}$, Current changes with time according to relation $i=t^{2} e^{-t}$. At what time emf is zero?
1 $4 \mathrm{~s}$
2 $3 \mathrm{~s}$
3 $2 \mathrm{~s}$
4 $1 \mathrm{~s}$
Explanation:
C Given, $\mathrm{L}=2 \mathrm{mH}$, and $\mathrm{i}=\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}$ We know that. $\text { emf }=\text { L. } \frac{\mathrm{di}}{\mathrm{dt}}$ Let at time $\mathrm{t}$ emf is zero. $\therefore \quad \mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}=0$ \(0=2 \times 10^{-3}\left(\frac{d}{d t}\left(t^{\frac{I I}{2}} \cdot e^{-\mathrm{t}}\right)\right)\) $0=2 \mathrm{te}^{-\mathrm{t}}+\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}(-1)$ $2 \mathrm{te}^{-\mathrm{t}}-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}=0$ $\mathrm{te}^{-\mathrm{t}}(2-\mathrm{t})=0$ $\therefore \quad 2-\mathrm{t}=0$ $\mathrm{t}=2 \mathrm{sec}$
154860
An inductor coil of inductance $L$ is cut into two equal parts and both the parts are connected in parallel. The net inductance is :
1 $\mathrm{L}$
2 $\mathrm{L} / 2$
3 $\mathrm{L} / 4$
4 $2 \mathrm{~L}$.
Explanation:
C If two inductors are connected in parallel combination then, $\mathrm{L}_{\mathrm{eq}}=\frac{\mathrm{L}_{1} \mathrm{~L}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}$ Since, we know that inductance of conductor directly proportional to its length $\therefore \quad \mathrm{L}_{1}=\mathrm{L}_{2}=\mathrm{L} / 2$ $\text { So, } \quad \mathrm{L}_{\text {eq }}=\frac{\frac{\mathrm{L}}{2} \times \frac{\mathrm{L}}{2}}{\frac{\mathrm{L}}{2}+\frac{\mathrm{L}}{2}}$ $\mathrm{~L}_{\mathrm{eq}}=\frac{\mathrm{L}}{4}$
AIIMS-2011
Electro Magnetic Induction
154861
With the decrease of current in the primary coil from 2 amperes to zero value in $0.01 \mathrm{~s}$ the emf generated in the secondary coil is $\mathbf{1 0 0 0}$ volts. The mutual inductance of the two coils is
1 $1.25 \mathrm{H}$
2 $2.50 \mathrm{H}$
3 $5.00 \mathrm{H}$
4 $10.00 \mathrm{H}$
Explanation:
C Given that, Primary current $\mathrm{I}_{1}=2 \mathrm{~A}$ Secondary current $\mathrm{I}_{2}=0 \mathrm{~A}$ Time $\Delta \mathrm{t}=0.01 \mathrm{sec}$ emf $(\varepsilon)=1000$ volts Mutual inductance $(\mathrm{M})=-\frac{\varepsilon}{(\mathrm{dI} / \mathrm{dt})}=-\frac{1000}{(0-2) / 0.01}$ $\mathrm{M}=5 \mathrm{H}$
AIIMS-2007
Electro Magnetic Induction
154862
The coefficient of mutual inductance, when magnetic flux changes by $2 \times 10^{-2}$ Wb and current changes by $0.01 \mathrm{~A}$ is
1 8 Henry
2 4 Henry
3 3 Henry
4 2 Henry
Explanation:
D Given, Change in Magnetic flux, $\mathrm{d} \phi=2 \times 10^{-2} \mathrm{~Wb}$ Change in current $\mathrm{dI}=0.01 \mathrm{~A}$ $\mathrm{M}=?$ $\mathrm{M}=\frac{\mathrm{d} \phi}{\mathrm{dI}}=\frac{2 \times 10^{-2}}{0.01}$ $\mathrm{M}=2 \text { Henry }$
AIIMS-2002
Electro Magnetic Induction
154864
In an inductor of self-inductance $L=\mathbf{2} \mathrm{mH}$, Current changes with time according to relation $i=t^{2} e^{-t}$. At what time emf is zero?
1 $4 \mathrm{~s}$
2 $3 \mathrm{~s}$
3 $2 \mathrm{~s}$
4 $1 \mathrm{~s}$
Explanation:
C Given, $\mathrm{L}=2 \mathrm{mH}$, and $\mathrm{i}=\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}$ We know that. $\text { emf }=\text { L. } \frac{\mathrm{di}}{\mathrm{dt}}$ Let at time $\mathrm{t}$ emf is zero. $\therefore \quad \mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}=0$ \(0=2 \times 10^{-3}\left(\frac{d}{d t}\left(t^{\frac{I I}{2}} \cdot e^{-\mathrm{t}}\right)\right)\) $0=2 \mathrm{te}^{-\mathrm{t}}+\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}(-1)$ $2 \mathrm{te}^{-\mathrm{t}}-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}=0$ $\mathrm{te}^{-\mathrm{t}}(2-\mathrm{t})=0$ $\therefore \quad 2-\mathrm{t}=0$ $\mathrm{t}=2 \mathrm{sec}$
154860
An inductor coil of inductance $L$ is cut into two equal parts and both the parts are connected in parallel. The net inductance is :
1 $\mathrm{L}$
2 $\mathrm{L} / 2$
3 $\mathrm{L} / 4$
4 $2 \mathrm{~L}$.
Explanation:
C If two inductors are connected in parallel combination then, $\mathrm{L}_{\mathrm{eq}}=\frac{\mathrm{L}_{1} \mathrm{~L}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}$ Since, we know that inductance of conductor directly proportional to its length $\therefore \quad \mathrm{L}_{1}=\mathrm{L}_{2}=\mathrm{L} / 2$ $\text { So, } \quad \mathrm{L}_{\text {eq }}=\frac{\frac{\mathrm{L}}{2} \times \frac{\mathrm{L}}{2}}{\frac{\mathrm{L}}{2}+\frac{\mathrm{L}}{2}}$ $\mathrm{~L}_{\mathrm{eq}}=\frac{\mathrm{L}}{4}$
AIIMS-2011
Electro Magnetic Induction
154861
With the decrease of current in the primary coil from 2 amperes to zero value in $0.01 \mathrm{~s}$ the emf generated in the secondary coil is $\mathbf{1 0 0 0}$ volts. The mutual inductance of the two coils is
1 $1.25 \mathrm{H}$
2 $2.50 \mathrm{H}$
3 $5.00 \mathrm{H}$
4 $10.00 \mathrm{H}$
Explanation:
C Given that, Primary current $\mathrm{I}_{1}=2 \mathrm{~A}$ Secondary current $\mathrm{I}_{2}=0 \mathrm{~A}$ Time $\Delta \mathrm{t}=0.01 \mathrm{sec}$ emf $(\varepsilon)=1000$ volts Mutual inductance $(\mathrm{M})=-\frac{\varepsilon}{(\mathrm{dI} / \mathrm{dt})}=-\frac{1000}{(0-2) / 0.01}$ $\mathrm{M}=5 \mathrm{H}$
AIIMS-2007
Electro Magnetic Induction
154862
The coefficient of mutual inductance, when magnetic flux changes by $2 \times 10^{-2}$ Wb and current changes by $0.01 \mathrm{~A}$ is
1 8 Henry
2 4 Henry
3 3 Henry
4 2 Henry
Explanation:
D Given, Change in Magnetic flux, $\mathrm{d} \phi=2 \times 10^{-2} \mathrm{~Wb}$ Change in current $\mathrm{dI}=0.01 \mathrm{~A}$ $\mathrm{M}=?$ $\mathrm{M}=\frac{\mathrm{d} \phi}{\mathrm{dI}}=\frac{2 \times 10^{-2}}{0.01}$ $\mathrm{M}=2 \text { Henry }$
AIIMS-2002
Electro Magnetic Induction
154864
In an inductor of self-inductance $L=\mathbf{2} \mathrm{mH}$, Current changes with time according to relation $i=t^{2} e^{-t}$. At what time emf is zero?
1 $4 \mathrm{~s}$
2 $3 \mathrm{~s}$
3 $2 \mathrm{~s}$
4 $1 \mathrm{~s}$
Explanation:
C Given, $\mathrm{L}=2 \mathrm{mH}$, and $\mathrm{i}=\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}$ We know that. $\text { emf }=\text { L. } \frac{\mathrm{di}}{\mathrm{dt}}$ Let at time $\mathrm{t}$ emf is zero. $\therefore \quad \mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}=0$ \(0=2 \times 10^{-3}\left(\frac{d}{d t}\left(t^{\frac{I I}{2}} \cdot e^{-\mathrm{t}}\right)\right)\) $0=2 \mathrm{te}^{-\mathrm{t}}+\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}(-1)$ $2 \mathrm{te}^{-\mathrm{t}}-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}=0$ $\mathrm{te}^{-\mathrm{t}}(2-\mathrm{t})=0$ $\therefore \quad 2-\mathrm{t}=0$ $\mathrm{t}=2 \mathrm{sec}$
