NEET Test Series from KOTA - 10 Papers In MS WORD
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Magnetism and Matter
154259
Two tangent galvanometers $A$ and $B$ are identical except in their number of turns. They are connected in series. On passing a current through them, deflections of $60^{\circ}$ and $30^{\circ}$ are produced. The ratio of the number of turns in $A$ and $B$ is :
1 $1: 3$
2 $3: 1$
3 $1: 2$
4 $2: 1$
Explanation:
B Given, $\theta_{1}=60^{\circ}, \theta_{2}=30^{\circ}$ We know that, $\tan \theta=\mathrm{KNI}$ Where, $\mathrm{K}=$ Constant $\mathrm{N}=\text { Number of turns }$ $\mathrm{I}=\text { Current }$ $\frac{\tan 60^{\circ}}{\tan 30^{\circ}}=\frac{\mathrm{KN}_{1} \mathrm{I}}{\mathrm{KN}_{2} \mathrm{I}}$ $\frac{\sqrt{3}}{1 / \sqrt{3}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{3}{1}$ Hence, ratio of the number of the turns $=3: 1$
Karnataka CET-2008
Magnetism and Matter
154260
In the circuit shown below, the ammeter and the voltmeter readings are $3 \mathrm{~A}$ and $6 \mathrm{~V}$ respectively. Then, the value of the resistance $R$ is :
1 $ \lt 2 \Omega$
2 $2 \Omega$
3 $\geq 2 \Omega$
4 $>2 \Omega$
Explanation:
A Given, $\mathrm{I}=3 \mathrm{~A}, \mathrm{~V}=6 \mathrm{~V}$ We know that, $\mathrm{V}=\mathrm{IR}$ $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{6}{3}=2 \Omega$ For a non-ideal ammeter, resistance of ammeter is not zero hence measured voltage is the sum of voltage across ammeter and resistance. In other word, measured voltage is more than the actual voltage across the resistor $\left(\mathrm{V}_{\mathrm{R}}\right)$ or $\mathrm{V}>\mathrm{V}_{\mathrm{R}}$, Measured resistance $(2 \Omega)$ is more than actual resistance. Therefore actual resistance $ \lt 2 \Omega$
Karnataka CET-2015
Magnetism and Matter
154265
In a moving coil galvanometer, the coil having 500 turns and an average area of $4 \mathrm{~cm}^{2}$ carries a current of $0.1 \mathrm{~A}$ and is set parallel to a magnetic field with a torque of $0.15 \mathrm{~N} \mathrm{~m}$. The strength of the magnetic field is
154267
In the circuit shown, the galvanometer $G$ of resistance $60 \Omega$ is shunted by a resistance $r=$ $0.02 \Omega$. The current through $R$ is nearly $1 \mathrm{~A}$. The value of resistance $R$ (in ohms) is nearly
1 $1.00 \Omega$
2 $5.00 \Omega$
3 $11.0 \Omega$
4 $60.00 \Omega$
Explanation:
B Given, Galvanometer resistance $(\mathrm{G})=60 \Omega$ Shunt $(\mathrm{S})=0.02 \Omega$ Current $(\mathrm{I})=1 \mathrm{~A}$ Equivalent resistance of $\mathrm{G}$ and shunt $\mathrm{R}^{\prime}=\frac{\mathrm{GS}}{\mathrm{G}+\mathrm{S}}$ $\mathrm{R}^{\prime}=\frac{60 \times 0.02}{60+0.02}=0.0199$ Total resistance of circuit $\left(R_{\text {eq }}\right)=R+R^{\prime}=R+0.0199$ $\mathrm{V}=\mathrm{IR}_{\text {eq }}$ $5=1 \times(\mathrm{R}+0.0199)$ $\mathrm{R}=5-0.0199$ $\mathrm{R}=4.98 = 5.00 \Omega$
154259
Two tangent galvanometers $A$ and $B$ are identical except in their number of turns. They are connected in series. On passing a current through them, deflections of $60^{\circ}$ and $30^{\circ}$ are produced. The ratio of the number of turns in $A$ and $B$ is :
1 $1: 3$
2 $3: 1$
3 $1: 2$
4 $2: 1$
Explanation:
B Given, $\theta_{1}=60^{\circ}, \theta_{2}=30^{\circ}$ We know that, $\tan \theta=\mathrm{KNI}$ Where, $\mathrm{K}=$ Constant $\mathrm{N}=\text { Number of turns }$ $\mathrm{I}=\text { Current }$ $\frac{\tan 60^{\circ}}{\tan 30^{\circ}}=\frac{\mathrm{KN}_{1} \mathrm{I}}{\mathrm{KN}_{2} \mathrm{I}}$ $\frac{\sqrt{3}}{1 / \sqrt{3}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{3}{1}$ Hence, ratio of the number of the turns $=3: 1$
Karnataka CET-2008
Magnetism and Matter
154260
In the circuit shown below, the ammeter and the voltmeter readings are $3 \mathrm{~A}$ and $6 \mathrm{~V}$ respectively. Then, the value of the resistance $R$ is :
1 $ \lt 2 \Omega$
2 $2 \Omega$
3 $\geq 2 \Omega$
4 $>2 \Omega$
Explanation:
A Given, $\mathrm{I}=3 \mathrm{~A}, \mathrm{~V}=6 \mathrm{~V}$ We know that, $\mathrm{V}=\mathrm{IR}$ $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{6}{3}=2 \Omega$ For a non-ideal ammeter, resistance of ammeter is not zero hence measured voltage is the sum of voltage across ammeter and resistance. In other word, measured voltage is more than the actual voltage across the resistor $\left(\mathrm{V}_{\mathrm{R}}\right)$ or $\mathrm{V}>\mathrm{V}_{\mathrm{R}}$, Measured resistance $(2 \Omega)$ is more than actual resistance. Therefore actual resistance $ \lt 2 \Omega$
Karnataka CET-2015
Magnetism and Matter
154265
In a moving coil galvanometer, the coil having 500 turns and an average area of $4 \mathrm{~cm}^{2}$ carries a current of $0.1 \mathrm{~A}$ and is set parallel to a magnetic field with a torque of $0.15 \mathrm{~N} \mathrm{~m}$. The strength of the magnetic field is
154267
In the circuit shown, the galvanometer $G$ of resistance $60 \Omega$ is shunted by a resistance $r=$ $0.02 \Omega$. The current through $R$ is nearly $1 \mathrm{~A}$. The value of resistance $R$ (in ohms) is nearly
1 $1.00 \Omega$
2 $5.00 \Omega$
3 $11.0 \Omega$
4 $60.00 \Omega$
Explanation:
B Given, Galvanometer resistance $(\mathrm{G})=60 \Omega$ Shunt $(\mathrm{S})=0.02 \Omega$ Current $(\mathrm{I})=1 \mathrm{~A}$ Equivalent resistance of $\mathrm{G}$ and shunt $\mathrm{R}^{\prime}=\frac{\mathrm{GS}}{\mathrm{G}+\mathrm{S}}$ $\mathrm{R}^{\prime}=\frac{60 \times 0.02}{60+0.02}=0.0199$ Total resistance of circuit $\left(R_{\text {eq }}\right)=R+R^{\prime}=R+0.0199$ $\mathrm{V}=\mathrm{IR}_{\text {eq }}$ $5=1 \times(\mathrm{R}+0.0199)$ $\mathrm{R}=5-0.0199$ $\mathrm{R}=4.98 = 5.00 \Omega$
154259
Two tangent galvanometers $A$ and $B$ are identical except in their number of turns. They are connected in series. On passing a current through them, deflections of $60^{\circ}$ and $30^{\circ}$ are produced. The ratio of the number of turns in $A$ and $B$ is :
1 $1: 3$
2 $3: 1$
3 $1: 2$
4 $2: 1$
Explanation:
B Given, $\theta_{1}=60^{\circ}, \theta_{2}=30^{\circ}$ We know that, $\tan \theta=\mathrm{KNI}$ Where, $\mathrm{K}=$ Constant $\mathrm{N}=\text { Number of turns }$ $\mathrm{I}=\text { Current }$ $\frac{\tan 60^{\circ}}{\tan 30^{\circ}}=\frac{\mathrm{KN}_{1} \mathrm{I}}{\mathrm{KN}_{2} \mathrm{I}}$ $\frac{\sqrt{3}}{1 / \sqrt{3}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{3}{1}$ Hence, ratio of the number of the turns $=3: 1$
Karnataka CET-2008
Magnetism and Matter
154260
In the circuit shown below, the ammeter and the voltmeter readings are $3 \mathrm{~A}$ and $6 \mathrm{~V}$ respectively. Then, the value of the resistance $R$ is :
1 $ \lt 2 \Omega$
2 $2 \Omega$
3 $\geq 2 \Omega$
4 $>2 \Omega$
Explanation:
A Given, $\mathrm{I}=3 \mathrm{~A}, \mathrm{~V}=6 \mathrm{~V}$ We know that, $\mathrm{V}=\mathrm{IR}$ $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{6}{3}=2 \Omega$ For a non-ideal ammeter, resistance of ammeter is not zero hence measured voltage is the sum of voltage across ammeter and resistance. In other word, measured voltage is more than the actual voltage across the resistor $\left(\mathrm{V}_{\mathrm{R}}\right)$ or $\mathrm{V}>\mathrm{V}_{\mathrm{R}}$, Measured resistance $(2 \Omega)$ is more than actual resistance. Therefore actual resistance $ \lt 2 \Omega$
Karnataka CET-2015
Magnetism and Matter
154265
In a moving coil galvanometer, the coil having 500 turns and an average area of $4 \mathrm{~cm}^{2}$ carries a current of $0.1 \mathrm{~A}$ and is set parallel to a magnetic field with a torque of $0.15 \mathrm{~N} \mathrm{~m}$. The strength of the magnetic field is
154267
In the circuit shown, the galvanometer $G$ of resistance $60 \Omega$ is shunted by a resistance $r=$ $0.02 \Omega$. The current through $R$ is nearly $1 \mathrm{~A}$. The value of resistance $R$ (in ohms) is nearly
1 $1.00 \Omega$
2 $5.00 \Omega$
3 $11.0 \Omega$
4 $60.00 \Omega$
Explanation:
B Given, Galvanometer resistance $(\mathrm{G})=60 \Omega$ Shunt $(\mathrm{S})=0.02 \Omega$ Current $(\mathrm{I})=1 \mathrm{~A}$ Equivalent resistance of $\mathrm{G}$ and shunt $\mathrm{R}^{\prime}=\frac{\mathrm{GS}}{\mathrm{G}+\mathrm{S}}$ $\mathrm{R}^{\prime}=\frac{60 \times 0.02}{60+0.02}=0.0199$ Total resistance of circuit $\left(R_{\text {eq }}\right)=R+R^{\prime}=R+0.0199$ $\mathrm{V}=\mathrm{IR}_{\text {eq }}$ $5=1 \times(\mathrm{R}+0.0199)$ $\mathrm{R}=5-0.0199$ $\mathrm{R}=4.98 = 5.00 \Omega$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Magnetism and Matter
154259
Two tangent galvanometers $A$ and $B$ are identical except in their number of turns. They are connected in series. On passing a current through them, deflections of $60^{\circ}$ and $30^{\circ}$ are produced. The ratio of the number of turns in $A$ and $B$ is :
1 $1: 3$
2 $3: 1$
3 $1: 2$
4 $2: 1$
Explanation:
B Given, $\theta_{1}=60^{\circ}, \theta_{2}=30^{\circ}$ We know that, $\tan \theta=\mathrm{KNI}$ Where, $\mathrm{K}=$ Constant $\mathrm{N}=\text { Number of turns }$ $\mathrm{I}=\text { Current }$ $\frac{\tan 60^{\circ}}{\tan 30^{\circ}}=\frac{\mathrm{KN}_{1} \mathrm{I}}{\mathrm{KN}_{2} \mathrm{I}}$ $\frac{\sqrt{3}}{1 / \sqrt{3}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{3}{1}$ Hence, ratio of the number of the turns $=3: 1$
Karnataka CET-2008
Magnetism and Matter
154260
In the circuit shown below, the ammeter and the voltmeter readings are $3 \mathrm{~A}$ and $6 \mathrm{~V}$ respectively. Then, the value of the resistance $R$ is :
1 $ \lt 2 \Omega$
2 $2 \Omega$
3 $\geq 2 \Omega$
4 $>2 \Omega$
Explanation:
A Given, $\mathrm{I}=3 \mathrm{~A}, \mathrm{~V}=6 \mathrm{~V}$ We know that, $\mathrm{V}=\mathrm{IR}$ $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{6}{3}=2 \Omega$ For a non-ideal ammeter, resistance of ammeter is not zero hence measured voltage is the sum of voltage across ammeter and resistance. In other word, measured voltage is more than the actual voltage across the resistor $\left(\mathrm{V}_{\mathrm{R}}\right)$ or $\mathrm{V}>\mathrm{V}_{\mathrm{R}}$, Measured resistance $(2 \Omega)$ is more than actual resistance. Therefore actual resistance $ \lt 2 \Omega$
Karnataka CET-2015
Magnetism and Matter
154265
In a moving coil galvanometer, the coil having 500 turns and an average area of $4 \mathrm{~cm}^{2}$ carries a current of $0.1 \mathrm{~A}$ and is set parallel to a magnetic field with a torque of $0.15 \mathrm{~N} \mathrm{~m}$. The strength of the magnetic field is
154267
In the circuit shown, the galvanometer $G$ of resistance $60 \Omega$ is shunted by a resistance $r=$ $0.02 \Omega$. The current through $R$ is nearly $1 \mathrm{~A}$. The value of resistance $R$ (in ohms) is nearly
1 $1.00 \Omega$
2 $5.00 \Omega$
3 $11.0 \Omega$
4 $60.00 \Omega$
Explanation:
B Given, Galvanometer resistance $(\mathrm{G})=60 \Omega$ Shunt $(\mathrm{S})=0.02 \Omega$ Current $(\mathrm{I})=1 \mathrm{~A}$ Equivalent resistance of $\mathrm{G}$ and shunt $\mathrm{R}^{\prime}=\frac{\mathrm{GS}}{\mathrm{G}+\mathrm{S}}$ $\mathrm{R}^{\prime}=\frac{60 \times 0.02}{60+0.02}=0.0199$ Total resistance of circuit $\left(R_{\text {eq }}\right)=R+R^{\prime}=R+0.0199$ $\mathrm{V}=\mathrm{IR}_{\text {eq }}$ $5=1 \times(\mathrm{R}+0.0199)$ $\mathrm{R}=5-0.0199$ $\mathrm{R}=4.98 = 5.00 \Omega$
