NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
152875
A galvanometer having a resistance of $8 \Omega$ is shunted by a wire of resistance $2 \Omega$. If the total current is $1 \mathrm{~A}$, the part of it passing through the shunt will be
1 $1.2 \mathrm{~A}$
2 $0.8 \mathrm{~A}$
3 $0.5 \mathrm{~A}$
4 $0.3 \mathrm{~A}$
Explanation:
B Applying Kirchhoff Voltage law, $\mathrm{I}_{\mathrm{g}} \mathrm{G}=\mathrm{I}_{\mathrm{S}} \cdot \mathrm{S}$ $\left(\mathrm{I}-\mathrm{I}_{\mathrm{S}}\right) \mathrm{G}=\mathrm{I}_{\mathrm{S}} \cdot \mathrm{S}$ $\mathrm{I}_{\mathrm{S}}=\frac{\mathrm{IG}}{\mathrm{G}+\mathrm{S}}=\frac{1 \times 8}{8+2}$ $\mathrm{I}_{\mathrm{S}}=\frac{8}{10}=0.8$
TS EAMCET 08.05.2019
Current Electricity
152876
The sensitivity of tangent galvanometer is increased, if
1 number of turns decreases
2 number of turns increases
3 fields increases
4 None of the above
Explanation:
B The sensitivity of a tangent galvanometer is defined as the ratio of change in deflection to the current responsible for the deflection. $\therefore$ Sensitivity of tangent galvanometer $(\mathrm{S})=\frac{\theta}{\mathrm{I}}$ We know that, current $(\mathrm{I})=\mathrm{k} \tan \theta$ $\text { (I) }=\frac{2 \mathrm{RB}_{\mathrm{H}}}{\mu_{\mathrm{o}} \mathrm{N}} \tan \theta\left(\because \mathrm{k}=\frac{2 \mathrm{RB}_{\mathrm{H}}}{\mu_{0} \mathrm{~N}}\right)$ Therefore, $S=\frac{\theta}{\left(\frac{2 \mathrm{RB}_{\mathrm{H}}}{\mu_{\mathrm{o}} \mathrm{N}}\right) \tan \theta}$ Hence, the sensitivity of a tangent galvanometer increases, if the number of turns in the coil increases.
CG PET 2019
Current Electricity
152877
A galvanometer of $50 \pi$ resistance has 25 divisions. A current of $4 \times 10^{-4} \mathrm{~A}$ gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25 volt, it should be connected with a resistance of-
1 $245 \pi$ as shunt
2 $2450 \pi$ as series
3 $2500 \pi$ as shunt
4 $2550 \pi$ as series
Explanation:
B The total current shown by the galvanometer $=25 \times 4 \times 10^{-4} \mathrm{~A}$ $\mathrm{I}_{\mathrm{g}}=10^{-2} \mathrm{Amp}$ The value of resistance connected in series to convert galvanometer into voltmeter of $25 \mathrm{~V}$ is $\mathrm{R} =\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{G}=\frac{25}{10^{-2}}-50 \pi$ $=2450 \pi \text { as series }$
JIPMER - 2019
Current Electricity
152878
$A B$ is a wire of uniform resistance. The galvanometer $G$ shows no current when the length $A C=20 \mathrm{~cm}$ and $C B=80 \mathrm{~cm}$. Then, the resistance $R$ is equal to
1 $12 \Omega$
2 $16 \Omega$
3 $20 \Omega$
4 $40 \Omega$
Explanation:
C The balance condition for this meter bridge, $\frac{\mathrm{R}}{80}=\frac{\mathrm{AC}}{\mathrm{BC}}$ $\frac{\mathrm{R}}{80}=\frac{20}{80}$ $\mathrm{R}=20 \Omega$
152875
A galvanometer having a resistance of $8 \Omega$ is shunted by a wire of resistance $2 \Omega$. If the total current is $1 \mathrm{~A}$, the part of it passing through the shunt will be
1 $1.2 \mathrm{~A}$
2 $0.8 \mathrm{~A}$
3 $0.5 \mathrm{~A}$
4 $0.3 \mathrm{~A}$
Explanation:
B Applying Kirchhoff Voltage law, $\mathrm{I}_{\mathrm{g}} \mathrm{G}=\mathrm{I}_{\mathrm{S}} \cdot \mathrm{S}$ $\left(\mathrm{I}-\mathrm{I}_{\mathrm{S}}\right) \mathrm{G}=\mathrm{I}_{\mathrm{S}} \cdot \mathrm{S}$ $\mathrm{I}_{\mathrm{S}}=\frac{\mathrm{IG}}{\mathrm{G}+\mathrm{S}}=\frac{1 \times 8}{8+2}$ $\mathrm{I}_{\mathrm{S}}=\frac{8}{10}=0.8$
TS EAMCET 08.05.2019
Current Electricity
152876
The sensitivity of tangent galvanometer is increased, if
1 number of turns decreases
2 number of turns increases
3 fields increases
4 None of the above
Explanation:
B The sensitivity of a tangent galvanometer is defined as the ratio of change in deflection to the current responsible for the deflection. $\therefore$ Sensitivity of tangent galvanometer $(\mathrm{S})=\frac{\theta}{\mathrm{I}}$ We know that, current $(\mathrm{I})=\mathrm{k} \tan \theta$ $\text { (I) }=\frac{2 \mathrm{RB}_{\mathrm{H}}}{\mu_{\mathrm{o}} \mathrm{N}} \tan \theta\left(\because \mathrm{k}=\frac{2 \mathrm{RB}_{\mathrm{H}}}{\mu_{0} \mathrm{~N}}\right)$ Therefore, $S=\frac{\theta}{\left(\frac{2 \mathrm{RB}_{\mathrm{H}}}{\mu_{\mathrm{o}} \mathrm{N}}\right) \tan \theta}$ Hence, the sensitivity of a tangent galvanometer increases, if the number of turns in the coil increases.
CG PET 2019
Current Electricity
152877
A galvanometer of $50 \pi$ resistance has 25 divisions. A current of $4 \times 10^{-4} \mathrm{~A}$ gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25 volt, it should be connected with a resistance of-
1 $245 \pi$ as shunt
2 $2450 \pi$ as series
3 $2500 \pi$ as shunt
4 $2550 \pi$ as series
Explanation:
B The total current shown by the galvanometer $=25 \times 4 \times 10^{-4} \mathrm{~A}$ $\mathrm{I}_{\mathrm{g}}=10^{-2} \mathrm{Amp}$ The value of resistance connected in series to convert galvanometer into voltmeter of $25 \mathrm{~V}$ is $\mathrm{R} =\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{G}=\frac{25}{10^{-2}}-50 \pi$ $=2450 \pi \text { as series }$
JIPMER - 2019
Current Electricity
152878
$A B$ is a wire of uniform resistance. The galvanometer $G$ shows no current when the length $A C=20 \mathrm{~cm}$ and $C B=80 \mathrm{~cm}$. Then, the resistance $R$ is equal to
1 $12 \Omega$
2 $16 \Omega$
3 $20 \Omega$
4 $40 \Omega$
Explanation:
C The balance condition for this meter bridge, $\frac{\mathrm{R}}{80}=\frac{\mathrm{AC}}{\mathrm{BC}}$ $\frac{\mathrm{R}}{80}=\frac{20}{80}$ $\mathrm{R}=20 \Omega$
152875
A galvanometer having a resistance of $8 \Omega$ is shunted by a wire of resistance $2 \Omega$. If the total current is $1 \mathrm{~A}$, the part of it passing through the shunt will be
1 $1.2 \mathrm{~A}$
2 $0.8 \mathrm{~A}$
3 $0.5 \mathrm{~A}$
4 $0.3 \mathrm{~A}$
Explanation:
B Applying Kirchhoff Voltage law, $\mathrm{I}_{\mathrm{g}} \mathrm{G}=\mathrm{I}_{\mathrm{S}} \cdot \mathrm{S}$ $\left(\mathrm{I}-\mathrm{I}_{\mathrm{S}}\right) \mathrm{G}=\mathrm{I}_{\mathrm{S}} \cdot \mathrm{S}$ $\mathrm{I}_{\mathrm{S}}=\frac{\mathrm{IG}}{\mathrm{G}+\mathrm{S}}=\frac{1 \times 8}{8+2}$ $\mathrm{I}_{\mathrm{S}}=\frac{8}{10}=0.8$
TS EAMCET 08.05.2019
Current Electricity
152876
The sensitivity of tangent galvanometer is increased, if
1 number of turns decreases
2 number of turns increases
3 fields increases
4 None of the above
Explanation:
B The sensitivity of a tangent galvanometer is defined as the ratio of change in deflection to the current responsible for the deflection. $\therefore$ Sensitivity of tangent galvanometer $(\mathrm{S})=\frac{\theta}{\mathrm{I}}$ We know that, current $(\mathrm{I})=\mathrm{k} \tan \theta$ $\text { (I) }=\frac{2 \mathrm{RB}_{\mathrm{H}}}{\mu_{\mathrm{o}} \mathrm{N}} \tan \theta\left(\because \mathrm{k}=\frac{2 \mathrm{RB}_{\mathrm{H}}}{\mu_{0} \mathrm{~N}}\right)$ Therefore, $S=\frac{\theta}{\left(\frac{2 \mathrm{RB}_{\mathrm{H}}}{\mu_{\mathrm{o}} \mathrm{N}}\right) \tan \theta}$ Hence, the sensitivity of a tangent galvanometer increases, if the number of turns in the coil increases.
CG PET 2019
Current Electricity
152877
A galvanometer of $50 \pi$ resistance has 25 divisions. A current of $4 \times 10^{-4} \mathrm{~A}$ gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25 volt, it should be connected with a resistance of-
1 $245 \pi$ as shunt
2 $2450 \pi$ as series
3 $2500 \pi$ as shunt
4 $2550 \pi$ as series
Explanation:
B The total current shown by the galvanometer $=25 \times 4 \times 10^{-4} \mathrm{~A}$ $\mathrm{I}_{\mathrm{g}}=10^{-2} \mathrm{Amp}$ The value of resistance connected in series to convert galvanometer into voltmeter of $25 \mathrm{~V}$ is $\mathrm{R} =\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{G}=\frac{25}{10^{-2}}-50 \pi$ $=2450 \pi \text { as series }$
JIPMER - 2019
Current Electricity
152878
$A B$ is a wire of uniform resistance. The galvanometer $G$ shows no current when the length $A C=20 \mathrm{~cm}$ and $C B=80 \mathrm{~cm}$. Then, the resistance $R$ is equal to
1 $12 \Omega$
2 $16 \Omega$
3 $20 \Omega$
4 $40 \Omega$
Explanation:
C The balance condition for this meter bridge, $\frac{\mathrm{R}}{80}=\frac{\mathrm{AC}}{\mathrm{BC}}$ $\frac{\mathrm{R}}{80}=\frac{20}{80}$ $\mathrm{R}=20 \Omega$
152875
A galvanometer having a resistance of $8 \Omega$ is shunted by a wire of resistance $2 \Omega$. If the total current is $1 \mathrm{~A}$, the part of it passing through the shunt will be
1 $1.2 \mathrm{~A}$
2 $0.8 \mathrm{~A}$
3 $0.5 \mathrm{~A}$
4 $0.3 \mathrm{~A}$
Explanation:
B Applying Kirchhoff Voltage law, $\mathrm{I}_{\mathrm{g}} \mathrm{G}=\mathrm{I}_{\mathrm{S}} \cdot \mathrm{S}$ $\left(\mathrm{I}-\mathrm{I}_{\mathrm{S}}\right) \mathrm{G}=\mathrm{I}_{\mathrm{S}} \cdot \mathrm{S}$ $\mathrm{I}_{\mathrm{S}}=\frac{\mathrm{IG}}{\mathrm{G}+\mathrm{S}}=\frac{1 \times 8}{8+2}$ $\mathrm{I}_{\mathrm{S}}=\frac{8}{10}=0.8$
TS EAMCET 08.05.2019
Current Electricity
152876
The sensitivity of tangent galvanometer is increased, if
1 number of turns decreases
2 number of turns increases
3 fields increases
4 None of the above
Explanation:
B The sensitivity of a tangent galvanometer is defined as the ratio of change in deflection to the current responsible for the deflection. $\therefore$ Sensitivity of tangent galvanometer $(\mathrm{S})=\frac{\theta}{\mathrm{I}}$ We know that, current $(\mathrm{I})=\mathrm{k} \tan \theta$ $\text { (I) }=\frac{2 \mathrm{RB}_{\mathrm{H}}}{\mu_{\mathrm{o}} \mathrm{N}} \tan \theta\left(\because \mathrm{k}=\frac{2 \mathrm{RB}_{\mathrm{H}}}{\mu_{0} \mathrm{~N}}\right)$ Therefore, $S=\frac{\theta}{\left(\frac{2 \mathrm{RB}_{\mathrm{H}}}{\mu_{\mathrm{o}} \mathrm{N}}\right) \tan \theta}$ Hence, the sensitivity of a tangent galvanometer increases, if the number of turns in the coil increases.
CG PET 2019
Current Electricity
152877
A galvanometer of $50 \pi$ resistance has 25 divisions. A current of $4 \times 10^{-4} \mathrm{~A}$ gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25 volt, it should be connected with a resistance of-
1 $245 \pi$ as shunt
2 $2450 \pi$ as series
3 $2500 \pi$ as shunt
4 $2550 \pi$ as series
Explanation:
B The total current shown by the galvanometer $=25 \times 4 \times 10^{-4} \mathrm{~A}$ $\mathrm{I}_{\mathrm{g}}=10^{-2} \mathrm{Amp}$ The value of resistance connected in series to convert galvanometer into voltmeter of $25 \mathrm{~V}$ is $\mathrm{R} =\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{G}=\frac{25}{10^{-2}}-50 \pi$ $=2450 \pi \text { as series }$
JIPMER - 2019
Current Electricity
152878
$A B$ is a wire of uniform resistance. The galvanometer $G$ shows no current when the length $A C=20 \mathrm{~cm}$ and $C B=80 \mathrm{~cm}$. Then, the resistance $R$ is equal to
1 $12 \Omega$
2 $16 \Omega$
3 $20 \Omega$
4 $40 \Omega$
Explanation:
C The balance condition for this meter bridge, $\frac{\mathrm{R}}{80}=\frac{\mathrm{AC}}{\mathrm{BC}}$ $\frac{\mathrm{R}}{80}=\frac{20}{80}$ $\mathrm{R}=20 \Omega$
