154211
The current sensitivity of moving coil galvanometer of resistance $100 \Omega$ is $1 \mathrm{div} / \mathrm{mA}$. Its voltage sensitivity is
1 $12 \operatorname{div} / \mathrm{V}$
2 $10 \operatorname{div} / \mathrm{V}$
3 $5 \mathrm{div} / \mathrm{V}$
4 $15 \operatorname{div} / \mathrm{V}$
Explanation:
B Given, $\mathrm{R}=100 \Omega$ Current Sensitivity $\mathrm{I}_{\mathrm{S}}=\frac{\mathrm{NBA}}{\mathrm{k}}=1 \frac{\mathrm{div}}{\mathrm{mA}}=1000 \frac{\mathrm{div}}{\mathrm{A}}$ Voltage Sensitivity $\mathrm{V}_{\mathrm{S}}=\frac{\mathrm{NBA}}{\mathrm{kR}}=\frac{1000}{100} \frac{\mathrm{div}}{\mathrm{V}}$ $\mathrm{V}_{\mathrm{S}}=10 \mathrm{div} / \mathrm{V}$
MHT-CET 2020
Magnetism and Matter
154214
Two galvanometers ' $G$ ' ' and ' $G{ }_{2}$ ' require $2 \mathrm{~mA}$ and $3 \mathrm{~mA}$ respectively to produce the same deflection. Then
1 Sensitivity of $\mathrm{G}_{2}$ is $\frac{3}{2}$ times that of $\mathrm{G}_{1}$
2 $\mathrm{G}_{1}$ and $\mathrm{G}_{2}$ are equally sensitive
3 $G_{1}$ is less sensitive then $G_{2}$
4 $\mathrm{G}_{1}$ is more sensitive than $\mathrm{G}_{2}$
Explanation:
D In first galvanometer, $\mathrm{G}_{1}=\frac{\theta}{\mathrm{I}_{1}}=\frac{\theta}{2 \mathrm{~mA}}$ And in second galvanometer, $\mathrm{G}_{2}=\frac{\theta}{\mathrm{I}_{2}}=\frac{\theta}{3 \mathrm{~mA}}$ On comparing equation (i) and equation (ii), we get$\mathrm{G}_{1}>\mathrm{G}_{2}$ $\mathrm{G}_{1}$ is more sensitive than $\mathrm{G}_{2}$.
MHT-CET 2020
Magnetism and Matter
154215
The current flowing through moving coil galvanometer is $20 \%$ of the current to be measured. The resistance of moving coil galvanometer is $48 \Omega$, then shunt required is
1 $48 \Omega$
2 $12 \Omega$
3 $96 \Omega$
4 $24 \Omega$
Explanation:
B Given, Resistance of moving coil galvanometer, $G=48 \Omega$ According to question, $\mathrm{I}_{\mathrm{g}}=20 \% \text { of } \mathrm{I}=\frac{20}{100} \mathrm{I}=0.2 \mathrm{I}$ We know that, $I_{g} G=\left(I-I_{g}\right) S$ $S=\frac{I_{g} G}{I-I_{g}}=\frac{0.2 I G}{I-0.2 I}$ $S=\frac{0.2 I G}{I(1-0.2)}$ $S=\frac{0.2 G}{0.8}$ $S=\frac{1}{4} \times 48$ $S=12 \Omega$
MHT-CET 2020
Magnetism and Matter
154216
A galvanometer having a resistance of $18 \Omega$ shunted by a wire of resistance $2 \Omega$. If the total current passing through the combination is 2 $A$, then current through shunt will be
1 $1.8 \mathrm{~A}$
2 $0.9 \mathrm{~A}$
3 $12 \mathrm{~A}$
4 $12.2 \mathrm{~A}$
Explanation:
A Given, $\mathrm{G}=18 \Omega$ $\mathrm{S}=2 \Omega$ $\mathrm{I}=2 \mathrm{Amp}$ We know that, $\left(I-I_{S}\right) G=I_{S} S$ $I_{S}=\frac{I G}{G+S}=\frac{2 \times 18}{18+2}=\frac{36}{20}=\frac{18}{10}=\frac{9}{5}$ $I_{S}=1.8 \mathrm{Amp}$
154211
The current sensitivity of moving coil galvanometer of resistance $100 \Omega$ is $1 \mathrm{div} / \mathrm{mA}$. Its voltage sensitivity is
1 $12 \operatorname{div} / \mathrm{V}$
2 $10 \operatorname{div} / \mathrm{V}$
3 $5 \mathrm{div} / \mathrm{V}$
4 $15 \operatorname{div} / \mathrm{V}$
Explanation:
B Given, $\mathrm{R}=100 \Omega$ Current Sensitivity $\mathrm{I}_{\mathrm{S}}=\frac{\mathrm{NBA}}{\mathrm{k}}=1 \frac{\mathrm{div}}{\mathrm{mA}}=1000 \frac{\mathrm{div}}{\mathrm{A}}$ Voltage Sensitivity $\mathrm{V}_{\mathrm{S}}=\frac{\mathrm{NBA}}{\mathrm{kR}}=\frac{1000}{100} \frac{\mathrm{div}}{\mathrm{V}}$ $\mathrm{V}_{\mathrm{S}}=10 \mathrm{div} / \mathrm{V}$
MHT-CET 2020
Magnetism and Matter
154214
Two galvanometers ' $G$ ' ' and ' $G{ }_{2}$ ' require $2 \mathrm{~mA}$ and $3 \mathrm{~mA}$ respectively to produce the same deflection. Then
1 Sensitivity of $\mathrm{G}_{2}$ is $\frac{3}{2}$ times that of $\mathrm{G}_{1}$
2 $\mathrm{G}_{1}$ and $\mathrm{G}_{2}$ are equally sensitive
3 $G_{1}$ is less sensitive then $G_{2}$
4 $\mathrm{G}_{1}$ is more sensitive than $\mathrm{G}_{2}$
Explanation:
D In first galvanometer, $\mathrm{G}_{1}=\frac{\theta}{\mathrm{I}_{1}}=\frac{\theta}{2 \mathrm{~mA}}$ And in second galvanometer, $\mathrm{G}_{2}=\frac{\theta}{\mathrm{I}_{2}}=\frac{\theta}{3 \mathrm{~mA}}$ On comparing equation (i) and equation (ii), we get$\mathrm{G}_{1}>\mathrm{G}_{2}$ $\mathrm{G}_{1}$ is more sensitive than $\mathrm{G}_{2}$.
MHT-CET 2020
Magnetism and Matter
154215
The current flowing through moving coil galvanometer is $20 \%$ of the current to be measured. The resistance of moving coil galvanometer is $48 \Omega$, then shunt required is
1 $48 \Omega$
2 $12 \Omega$
3 $96 \Omega$
4 $24 \Omega$
Explanation:
B Given, Resistance of moving coil galvanometer, $G=48 \Omega$ According to question, $\mathrm{I}_{\mathrm{g}}=20 \% \text { of } \mathrm{I}=\frac{20}{100} \mathrm{I}=0.2 \mathrm{I}$ We know that, $I_{g} G=\left(I-I_{g}\right) S$ $S=\frac{I_{g} G}{I-I_{g}}=\frac{0.2 I G}{I-0.2 I}$ $S=\frac{0.2 I G}{I(1-0.2)}$ $S=\frac{0.2 G}{0.8}$ $S=\frac{1}{4} \times 48$ $S=12 \Omega$
MHT-CET 2020
Magnetism and Matter
154216
A galvanometer having a resistance of $18 \Omega$ shunted by a wire of resistance $2 \Omega$. If the total current passing through the combination is 2 $A$, then current through shunt will be
1 $1.8 \mathrm{~A}$
2 $0.9 \mathrm{~A}$
3 $12 \mathrm{~A}$
4 $12.2 \mathrm{~A}$
Explanation:
A Given, $\mathrm{G}=18 \Omega$ $\mathrm{S}=2 \Omega$ $\mathrm{I}=2 \mathrm{Amp}$ We know that, $\left(I-I_{S}\right) G=I_{S} S$ $I_{S}=\frac{I G}{G+S}=\frac{2 \times 18}{18+2}=\frac{36}{20}=\frac{18}{10}=\frac{9}{5}$ $I_{S}=1.8 \mathrm{Amp}$
154211
The current sensitivity of moving coil galvanometer of resistance $100 \Omega$ is $1 \mathrm{div} / \mathrm{mA}$. Its voltage sensitivity is
1 $12 \operatorname{div} / \mathrm{V}$
2 $10 \operatorname{div} / \mathrm{V}$
3 $5 \mathrm{div} / \mathrm{V}$
4 $15 \operatorname{div} / \mathrm{V}$
Explanation:
B Given, $\mathrm{R}=100 \Omega$ Current Sensitivity $\mathrm{I}_{\mathrm{S}}=\frac{\mathrm{NBA}}{\mathrm{k}}=1 \frac{\mathrm{div}}{\mathrm{mA}}=1000 \frac{\mathrm{div}}{\mathrm{A}}$ Voltage Sensitivity $\mathrm{V}_{\mathrm{S}}=\frac{\mathrm{NBA}}{\mathrm{kR}}=\frac{1000}{100} \frac{\mathrm{div}}{\mathrm{V}}$ $\mathrm{V}_{\mathrm{S}}=10 \mathrm{div} / \mathrm{V}$
MHT-CET 2020
Magnetism and Matter
154214
Two galvanometers ' $G$ ' ' and ' $G{ }_{2}$ ' require $2 \mathrm{~mA}$ and $3 \mathrm{~mA}$ respectively to produce the same deflection. Then
1 Sensitivity of $\mathrm{G}_{2}$ is $\frac{3}{2}$ times that of $\mathrm{G}_{1}$
2 $\mathrm{G}_{1}$ and $\mathrm{G}_{2}$ are equally sensitive
3 $G_{1}$ is less sensitive then $G_{2}$
4 $\mathrm{G}_{1}$ is more sensitive than $\mathrm{G}_{2}$
Explanation:
D In first galvanometer, $\mathrm{G}_{1}=\frac{\theta}{\mathrm{I}_{1}}=\frac{\theta}{2 \mathrm{~mA}}$ And in second galvanometer, $\mathrm{G}_{2}=\frac{\theta}{\mathrm{I}_{2}}=\frac{\theta}{3 \mathrm{~mA}}$ On comparing equation (i) and equation (ii), we get$\mathrm{G}_{1}>\mathrm{G}_{2}$ $\mathrm{G}_{1}$ is more sensitive than $\mathrm{G}_{2}$.
MHT-CET 2020
Magnetism and Matter
154215
The current flowing through moving coil galvanometer is $20 \%$ of the current to be measured. The resistance of moving coil galvanometer is $48 \Omega$, then shunt required is
1 $48 \Omega$
2 $12 \Omega$
3 $96 \Omega$
4 $24 \Omega$
Explanation:
B Given, Resistance of moving coil galvanometer, $G=48 \Omega$ According to question, $\mathrm{I}_{\mathrm{g}}=20 \% \text { of } \mathrm{I}=\frac{20}{100} \mathrm{I}=0.2 \mathrm{I}$ We know that, $I_{g} G=\left(I-I_{g}\right) S$ $S=\frac{I_{g} G}{I-I_{g}}=\frac{0.2 I G}{I-0.2 I}$ $S=\frac{0.2 I G}{I(1-0.2)}$ $S=\frac{0.2 G}{0.8}$ $S=\frac{1}{4} \times 48$ $S=12 \Omega$
MHT-CET 2020
Magnetism and Matter
154216
A galvanometer having a resistance of $18 \Omega$ shunted by a wire of resistance $2 \Omega$. If the total current passing through the combination is 2 $A$, then current through shunt will be
1 $1.8 \mathrm{~A}$
2 $0.9 \mathrm{~A}$
3 $12 \mathrm{~A}$
4 $12.2 \mathrm{~A}$
Explanation:
A Given, $\mathrm{G}=18 \Omega$ $\mathrm{S}=2 \Omega$ $\mathrm{I}=2 \mathrm{Amp}$ We know that, $\left(I-I_{S}\right) G=I_{S} S$ $I_{S}=\frac{I G}{G+S}=\frac{2 \times 18}{18+2}=\frac{36}{20}=\frac{18}{10}=\frac{9}{5}$ $I_{S}=1.8 \mathrm{Amp}$
154211
The current sensitivity of moving coil galvanometer of resistance $100 \Omega$ is $1 \mathrm{div} / \mathrm{mA}$. Its voltage sensitivity is
1 $12 \operatorname{div} / \mathrm{V}$
2 $10 \operatorname{div} / \mathrm{V}$
3 $5 \mathrm{div} / \mathrm{V}$
4 $15 \operatorname{div} / \mathrm{V}$
Explanation:
B Given, $\mathrm{R}=100 \Omega$ Current Sensitivity $\mathrm{I}_{\mathrm{S}}=\frac{\mathrm{NBA}}{\mathrm{k}}=1 \frac{\mathrm{div}}{\mathrm{mA}}=1000 \frac{\mathrm{div}}{\mathrm{A}}$ Voltage Sensitivity $\mathrm{V}_{\mathrm{S}}=\frac{\mathrm{NBA}}{\mathrm{kR}}=\frac{1000}{100} \frac{\mathrm{div}}{\mathrm{V}}$ $\mathrm{V}_{\mathrm{S}}=10 \mathrm{div} / \mathrm{V}$
MHT-CET 2020
Magnetism and Matter
154214
Two galvanometers ' $G$ ' ' and ' $G{ }_{2}$ ' require $2 \mathrm{~mA}$ and $3 \mathrm{~mA}$ respectively to produce the same deflection. Then
1 Sensitivity of $\mathrm{G}_{2}$ is $\frac{3}{2}$ times that of $\mathrm{G}_{1}$
2 $\mathrm{G}_{1}$ and $\mathrm{G}_{2}$ are equally sensitive
3 $G_{1}$ is less sensitive then $G_{2}$
4 $\mathrm{G}_{1}$ is more sensitive than $\mathrm{G}_{2}$
Explanation:
D In first galvanometer, $\mathrm{G}_{1}=\frac{\theta}{\mathrm{I}_{1}}=\frac{\theta}{2 \mathrm{~mA}}$ And in second galvanometer, $\mathrm{G}_{2}=\frac{\theta}{\mathrm{I}_{2}}=\frac{\theta}{3 \mathrm{~mA}}$ On comparing equation (i) and equation (ii), we get$\mathrm{G}_{1}>\mathrm{G}_{2}$ $\mathrm{G}_{1}$ is more sensitive than $\mathrm{G}_{2}$.
MHT-CET 2020
Magnetism and Matter
154215
The current flowing through moving coil galvanometer is $20 \%$ of the current to be measured. The resistance of moving coil galvanometer is $48 \Omega$, then shunt required is
1 $48 \Omega$
2 $12 \Omega$
3 $96 \Omega$
4 $24 \Omega$
Explanation:
B Given, Resistance of moving coil galvanometer, $G=48 \Omega$ According to question, $\mathrm{I}_{\mathrm{g}}=20 \% \text { of } \mathrm{I}=\frac{20}{100} \mathrm{I}=0.2 \mathrm{I}$ We know that, $I_{g} G=\left(I-I_{g}\right) S$ $S=\frac{I_{g} G}{I-I_{g}}=\frac{0.2 I G}{I-0.2 I}$ $S=\frac{0.2 I G}{I(1-0.2)}$ $S=\frac{0.2 G}{0.8}$ $S=\frac{1}{4} \times 48$ $S=12 \Omega$
MHT-CET 2020
Magnetism and Matter
154216
A galvanometer having a resistance of $18 \Omega$ shunted by a wire of resistance $2 \Omega$. If the total current passing through the combination is 2 $A$, then current through shunt will be
1 $1.8 \mathrm{~A}$
2 $0.9 \mathrm{~A}$
3 $12 \mathrm{~A}$
4 $12.2 \mathrm{~A}$
Explanation:
A Given, $\mathrm{G}=18 \Omega$ $\mathrm{S}=2 \Omega$ $\mathrm{I}=2 \mathrm{Amp}$ We know that, $\left(I-I_{S}\right) G=I_{S} S$ $I_{S}=\frac{I G}{G+S}=\frac{2 \times 18}{18+2}=\frac{36}{20}=\frac{18}{10}=\frac{9}{5}$ $I_{S}=1.8 \mathrm{Amp}$
