154834
The equivalent inductance of two inductor is $2.4 \mathrm{H}$ when connected in parallel and $10 \mathrm{H}$ when connected in series. The difference between the two inductance is
1 $2 \mathrm{H}$
2 $3 \mathrm{H}$
3 $4 \mathrm{H}$
4 $5 \mathrm{H}$
Explanation:
A Given, $\mathrm{L}_{\text {parallel }}=2.4 \mathrm{H}$ $\mathrm{L}_{\text {series }}=10 \mathrm{H}$ We know that, when two inducator connected in series- $\mathrm{L}_{\text {series }}=\mathrm{L}_{1}+\mathrm{L}_{2}$ $10=\mathrm{L}_{1}+\mathrm{L}_{2}$ $\mathrm{~L}_{1}=10-\mathrm{L}_{2}$ When inductor connected in parallel- $\mathrm{L}_{\text {parallel }}=\frac{\mathrm{L}_{1} \times \mathrm{L}_{2}}{\left(\mathrm{~L}_{1}+\mathrm{L}_{2}\right)}$ $2.4=\frac{\left(10-\mathrm{L}_{2}\right) \mathrm{L}_{2}}{10}$ $\mathrm{~L}_{2}^{2}-10 \mathrm{~L}_{2}+24=0$ $\left(\mathrm{~L}_{2}-6\right)\left(\mathrm{L}_{2}-4\right)=0$ $\mathrm{~L}_{2}=6,4 .$ Putting the value in equation (i), we get- $\mathrm{L}_{1}+\mathrm{L}_{2}=10$ $6+\mathrm{L}_{2}=10$ $\mathrm{~L}_{2}=4 \mathrm{H}$ $\mathrm{L}_{1}+\mathrm{L}_{2}=10$ $4+\mathrm{L}_{2}=10$ $\mathrm{~L}_{2}=6 \mathrm{H}$ The two inductors is $6 \mathrm{H}$ and $4 \mathrm{H}$. Hence, the difference between two inductors is- $=6-4=2 \mathrm{H}$
CG PET- 2004
Electro Magnetic Induction
154835
The inductance of coil is $60 \mu \mathrm{H}$. A current in this coil increases from 1.0 $\mathrm{A}$ to $1.5 \mathrm{~A}$ in $0.1 \mathrm{~s}$. The magnitude of the induced emf is
1 $60 \times 10^{-6} \mathrm{~V}$
2 $300 \times 10^{-4} \mathrm{~V}$
3 $30 \times 10^{-4} \mathrm{~V}$
4 $3 \times 10^{-4} \mathrm{~V}$
Explanation:
D Given, Inductance of coil $(\mathrm{L})=60 \mu \mathrm{H}=60 \times 10^{-6} \mathrm{H}$ Current $(\mathrm{dI})=1.5-1.0=0.5 \mathrm{~A}$ Time, $(\mathrm{dt})=0.1 \mathrm{~s}$ We know that, Induced emf $(\varepsilon)=\mathrm{L} \cdot \frac{\mathrm{dI}}{\mathrm{dt}}$ $\varepsilon=60 \times 10^{-6} \times \frac{0.5}{0.1}$ $\varepsilon=300 \times 10^{-6} \mathrm{~V}$ $\varepsilon=3 \times 10^{-4} \mathrm{~V}$
CG PET- 2004
Electro Magnetic Induction
154836
A loop of area $0.1 \mathrm{~m}^{2}$ rotates with a speed of 60 $\mathrm{rev} / \mathrm{s}$ with the axis of rotation perpendicular to a magnetic field $B=0.4 T$. If there are 100 turns in the loop, the maximum voltage induced in the loop is
1 $15.07 \mathrm{~V}$
2 $150.7 \mathrm{~V}$
3 $1507 \mathrm{~V}$
4 $240 \mathrm{~V}$
Explanation:
C Given, Number of turns $=100$ Area $(A)=0.1 \mathrm{~m}^{2}$ Speed $(\mathrm{N})=60 \mathrm{rev} / \mathrm{s}$ Magnetic field $(B)=0.4 \mathrm{~T}$ We know that, Induced emf in the coil - $\varepsilon=\mathrm{NBA} \omega \sin \omega \mathrm{t}$ For maximum induced voltage - $\sin \omega \mathrm{t}=1$ So, $\varepsilon_{0} =N B A \omega$ $\varepsilon_{0} =100 \times 0.4 \times 0.1 \times(2 \pi \times 60)$ $\varepsilon_{0} =480 \pi$ $\varepsilon_{0} =1507.2 \mathrm{~V} = 1507 \mathrm{~V}$
CG PET- 2004
Electro Magnetic Induction
154839
Two solenoids of same cross-sectional area have their lengths and number of turns in ratio of 1:2. The ratio of self-inductance of two solenoids is
1 $1: 1$
2 $1: 2$
3 $2: 1$
4 $1: 4$
Explanation:
B Given, The ratio of number of turns $=1: 2$ We know that, $\text { Self inductance }(\mathrm{L}) =\frac{\mu_{0} \mathrm{~N}^{2} \mathrm{IA}}{l}$ $\mathrm{~L} \propto \frac{\mathrm{N}^{2}}{l}$ So, $\quad \frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\frac{\mu_{\circ} \mathrm{N}_{1}^{2} \mathrm{IA} / l_{1}}{\mu_{\circ} \mathrm{N}_{2}^{2} \mathrm{IA} / l_{2}}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\left(\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}\right)^{2} \times\left(\frac{l_{2}}{l_{1}}\right)$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\left(\frac{1}{2}\right)^{2} \times\left(\frac{2}{1}\right) \quad\left(\because \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}} \& \frac{l_{1}}{l_{2}}=\frac{1}{2}\right)$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\frac{1}{2}$
154834
The equivalent inductance of two inductor is $2.4 \mathrm{H}$ when connected in parallel and $10 \mathrm{H}$ when connected in series. The difference between the two inductance is
1 $2 \mathrm{H}$
2 $3 \mathrm{H}$
3 $4 \mathrm{H}$
4 $5 \mathrm{H}$
Explanation:
A Given, $\mathrm{L}_{\text {parallel }}=2.4 \mathrm{H}$ $\mathrm{L}_{\text {series }}=10 \mathrm{H}$ We know that, when two inducator connected in series- $\mathrm{L}_{\text {series }}=\mathrm{L}_{1}+\mathrm{L}_{2}$ $10=\mathrm{L}_{1}+\mathrm{L}_{2}$ $\mathrm{~L}_{1}=10-\mathrm{L}_{2}$ When inductor connected in parallel- $\mathrm{L}_{\text {parallel }}=\frac{\mathrm{L}_{1} \times \mathrm{L}_{2}}{\left(\mathrm{~L}_{1}+\mathrm{L}_{2}\right)}$ $2.4=\frac{\left(10-\mathrm{L}_{2}\right) \mathrm{L}_{2}}{10}$ $\mathrm{~L}_{2}^{2}-10 \mathrm{~L}_{2}+24=0$ $\left(\mathrm{~L}_{2}-6\right)\left(\mathrm{L}_{2}-4\right)=0$ $\mathrm{~L}_{2}=6,4 .$ Putting the value in equation (i), we get- $\mathrm{L}_{1}+\mathrm{L}_{2}=10$ $6+\mathrm{L}_{2}=10$ $\mathrm{~L}_{2}=4 \mathrm{H}$ $\mathrm{L}_{1}+\mathrm{L}_{2}=10$ $4+\mathrm{L}_{2}=10$ $\mathrm{~L}_{2}=6 \mathrm{H}$ The two inductors is $6 \mathrm{H}$ and $4 \mathrm{H}$. Hence, the difference between two inductors is- $=6-4=2 \mathrm{H}$
CG PET- 2004
Electro Magnetic Induction
154835
The inductance of coil is $60 \mu \mathrm{H}$. A current in this coil increases from 1.0 $\mathrm{A}$ to $1.5 \mathrm{~A}$ in $0.1 \mathrm{~s}$. The magnitude of the induced emf is
1 $60 \times 10^{-6} \mathrm{~V}$
2 $300 \times 10^{-4} \mathrm{~V}$
3 $30 \times 10^{-4} \mathrm{~V}$
4 $3 \times 10^{-4} \mathrm{~V}$
Explanation:
D Given, Inductance of coil $(\mathrm{L})=60 \mu \mathrm{H}=60 \times 10^{-6} \mathrm{H}$ Current $(\mathrm{dI})=1.5-1.0=0.5 \mathrm{~A}$ Time, $(\mathrm{dt})=0.1 \mathrm{~s}$ We know that, Induced emf $(\varepsilon)=\mathrm{L} \cdot \frac{\mathrm{dI}}{\mathrm{dt}}$ $\varepsilon=60 \times 10^{-6} \times \frac{0.5}{0.1}$ $\varepsilon=300 \times 10^{-6} \mathrm{~V}$ $\varepsilon=3 \times 10^{-4} \mathrm{~V}$
CG PET- 2004
Electro Magnetic Induction
154836
A loop of area $0.1 \mathrm{~m}^{2}$ rotates with a speed of 60 $\mathrm{rev} / \mathrm{s}$ with the axis of rotation perpendicular to a magnetic field $B=0.4 T$. If there are 100 turns in the loop, the maximum voltage induced in the loop is
1 $15.07 \mathrm{~V}$
2 $150.7 \mathrm{~V}$
3 $1507 \mathrm{~V}$
4 $240 \mathrm{~V}$
Explanation:
C Given, Number of turns $=100$ Area $(A)=0.1 \mathrm{~m}^{2}$ Speed $(\mathrm{N})=60 \mathrm{rev} / \mathrm{s}$ Magnetic field $(B)=0.4 \mathrm{~T}$ We know that, Induced emf in the coil - $\varepsilon=\mathrm{NBA} \omega \sin \omega \mathrm{t}$ For maximum induced voltage - $\sin \omega \mathrm{t}=1$ So, $\varepsilon_{0} =N B A \omega$ $\varepsilon_{0} =100 \times 0.4 \times 0.1 \times(2 \pi \times 60)$ $\varepsilon_{0} =480 \pi$ $\varepsilon_{0} =1507.2 \mathrm{~V} = 1507 \mathrm{~V}$
CG PET- 2004
Electro Magnetic Induction
154839
Two solenoids of same cross-sectional area have their lengths and number of turns in ratio of 1:2. The ratio of self-inductance of two solenoids is
1 $1: 1$
2 $1: 2$
3 $2: 1$
4 $1: 4$
Explanation:
B Given, The ratio of number of turns $=1: 2$ We know that, $\text { Self inductance }(\mathrm{L}) =\frac{\mu_{0} \mathrm{~N}^{2} \mathrm{IA}}{l}$ $\mathrm{~L} \propto \frac{\mathrm{N}^{2}}{l}$ So, $\quad \frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\frac{\mu_{\circ} \mathrm{N}_{1}^{2} \mathrm{IA} / l_{1}}{\mu_{\circ} \mathrm{N}_{2}^{2} \mathrm{IA} / l_{2}}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\left(\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}\right)^{2} \times\left(\frac{l_{2}}{l_{1}}\right)$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\left(\frac{1}{2}\right)^{2} \times\left(\frac{2}{1}\right) \quad\left(\because \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}} \& \frac{l_{1}}{l_{2}}=\frac{1}{2}\right)$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\frac{1}{2}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Electro Magnetic Induction
154834
The equivalent inductance of two inductor is $2.4 \mathrm{H}$ when connected in parallel and $10 \mathrm{H}$ when connected in series. The difference between the two inductance is
1 $2 \mathrm{H}$
2 $3 \mathrm{H}$
3 $4 \mathrm{H}$
4 $5 \mathrm{H}$
Explanation:
A Given, $\mathrm{L}_{\text {parallel }}=2.4 \mathrm{H}$ $\mathrm{L}_{\text {series }}=10 \mathrm{H}$ We know that, when two inducator connected in series- $\mathrm{L}_{\text {series }}=\mathrm{L}_{1}+\mathrm{L}_{2}$ $10=\mathrm{L}_{1}+\mathrm{L}_{2}$ $\mathrm{~L}_{1}=10-\mathrm{L}_{2}$ When inductor connected in parallel- $\mathrm{L}_{\text {parallel }}=\frac{\mathrm{L}_{1} \times \mathrm{L}_{2}}{\left(\mathrm{~L}_{1}+\mathrm{L}_{2}\right)}$ $2.4=\frac{\left(10-\mathrm{L}_{2}\right) \mathrm{L}_{2}}{10}$ $\mathrm{~L}_{2}^{2}-10 \mathrm{~L}_{2}+24=0$ $\left(\mathrm{~L}_{2}-6\right)\left(\mathrm{L}_{2}-4\right)=0$ $\mathrm{~L}_{2}=6,4 .$ Putting the value in equation (i), we get- $\mathrm{L}_{1}+\mathrm{L}_{2}=10$ $6+\mathrm{L}_{2}=10$ $\mathrm{~L}_{2}=4 \mathrm{H}$ $\mathrm{L}_{1}+\mathrm{L}_{2}=10$ $4+\mathrm{L}_{2}=10$ $\mathrm{~L}_{2}=6 \mathrm{H}$ The two inductors is $6 \mathrm{H}$ and $4 \mathrm{H}$. Hence, the difference between two inductors is- $=6-4=2 \mathrm{H}$
CG PET- 2004
Electro Magnetic Induction
154835
The inductance of coil is $60 \mu \mathrm{H}$. A current in this coil increases from 1.0 $\mathrm{A}$ to $1.5 \mathrm{~A}$ in $0.1 \mathrm{~s}$. The magnitude of the induced emf is
1 $60 \times 10^{-6} \mathrm{~V}$
2 $300 \times 10^{-4} \mathrm{~V}$
3 $30 \times 10^{-4} \mathrm{~V}$
4 $3 \times 10^{-4} \mathrm{~V}$
Explanation:
D Given, Inductance of coil $(\mathrm{L})=60 \mu \mathrm{H}=60 \times 10^{-6} \mathrm{H}$ Current $(\mathrm{dI})=1.5-1.0=0.5 \mathrm{~A}$ Time, $(\mathrm{dt})=0.1 \mathrm{~s}$ We know that, Induced emf $(\varepsilon)=\mathrm{L} \cdot \frac{\mathrm{dI}}{\mathrm{dt}}$ $\varepsilon=60 \times 10^{-6} \times \frac{0.5}{0.1}$ $\varepsilon=300 \times 10^{-6} \mathrm{~V}$ $\varepsilon=3 \times 10^{-4} \mathrm{~V}$
CG PET- 2004
Electro Magnetic Induction
154836
A loop of area $0.1 \mathrm{~m}^{2}$ rotates with a speed of 60 $\mathrm{rev} / \mathrm{s}$ with the axis of rotation perpendicular to a magnetic field $B=0.4 T$. If there are 100 turns in the loop, the maximum voltage induced in the loop is
1 $15.07 \mathrm{~V}$
2 $150.7 \mathrm{~V}$
3 $1507 \mathrm{~V}$
4 $240 \mathrm{~V}$
Explanation:
C Given, Number of turns $=100$ Area $(A)=0.1 \mathrm{~m}^{2}$ Speed $(\mathrm{N})=60 \mathrm{rev} / \mathrm{s}$ Magnetic field $(B)=0.4 \mathrm{~T}$ We know that, Induced emf in the coil - $\varepsilon=\mathrm{NBA} \omega \sin \omega \mathrm{t}$ For maximum induced voltage - $\sin \omega \mathrm{t}=1$ So, $\varepsilon_{0} =N B A \omega$ $\varepsilon_{0} =100 \times 0.4 \times 0.1 \times(2 \pi \times 60)$ $\varepsilon_{0} =480 \pi$ $\varepsilon_{0} =1507.2 \mathrm{~V} = 1507 \mathrm{~V}$
CG PET- 2004
Electro Magnetic Induction
154839
Two solenoids of same cross-sectional area have their lengths and number of turns in ratio of 1:2. The ratio of self-inductance of two solenoids is
1 $1: 1$
2 $1: 2$
3 $2: 1$
4 $1: 4$
Explanation:
B Given, The ratio of number of turns $=1: 2$ We know that, $\text { Self inductance }(\mathrm{L}) =\frac{\mu_{0} \mathrm{~N}^{2} \mathrm{IA}}{l}$ $\mathrm{~L} \propto \frac{\mathrm{N}^{2}}{l}$ So, $\quad \frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\frac{\mu_{\circ} \mathrm{N}_{1}^{2} \mathrm{IA} / l_{1}}{\mu_{\circ} \mathrm{N}_{2}^{2} \mathrm{IA} / l_{2}}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\left(\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}\right)^{2} \times\left(\frac{l_{2}}{l_{1}}\right)$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\left(\frac{1}{2}\right)^{2} \times\left(\frac{2}{1}\right) \quad\left(\because \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}} \& \frac{l_{1}}{l_{2}}=\frac{1}{2}\right)$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\frac{1}{2}$
154834
The equivalent inductance of two inductor is $2.4 \mathrm{H}$ when connected in parallel and $10 \mathrm{H}$ when connected in series. The difference between the two inductance is
1 $2 \mathrm{H}$
2 $3 \mathrm{H}$
3 $4 \mathrm{H}$
4 $5 \mathrm{H}$
Explanation:
A Given, $\mathrm{L}_{\text {parallel }}=2.4 \mathrm{H}$ $\mathrm{L}_{\text {series }}=10 \mathrm{H}$ We know that, when two inducator connected in series- $\mathrm{L}_{\text {series }}=\mathrm{L}_{1}+\mathrm{L}_{2}$ $10=\mathrm{L}_{1}+\mathrm{L}_{2}$ $\mathrm{~L}_{1}=10-\mathrm{L}_{2}$ When inductor connected in parallel- $\mathrm{L}_{\text {parallel }}=\frac{\mathrm{L}_{1} \times \mathrm{L}_{2}}{\left(\mathrm{~L}_{1}+\mathrm{L}_{2}\right)}$ $2.4=\frac{\left(10-\mathrm{L}_{2}\right) \mathrm{L}_{2}}{10}$ $\mathrm{~L}_{2}^{2}-10 \mathrm{~L}_{2}+24=0$ $\left(\mathrm{~L}_{2}-6\right)\left(\mathrm{L}_{2}-4\right)=0$ $\mathrm{~L}_{2}=6,4 .$ Putting the value in equation (i), we get- $\mathrm{L}_{1}+\mathrm{L}_{2}=10$ $6+\mathrm{L}_{2}=10$ $\mathrm{~L}_{2}=4 \mathrm{H}$ $\mathrm{L}_{1}+\mathrm{L}_{2}=10$ $4+\mathrm{L}_{2}=10$ $\mathrm{~L}_{2}=6 \mathrm{H}$ The two inductors is $6 \mathrm{H}$ and $4 \mathrm{H}$. Hence, the difference between two inductors is- $=6-4=2 \mathrm{H}$
CG PET- 2004
Electro Magnetic Induction
154835
The inductance of coil is $60 \mu \mathrm{H}$. A current in this coil increases from 1.0 $\mathrm{A}$ to $1.5 \mathrm{~A}$ in $0.1 \mathrm{~s}$. The magnitude of the induced emf is
1 $60 \times 10^{-6} \mathrm{~V}$
2 $300 \times 10^{-4} \mathrm{~V}$
3 $30 \times 10^{-4} \mathrm{~V}$
4 $3 \times 10^{-4} \mathrm{~V}$
Explanation:
D Given, Inductance of coil $(\mathrm{L})=60 \mu \mathrm{H}=60 \times 10^{-6} \mathrm{H}$ Current $(\mathrm{dI})=1.5-1.0=0.5 \mathrm{~A}$ Time, $(\mathrm{dt})=0.1 \mathrm{~s}$ We know that, Induced emf $(\varepsilon)=\mathrm{L} \cdot \frac{\mathrm{dI}}{\mathrm{dt}}$ $\varepsilon=60 \times 10^{-6} \times \frac{0.5}{0.1}$ $\varepsilon=300 \times 10^{-6} \mathrm{~V}$ $\varepsilon=3 \times 10^{-4} \mathrm{~V}$
CG PET- 2004
Electro Magnetic Induction
154836
A loop of area $0.1 \mathrm{~m}^{2}$ rotates with a speed of 60 $\mathrm{rev} / \mathrm{s}$ with the axis of rotation perpendicular to a magnetic field $B=0.4 T$. If there are 100 turns in the loop, the maximum voltage induced in the loop is
1 $15.07 \mathrm{~V}$
2 $150.7 \mathrm{~V}$
3 $1507 \mathrm{~V}$
4 $240 \mathrm{~V}$
Explanation:
C Given, Number of turns $=100$ Area $(A)=0.1 \mathrm{~m}^{2}$ Speed $(\mathrm{N})=60 \mathrm{rev} / \mathrm{s}$ Magnetic field $(B)=0.4 \mathrm{~T}$ We know that, Induced emf in the coil - $\varepsilon=\mathrm{NBA} \omega \sin \omega \mathrm{t}$ For maximum induced voltage - $\sin \omega \mathrm{t}=1$ So, $\varepsilon_{0} =N B A \omega$ $\varepsilon_{0} =100 \times 0.4 \times 0.1 \times(2 \pi \times 60)$ $\varepsilon_{0} =480 \pi$ $\varepsilon_{0} =1507.2 \mathrm{~V} = 1507 \mathrm{~V}$
CG PET- 2004
Electro Magnetic Induction
154839
Two solenoids of same cross-sectional area have their lengths and number of turns in ratio of 1:2. The ratio of self-inductance of two solenoids is
1 $1: 1$
2 $1: 2$
3 $2: 1$
4 $1: 4$
Explanation:
B Given, The ratio of number of turns $=1: 2$ We know that, $\text { Self inductance }(\mathrm{L}) =\frac{\mu_{0} \mathrm{~N}^{2} \mathrm{IA}}{l}$ $\mathrm{~L} \propto \frac{\mathrm{N}^{2}}{l}$ So, $\quad \frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\frac{\mu_{\circ} \mathrm{N}_{1}^{2} \mathrm{IA} / l_{1}}{\mu_{\circ} \mathrm{N}_{2}^{2} \mathrm{IA} / l_{2}}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\left(\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}\right)^{2} \times\left(\frac{l_{2}}{l_{1}}\right)$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\left(\frac{1}{2}\right)^{2} \times\left(\frac{2}{1}\right) \quad\left(\because \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}} \& \frac{l_{1}}{l_{2}}=\frac{1}{2}\right)$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\frac{1}{2}$