NEET Test Series from KOTA - 10 Papers In MS WORD
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Electro Magnetic Induction
154908
A coil of inductance $300 \mathrm{mH}$ and resistance $2 \Omega$ is connected to a source of voltage $2 \mathrm{~V}$. The current reaches half of its steady state value in
1 $0.05 \mathrm{~s}$
2 $0.1 \mathrm{~s}$
3 $0.15 \mathrm{~s}$
4 $0.3 \mathrm{~s}$
Explanation:
B Given, Inductance $\mathrm{L}=300 \mathrm{mH}$ Resistance $\mathrm{R}=2 \Omega$ Voltage $\mathrm{e}=2 \mathrm{~V}$ $\mathrm{I}=\frac{\mathrm{I}_{0}}{2}$ For charging of inductance $\mathrm{I}=\mathrm{I}_{0}\left[1-\mathrm{e}^{\frac{-\mathrm{Rt}}{\mathrm{L}}}\right]$ $\therefore \quad \frac{\mathrm{I}_{0}}{2}=\mathrm{I}_{0}\left[1-\mathrm{e}^{\frac{-\mathrm{Rt}}{\mathrm{L}}}\right]$ $1-\mathrm{e}^{-\frac{\mathrm{Rt}}{\mathrm{L}}}=\frac{1}{2}$ $\mathrm{e}^{-\frac{\mathrm{Rt}}{\mathrm{L}}}=1-\frac{1}{2}=\frac{1}{2}$ $\mathrm{e}^{-\frac{\mathrm{Rt}}{\mathrm{L}}}=\frac{1}{2}$ $-\frac{\mathrm{Rt}}{\mathrm{L}}=\log _{\mathrm{e}} \frac{1}{2}$ $\frac{\mathrm{Rt}}{\mathrm{L}}=\log _{\mathrm{e}} 2$ $\mathrm{t}=\frac{\mathrm{L}}{\mathrm{R}} \log _{\mathrm{e}} 2$ $\mathrm{t}=\frac{300 \times 10^{-3}}{2} \times \log _{\mathrm{e}} 2=0.1039$
UP CPMT-2008
Electro Magnetic Induction
154909
A current $i=10 \sin (100 \pi t)$ amp is passed in first coil, which induces a maximum emf. $5 \pi$ volt in second coil. The mutual inductance between the coils is
154911
When current in a coil is changed from $10 \mathrm{~A}$ in one direction to $10 \mathrm{a}$ in opposite direction in 0.5 $\mathrm{s}$, the induced emf is $1 \mathrm{~V}$. The self-inductance of the coil is
1 $25 \mathrm{mH}$
2 $50 \mathrm{mH}$
3 $75 \mathrm{mH}$
4 $100 \mathrm{mH}$
Explanation:
A Given current $\mathrm{I}_{1}=10 \mathrm{~A}$ $\mathrm{I}_{2}=-10 \mathrm{~A}$ Change in current $\Delta \mathrm{I}=10-(-10)$ $\Delta \mathrm{I}=20 \mathrm{~A}$ Time $\Delta \mathrm{t}=0.5 \mathrm{~s}$ Induced voltage $=1 \mathrm{~V}$ Induced emf $(\varepsilon)=M \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $\mathrm{M}=\frac{\varepsilon \Delta \mathrm{t}}{\Delta \mathrm{I}}=\frac{1 \times 0.5}{20}$. $\mathrm{M}=0.025$ $\mathrm{M}=25 \mathrm{mH}$
UP CPMT-2001
Electro Magnetic Induction
154912
In a coil of self inductance $0.5 \mathrm{H}$, the current varies at a constant rate from zero to $10 \mathrm{~A}$ in 2 sec. The emf generated in the coil is
1 $1.5 \mathrm{~V}$
2 $2.5 \mathrm{~V}$
3 $5.0 \mathrm{~V}$
4 $10.0 \mathrm{~V}$
Explanation:
B Given, Self inductance $(\mathrm{L})=0.5 \mathrm{H}$ Change in current $(\Delta \mathrm{I})=10-0=10 \mathrm{~A}$ Change in time $(\Delta \mathrm{t})=2 \mathrm{sec}$ We know,induced emf $\varepsilon=\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $\varepsilon=0.5 \times \frac{10}{2}=2.5$ $\varepsilon=2.5 \mathrm{~V}$
154908
A coil of inductance $300 \mathrm{mH}$ and resistance $2 \Omega$ is connected to a source of voltage $2 \mathrm{~V}$. The current reaches half of its steady state value in
1 $0.05 \mathrm{~s}$
2 $0.1 \mathrm{~s}$
3 $0.15 \mathrm{~s}$
4 $0.3 \mathrm{~s}$
Explanation:
B Given, Inductance $\mathrm{L}=300 \mathrm{mH}$ Resistance $\mathrm{R}=2 \Omega$ Voltage $\mathrm{e}=2 \mathrm{~V}$ $\mathrm{I}=\frac{\mathrm{I}_{0}}{2}$ For charging of inductance $\mathrm{I}=\mathrm{I}_{0}\left[1-\mathrm{e}^{\frac{-\mathrm{Rt}}{\mathrm{L}}}\right]$ $\therefore \quad \frac{\mathrm{I}_{0}}{2}=\mathrm{I}_{0}\left[1-\mathrm{e}^{\frac{-\mathrm{Rt}}{\mathrm{L}}}\right]$ $1-\mathrm{e}^{-\frac{\mathrm{Rt}}{\mathrm{L}}}=\frac{1}{2}$ $\mathrm{e}^{-\frac{\mathrm{Rt}}{\mathrm{L}}}=1-\frac{1}{2}=\frac{1}{2}$ $\mathrm{e}^{-\frac{\mathrm{Rt}}{\mathrm{L}}}=\frac{1}{2}$ $-\frac{\mathrm{Rt}}{\mathrm{L}}=\log _{\mathrm{e}} \frac{1}{2}$ $\frac{\mathrm{Rt}}{\mathrm{L}}=\log _{\mathrm{e}} 2$ $\mathrm{t}=\frac{\mathrm{L}}{\mathrm{R}} \log _{\mathrm{e}} 2$ $\mathrm{t}=\frac{300 \times 10^{-3}}{2} \times \log _{\mathrm{e}} 2=0.1039$
UP CPMT-2008
Electro Magnetic Induction
154909
A current $i=10 \sin (100 \pi t)$ amp is passed in first coil, which induces a maximum emf. $5 \pi$ volt in second coil. The mutual inductance between the coils is
154911
When current in a coil is changed from $10 \mathrm{~A}$ in one direction to $10 \mathrm{a}$ in opposite direction in 0.5 $\mathrm{s}$, the induced emf is $1 \mathrm{~V}$. The self-inductance of the coil is
1 $25 \mathrm{mH}$
2 $50 \mathrm{mH}$
3 $75 \mathrm{mH}$
4 $100 \mathrm{mH}$
Explanation:
A Given current $\mathrm{I}_{1}=10 \mathrm{~A}$ $\mathrm{I}_{2}=-10 \mathrm{~A}$ Change in current $\Delta \mathrm{I}=10-(-10)$ $\Delta \mathrm{I}=20 \mathrm{~A}$ Time $\Delta \mathrm{t}=0.5 \mathrm{~s}$ Induced voltage $=1 \mathrm{~V}$ Induced emf $(\varepsilon)=M \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $\mathrm{M}=\frac{\varepsilon \Delta \mathrm{t}}{\Delta \mathrm{I}}=\frac{1 \times 0.5}{20}$. $\mathrm{M}=0.025$ $\mathrm{M}=25 \mathrm{mH}$
UP CPMT-2001
Electro Magnetic Induction
154912
In a coil of self inductance $0.5 \mathrm{H}$, the current varies at a constant rate from zero to $10 \mathrm{~A}$ in 2 sec. The emf generated in the coil is
1 $1.5 \mathrm{~V}$
2 $2.5 \mathrm{~V}$
3 $5.0 \mathrm{~V}$
4 $10.0 \mathrm{~V}$
Explanation:
B Given, Self inductance $(\mathrm{L})=0.5 \mathrm{H}$ Change in current $(\Delta \mathrm{I})=10-0=10 \mathrm{~A}$ Change in time $(\Delta \mathrm{t})=2 \mathrm{sec}$ We know,induced emf $\varepsilon=\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $\varepsilon=0.5 \times \frac{10}{2}=2.5$ $\varepsilon=2.5 \mathrm{~V}$
154908
A coil of inductance $300 \mathrm{mH}$ and resistance $2 \Omega$ is connected to a source of voltage $2 \mathrm{~V}$. The current reaches half of its steady state value in
1 $0.05 \mathrm{~s}$
2 $0.1 \mathrm{~s}$
3 $0.15 \mathrm{~s}$
4 $0.3 \mathrm{~s}$
Explanation:
B Given, Inductance $\mathrm{L}=300 \mathrm{mH}$ Resistance $\mathrm{R}=2 \Omega$ Voltage $\mathrm{e}=2 \mathrm{~V}$ $\mathrm{I}=\frac{\mathrm{I}_{0}}{2}$ For charging of inductance $\mathrm{I}=\mathrm{I}_{0}\left[1-\mathrm{e}^{\frac{-\mathrm{Rt}}{\mathrm{L}}}\right]$ $\therefore \quad \frac{\mathrm{I}_{0}}{2}=\mathrm{I}_{0}\left[1-\mathrm{e}^{\frac{-\mathrm{Rt}}{\mathrm{L}}}\right]$ $1-\mathrm{e}^{-\frac{\mathrm{Rt}}{\mathrm{L}}}=\frac{1}{2}$ $\mathrm{e}^{-\frac{\mathrm{Rt}}{\mathrm{L}}}=1-\frac{1}{2}=\frac{1}{2}$ $\mathrm{e}^{-\frac{\mathrm{Rt}}{\mathrm{L}}}=\frac{1}{2}$ $-\frac{\mathrm{Rt}}{\mathrm{L}}=\log _{\mathrm{e}} \frac{1}{2}$ $\frac{\mathrm{Rt}}{\mathrm{L}}=\log _{\mathrm{e}} 2$ $\mathrm{t}=\frac{\mathrm{L}}{\mathrm{R}} \log _{\mathrm{e}} 2$ $\mathrm{t}=\frac{300 \times 10^{-3}}{2} \times \log _{\mathrm{e}} 2=0.1039$
UP CPMT-2008
Electro Magnetic Induction
154909
A current $i=10 \sin (100 \pi t)$ amp is passed in first coil, which induces a maximum emf. $5 \pi$ volt in second coil. The mutual inductance between the coils is
154911
When current in a coil is changed from $10 \mathrm{~A}$ in one direction to $10 \mathrm{a}$ in opposite direction in 0.5 $\mathrm{s}$, the induced emf is $1 \mathrm{~V}$. The self-inductance of the coil is
1 $25 \mathrm{mH}$
2 $50 \mathrm{mH}$
3 $75 \mathrm{mH}$
4 $100 \mathrm{mH}$
Explanation:
A Given current $\mathrm{I}_{1}=10 \mathrm{~A}$ $\mathrm{I}_{2}=-10 \mathrm{~A}$ Change in current $\Delta \mathrm{I}=10-(-10)$ $\Delta \mathrm{I}=20 \mathrm{~A}$ Time $\Delta \mathrm{t}=0.5 \mathrm{~s}$ Induced voltage $=1 \mathrm{~V}$ Induced emf $(\varepsilon)=M \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $\mathrm{M}=\frac{\varepsilon \Delta \mathrm{t}}{\Delta \mathrm{I}}=\frac{1 \times 0.5}{20}$. $\mathrm{M}=0.025$ $\mathrm{M}=25 \mathrm{mH}$
UP CPMT-2001
Electro Magnetic Induction
154912
In a coil of self inductance $0.5 \mathrm{H}$, the current varies at a constant rate from zero to $10 \mathrm{~A}$ in 2 sec. The emf generated in the coil is
1 $1.5 \mathrm{~V}$
2 $2.5 \mathrm{~V}$
3 $5.0 \mathrm{~V}$
4 $10.0 \mathrm{~V}$
Explanation:
B Given, Self inductance $(\mathrm{L})=0.5 \mathrm{H}$ Change in current $(\Delta \mathrm{I})=10-0=10 \mathrm{~A}$ Change in time $(\Delta \mathrm{t})=2 \mathrm{sec}$ We know,induced emf $\varepsilon=\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $\varepsilon=0.5 \times \frac{10}{2}=2.5$ $\varepsilon=2.5 \mathrm{~V}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Electro Magnetic Induction
154908
A coil of inductance $300 \mathrm{mH}$ and resistance $2 \Omega$ is connected to a source of voltage $2 \mathrm{~V}$. The current reaches half of its steady state value in
1 $0.05 \mathrm{~s}$
2 $0.1 \mathrm{~s}$
3 $0.15 \mathrm{~s}$
4 $0.3 \mathrm{~s}$
Explanation:
B Given, Inductance $\mathrm{L}=300 \mathrm{mH}$ Resistance $\mathrm{R}=2 \Omega$ Voltage $\mathrm{e}=2 \mathrm{~V}$ $\mathrm{I}=\frac{\mathrm{I}_{0}}{2}$ For charging of inductance $\mathrm{I}=\mathrm{I}_{0}\left[1-\mathrm{e}^{\frac{-\mathrm{Rt}}{\mathrm{L}}}\right]$ $\therefore \quad \frac{\mathrm{I}_{0}}{2}=\mathrm{I}_{0}\left[1-\mathrm{e}^{\frac{-\mathrm{Rt}}{\mathrm{L}}}\right]$ $1-\mathrm{e}^{-\frac{\mathrm{Rt}}{\mathrm{L}}}=\frac{1}{2}$ $\mathrm{e}^{-\frac{\mathrm{Rt}}{\mathrm{L}}}=1-\frac{1}{2}=\frac{1}{2}$ $\mathrm{e}^{-\frac{\mathrm{Rt}}{\mathrm{L}}}=\frac{1}{2}$ $-\frac{\mathrm{Rt}}{\mathrm{L}}=\log _{\mathrm{e}} \frac{1}{2}$ $\frac{\mathrm{Rt}}{\mathrm{L}}=\log _{\mathrm{e}} 2$ $\mathrm{t}=\frac{\mathrm{L}}{\mathrm{R}} \log _{\mathrm{e}} 2$ $\mathrm{t}=\frac{300 \times 10^{-3}}{2} \times \log _{\mathrm{e}} 2=0.1039$
UP CPMT-2008
Electro Magnetic Induction
154909
A current $i=10 \sin (100 \pi t)$ amp is passed in first coil, which induces a maximum emf. $5 \pi$ volt in second coil. The mutual inductance between the coils is
154911
When current in a coil is changed from $10 \mathrm{~A}$ in one direction to $10 \mathrm{a}$ in opposite direction in 0.5 $\mathrm{s}$, the induced emf is $1 \mathrm{~V}$. The self-inductance of the coil is
1 $25 \mathrm{mH}$
2 $50 \mathrm{mH}$
3 $75 \mathrm{mH}$
4 $100 \mathrm{mH}$
Explanation:
A Given current $\mathrm{I}_{1}=10 \mathrm{~A}$ $\mathrm{I}_{2}=-10 \mathrm{~A}$ Change in current $\Delta \mathrm{I}=10-(-10)$ $\Delta \mathrm{I}=20 \mathrm{~A}$ Time $\Delta \mathrm{t}=0.5 \mathrm{~s}$ Induced voltage $=1 \mathrm{~V}$ Induced emf $(\varepsilon)=M \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $\mathrm{M}=\frac{\varepsilon \Delta \mathrm{t}}{\Delta \mathrm{I}}=\frac{1 \times 0.5}{20}$. $\mathrm{M}=0.025$ $\mathrm{M}=25 \mathrm{mH}$
UP CPMT-2001
Electro Magnetic Induction
154912
In a coil of self inductance $0.5 \mathrm{H}$, the current varies at a constant rate from zero to $10 \mathrm{~A}$ in 2 sec. The emf generated in the coil is
1 $1.5 \mathrm{~V}$
2 $2.5 \mathrm{~V}$
3 $5.0 \mathrm{~V}$
4 $10.0 \mathrm{~V}$
Explanation:
B Given, Self inductance $(\mathrm{L})=0.5 \mathrm{H}$ Change in current $(\Delta \mathrm{I})=10-0=10 \mathrm{~A}$ Change in time $(\Delta \mathrm{t})=2 \mathrm{sec}$ We know,induced emf $\varepsilon=\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $\varepsilon=0.5 \times \frac{10}{2}=2.5$ $\varepsilon=2.5 \mathrm{~V}$