268484
If a wire is stretched to make it\(0.1 \%\) longer, its rsistance will : [M ains-2011]
1 increase by\(0.2 \%\)
2 decrease by\(0.2 \%\)
3 decrease by\(0.05 \%\)
4 increase by\(0.05 \%\)
Explanation:
\(R \propto I^{2}\)
Current Electricity
268517
The sides of rectangular block are\(2 \mathrm{~cm}, 3 \mathrm{~cm}\) and \(4 \mathrm{~cm}\). The ratio of the maximum to mini mum resistance between its parallel faces is
268519
Two wires of the same material have length\(6 \mathrm{~cm}\) and \(10 \mathrm{~cm}\) and radii \(0.5 \mathrm{~mm}\) and \(1.5 \mathrm{~mm}\) respectively. They are connected in series across a battery of \(16 \mathrm{~V}\). The p.d. across the shorter wire is
1 \(5 \mathrm{~V}\)
2 \(13.5 \mathrm{~V}\)
3 \(27 \mathrm{~V}\)
4 \(10 \mathrm{~V}\)
Explanation:
\(I_{1}=6 \mathrm{~cm}, I_{2}=10 \mathrm{~cm}\). \(r_{1}=0.5 \times 10^{-3}, r_{2}=1.5 \times 10^{-3}\) In series combination \(\mathrm{i}=\) constant \(\frac{V_{1}}{V_{2}}=\frac{R_{1}}{R_{2}}=\frac{\frac{\rho l_{1}}{A_{1}}}{\frac{\rho l_{2}}{A_{2}}}=\frac{I_{1}}{I_{2}} \times \frac{A_{2}}{A_{1}} V_{1}+\mathrm{V}_{2}=16 \mathrm{~V}\) Solving for \(V_{1}=13.5 \mathrm{~V}\)
268484
If a wire is stretched to make it\(0.1 \%\) longer, its rsistance will : [M ains-2011]
1 increase by\(0.2 \%\)
2 decrease by\(0.2 \%\)
3 decrease by\(0.05 \%\)
4 increase by\(0.05 \%\)
Explanation:
\(R \propto I^{2}\)
Current Electricity
268517
The sides of rectangular block are\(2 \mathrm{~cm}, 3 \mathrm{~cm}\) and \(4 \mathrm{~cm}\). The ratio of the maximum to mini mum resistance between its parallel faces is
268519
Two wires of the same material have length\(6 \mathrm{~cm}\) and \(10 \mathrm{~cm}\) and radii \(0.5 \mathrm{~mm}\) and \(1.5 \mathrm{~mm}\) respectively. They are connected in series across a battery of \(16 \mathrm{~V}\). The p.d. across the shorter wire is
1 \(5 \mathrm{~V}\)
2 \(13.5 \mathrm{~V}\)
3 \(27 \mathrm{~V}\)
4 \(10 \mathrm{~V}\)
Explanation:
\(I_{1}=6 \mathrm{~cm}, I_{2}=10 \mathrm{~cm}\). \(r_{1}=0.5 \times 10^{-3}, r_{2}=1.5 \times 10^{-3}\) In series combination \(\mathrm{i}=\) constant \(\frac{V_{1}}{V_{2}}=\frac{R_{1}}{R_{2}}=\frac{\frac{\rho l_{1}}{A_{1}}}{\frac{\rho l_{2}}{A_{2}}}=\frac{I_{1}}{I_{2}} \times \frac{A_{2}}{A_{1}} V_{1}+\mathrm{V}_{2}=16 \mathrm{~V}\) Solving for \(V_{1}=13.5 \mathrm{~V}\)
268484
If a wire is stretched to make it\(0.1 \%\) longer, its rsistance will : [M ains-2011]
1 increase by\(0.2 \%\)
2 decrease by\(0.2 \%\)
3 decrease by\(0.05 \%\)
4 increase by\(0.05 \%\)
Explanation:
\(R \propto I^{2}\)
Current Electricity
268517
The sides of rectangular block are\(2 \mathrm{~cm}, 3 \mathrm{~cm}\) and \(4 \mathrm{~cm}\). The ratio of the maximum to mini mum resistance between its parallel faces is
268519
Two wires of the same material have length\(6 \mathrm{~cm}\) and \(10 \mathrm{~cm}\) and radii \(0.5 \mathrm{~mm}\) and \(1.5 \mathrm{~mm}\) respectively. They are connected in series across a battery of \(16 \mathrm{~V}\). The p.d. across the shorter wire is
1 \(5 \mathrm{~V}\)
2 \(13.5 \mathrm{~V}\)
3 \(27 \mathrm{~V}\)
4 \(10 \mathrm{~V}\)
Explanation:
\(I_{1}=6 \mathrm{~cm}, I_{2}=10 \mathrm{~cm}\). \(r_{1}=0.5 \times 10^{-3}, r_{2}=1.5 \times 10^{-3}\) In series combination \(\mathrm{i}=\) constant \(\frac{V_{1}}{V_{2}}=\frac{R_{1}}{R_{2}}=\frac{\frac{\rho l_{1}}{A_{1}}}{\frac{\rho l_{2}}{A_{2}}}=\frac{I_{1}}{I_{2}} \times \frac{A_{2}}{A_{1}} V_{1}+\mathrm{V}_{2}=16 \mathrm{~V}\) Solving for \(V_{1}=13.5 \mathrm{~V}\)
268484
If a wire is stretched to make it\(0.1 \%\) longer, its rsistance will : [M ains-2011]
1 increase by\(0.2 \%\)
2 decrease by\(0.2 \%\)
3 decrease by\(0.05 \%\)
4 increase by\(0.05 \%\)
Explanation:
\(R \propto I^{2}\)
Current Electricity
268517
The sides of rectangular block are\(2 \mathrm{~cm}, 3 \mathrm{~cm}\) and \(4 \mathrm{~cm}\). The ratio of the maximum to mini mum resistance between its parallel faces is
268519
Two wires of the same material have length\(6 \mathrm{~cm}\) and \(10 \mathrm{~cm}\) and radii \(0.5 \mathrm{~mm}\) and \(1.5 \mathrm{~mm}\) respectively. They are connected in series across a battery of \(16 \mathrm{~V}\). The p.d. across the shorter wire is
1 \(5 \mathrm{~V}\)
2 \(13.5 \mathrm{~V}\)
3 \(27 \mathrm{~V}\)
4 \(10 \mathrm{~V}\)
Explanation:
\(I_{1}=6 \mathrm{~cm}, I_{2}=10 \mathrm{~cm}\). \(r_{1}=0.5 \times 10^{-3}, r_{2}=1.5 \times 10^{-3}\) In series combination \(\mathrm{i}=\) constant \(\frac{V_{1}}{V_{2}}=\frac{R_{1}}{R_{2}}=\frac{\frac{\rho l_{1}}{A_{1}}}{\frac{\rho l_{2}}{A_{2}}}=\frac{I_{1}}{I_{2}} \times \frac{A_{2}}{A_{1}} V_{1}+\mathrm{V}_{2}=16 \mathrm{~V}\) Solving for \(V_{1}=13.5 \mathrm{~V}\)
268484
If a wire is stretched to make it\(0.1 \%\) longer, its rsistance will : [M ains-2011]
1 increase by\(0.2 \%\)
2 decrease by\(0.2 \%\)
3 decrease by\(0.05 \%\)
4 increase by\(0.05 \%\)
Explanation:
\(R \propto I^{2}\)
Current Electricity
268517
The sides of rectangular block are\(2 \mathrm{~cm}, 3 \mathrm{~cm}\) and \(4 \mathrm{~cm}\). The ratio of the maximum to mini mum resistance between its parallel faces is
268519
Two wires of the same material have length\(6 \mathrm{~cm}\) and \(10 \mathrm{~cm}\) and radii \(0.5 \mathrm{~mm}\) and \(1.5 \mathrm{~mm}\) respectively. They are connected in series across a battery of \(16 \mathrm{~V}\). The p.d. across the shorter wire is
1 \(5 \mathrm{~V}\)
2 \(13.5 \mathrm{~V}\)
3 \(27 \mathrm{~V}\)
4 \(10 \mathrm{~V}\)
Explanation:
\(I_{1}=6 \mathrm{~cm}, I_{2}=10 \mathrm{~cm}\). \(r_{1}=0.5 \times 10^{-3}, r_{2}=1.5 \times 10^{-3}\) In series combination \(\mathrm{i}=\) constant \(\frac{V_{1}}{V_{2}}=\frac{R_{1}}{R_{2}}=\frac{\frac{\rho l_{1}}{A_{1}}}{\frac{\rho l_{2}}{A_{2}}}=\frac{I_{1}}{I_{2}} \times \frac{A_{2}}{A_{1}} V_{1}+\mathrm{V}_{2}=16 \mathrm{~V}\) Solving for \(V_{1}=13.5 \mathrm{~V}\)
