01. Ohm's Law, Resistance, Conductivity and Thermal Dependency of Resistance
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Current Electricity

151894 The I-V graphs for two different electrical applicances $P$ and $Q$ are shown in the diagram. If $R_{P}$ and $R_{Q}$ be the resistances of the devices, then :

1 $R_{P}=R_{Q}$
2 $R_{P}>R_{Q}$
3 $\mathrm{R}_{\mathrm{P}}\lt\mathrm{R}_{\mathrm{Q}}$
4 $R_{P}=\frac{R_{Q}}{2}$
Karnataka CET-2018
Current Electricity

151896 The resistance of a wire is ' $R$ ' $o h m$. If it is melted and stretched to ' $n$ ' times its original length, its new resistance will be

1 $\frac{R}{n}$
2 $n^{2} R$
3 $\frac{\mathrm{R}}{\mathrm{n}^{2}}$
4 $\mathrm{nR}$
NEET-2017
Current Electricity

151897 A conducting wire has length ' $L_{1}$ ' and diameter ' $d_{1}$ '. After stretching the same wire length becomes ' $L_{2}$ ' and diameter ' $d_{2}$ '. The ratio of resistances before and after stretching is

1 $d_{2}^{4}: d_{1}^{4}$
2 $\mathrm{d}_{1}^{4}: \mathrm{d}_{2}^{4}$
3 $\mathrm{d}_{2}^{2}: \mathrm{d}_{1}^{2}$
4 $\mathrm{d}_{1}^{2}: \mathrm{d}_{2}^{2}$
MHT-CET 2018
Current Electricity

151899 The resistance of a wire at room temperature $30^{\circ} \mathrm{C}$ is found to be $10 \Omega$. Now to increase the resistance by $10 \%$. The temperature of the wire must be [The temperature coefficient of resistance of the material of the wire is $\mathbf{0 . 0 0 2}$ per ${ }^{0} \mathrm{Cl}$

1 $36{ }^{0} \mathrm{C}$
2 $83{ }^{\circ} \mathrm{C}$
3 $63{ }^{\circ} \mathrm{C}$
4 $33{ }^{\circ} \mathrm{C}$
UPSEE-2014
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Current Electricity

151894 The I-V graphs for two different electrical applicances $P$ and $Q$ are shown in the diagram. If $R_{P}$ and $R_{Q}$ be the resistances of the devices, then :

1 $R_{P}=R_{Q}$
2 $R_{P}>R_{Q}$
3 $\mathrm{R}_{\mathrm{P}}\lt\mathrm{R}_{\mathrm{Q}}$
4 $R_{P}=\frac{R_{Q}}{2}$
Karnataka CET-2018
Current Electricity

151896 The resistance of a wire is ' $R$ ' $o h m$. If it is melted and stretched to ' $n$ ' times its original length, its new resistance will be

1 $\frac{R}{n}$
2 $n^{2} R$
3 $\frac{\mathrm{R}}{\mathrm{n}^{2}}$
4 $\mathrm{nR}$
NEET-2017
Current Electricity

151897 A conducting wire has length ' $L_{1}$ ' and diameter ' $d_{1}$ '. After stretching the same wire length becomes ' $L_{2}$ ' and diameter ' $d_{2}$ '. The ratio of resistances before and after stretching is

1 $d_{2}^{4}: d_{1}^{4}$
2 $\mathrm{d}_{1}^{4}: \mathrm{d}_{2}^{4}$
3 $\mathrm{d}_{2}^{2}: \mathrm{d}_{1}^{2}$
4 $\mathrm{d}_{1}^{2}: \mathrm{d}_{2}^{2}$
MHT-CET 2018
Current Electricity

151899 The resistance of a wire at room temperature $30^{\circ} \mathrm{C}$ is found to be $10 \Omega$. Now to increase the resistance by $10 \%$. The temperature of the wire must be [The temperature coefficient of resistance of the material of the wire is $\mathbf{0 . 0 0 2}$ per ${ }^{0} \mathrm{Cl}$

1 $36{ }^{0} \mathrm{C}$
2 $83{ }^{\circ} \mathrm{C}$
3 $63{ }^{\circ} \mathrm{C}$
4 $33{ }^{\circ} \mathrm{C}$
UPSEE-2014
NEET DIGITAL LIBRARY
Current Electricity

151894 The I-V graphs for two different electrical applicances $P$ and $Q$ are shown in the diagram. If $R_{P}$ and $R_{Q}$ be the resistances of the devices, then :

1 $R_{P}=R_{Q}$
2 $R_{P}>R_{Q}$
3 $\mathrm{R}_{\mathrm{P}}\lt\mathrm{R}_{\mathrm{Q}}$
4 $R_{P}=\frac{R_{Q}}{2}$
Karnataka CET-2018
Current Electricity

151896 The resistance of a wire is ' $R$ ' $o h m$. If it is melted and stretched to ' $n$ ' times its original length, its new resistance will be

1 $\frac{R}{n}$
2 $n^{2} R$
3 $\frac{\mathrm{R}}{\mathrm{n}^{2}}$
4 $\mathrm{nR}$
NEET-2017
Current Electricity

151897 A conducting wire has length ' $L_{1}$ ' and diameter ' $d_{1}$ '. After stretching the same wire length becomes ' $L_{2}$ ' and diameter ' $d_{2}$ '. The ratio of resistances before and after stretching is

1 $d_{2}^{4}: d_{1}^{4}$
2 $\mathrm{d}_{1}^{4}: \mathrm{d}_{2}^{4}$
3 $\mathrm{d}_{2}^{2}: \mathrm{d}_{1}^{2}$
4 $\mathrm{d}_{1}^{2}: \mathrm{d}_{2}^{2}$
MHT-CET 2018
Current Electricity

151899 The resistance of a wire at room temperature $30^{\circ} \mathrm{C}$ is found to be $10 \Omega$. Now to increase the resistance by $10 \%$. The temperature of the wire must be [The temperature coefficient of resistance of the material of the wire is $\mathbf{0 . 0 0 2}$ per ${ }^{0} \mathrm{Cl}$

1 $36{ }^{0} \mathrm{C}$
2 $83{ }^{\circ} \mathrm{C}$
3 $63{ }^{\circ} \mathrm{C}$
4 $33{ }^{\circ} \mathrm{C}$
UPSEE-2014
Join With Us for More Updates
Current Electricity

151894 The I-V graphs for two different electrical applicances $P$ and $Q$ are shown in the diagram. If $R_{P}$ and $R_{Q}$ be the resistances of the devices, then :

1 $R_{P}=R_{Q}$
2 $R_{P}>R_{Q}$
3 $\mathrm{R}_{\mathrm{P}}\lt\mathrm{R}_{\mathrm{Q}}$
4 $R_{P}=\frac{R_{Q}}{2}$
Karnataka CET-2018
Current Electricity

151896 The resistance of a wire is ' $R$ ' $o h m$. If it is melted and stretched to ' $n$ ' times its original length, its new resistance will be

1 $\frac{R}{n}$
2 $n^{2} R$
3 $\frac{\mathrm{R}}{\mathrm{n}^{2}}$
4 $\mathrm{nR}$
NEET-2017
Current Electricity

151897 A conducting wire has length ' $L_{1}$ ' and diameter ' $d_{1}$ '. After stretching the same wire length becomes ' $L_{2}$ ' and diameter ' $d_{2}$ '. The ratio of resistances before and after stretching is

1 $d_{2}^{4}: d_{1}^{4}$
2 $\mathrm{d}_{1}^{4}: \mathrm{d}_{2}^{4}$
3 $\mathrm{d}_{2}^{2}: \mathrm{d}_{1}^{2}$
4 $\mathrm{d}_{1}^{2}: \mathrm{d}_{2}^{2}$
MHT-CET 2018
Current Electricity

151899 The resistance of a wire at room temperature $30^{\circ} \mathrm{C}$ is found to be $10 \Omega$. Now to increase the resistance by $10 \%$. The temperature of the wire must be [The temperature coefficient of resistance of the material of the wire is $\mathbf{0 . 0 0 2}$ per ${ }^{0} \mathrm{Cl}$

1 $36{ }^{0} \mathrm{C}$
2 $83{ }^{\circ} \mathrm{C}$
3 $63{ }^{\circ} \mathrm{C}$
4 $33{ }^{\circ} \mathrm{C}$
UPSEE-2014
For More Updates join with Us