357065
The length of a wire is doubled. Its conductance will be
1 Unchanged
2 Halved
3 Quadrupled
4 1/4 of the original value
Explanation:
Conductance \(C = \frac{1}{R} = \frac{A}{{\rho l}} = \frac{V}{{\rho {l^2}}}\) \(C \propto \frac{1}{{{l^2}}}\) ( \(V\) is the Volume of the conductor) \(\frac{{{C_1}}}{{{C_2}}} = {\left( {\frac{{{l_2}}}{{{l_1}}}} \right)^2} = {\left( {\frac{{2{l_1}}}{{{l_1}}}} \right)^2} = 4\) \( \Rightarrow {C_2} = \frac{{{C_1}}}{4}\)
PHXII03:CURRENT ELECTRICITY
357066
A steady current is passing through a linear conductor of nonuniform cross-section. The net quantity of charge crossing any cross-section per second is
1 Independent of area of cross-section
2 Directly proportional to the length of the conductor
3 Directly proportional to the area of cross section.
4 Inversely proportional to the area of the conductor
Explanation:
The net quantity of charge crossing any cross - section per second in a steady current is independent of the area of the cross - section. This is a fundamental principle of current flow in a conductor, as the same amount of charge passes through any cross - section in a given time, regardless of its area. The current remains constant throughout the conductor, so the charge crossing per second does not depend on the cross - sectional area. So option (1) is correct.
PHXII03:CURRENT ELECTRICITY
357067
A copper wire of cross-sectional area \(2.0\,m{m^2},\) resistivity \( = 1.7 \times {10^{ - 8}}\,\,\Omega m\), carries a current of 1 \(A\). The electric field in the copper wire is
357068
A wire of resistance \(R\) is elongated \(n\)-fold to make a new uniform wire. The resistance of new wire is:
1 \(n R\)
2 \(n^{2} R\)
3 \(2 n R\)
4 \(2 n^{2} R\)
Explanation:
As \(R \propto l\), so the length increases \(n\)-fold. Also, since the volume of the wire remains constant, so the area of the wire decreases \(n\) fold. \(\begin{aligned}& A_{1} l_{1}=A_{2}\left(n l_{1}\right) \Rightarrow A_{2}=\dfrac{A_{1}}{n} \\& R_{1}=\dfrac{\rho l_{1}}{A_{1}}, R_{2}=\dfrac{\rho l_{2}}{A_{2}}=n^{2}\left(\dfrac{\rho l_{1}}{A_{1}}\right)=n^{2} R_{1}=n^{2} R\end{aligned}\) Hence, the resistance increases to \(n^{2} R\).
357065
The length of a wire is doubled. Its conductance will be
1 Unchanged
2 Halved
3 Quadrupled
4 1/4 of the original value
Explanation:
Conductance \(C = \frac{1}{R} = \frac{A}{{\rho l}} = \frac{V}{{\rho {l^2}}}\) \(C \propto \frac{1}{{{l^2}}}\) ( \(V\) is the Volume of the conductor) \(\frac{{{C_1}}}{{{C_2}}} = {\left( {\frac{{{l_2}}}{{{l_1}}}} \right)^2} = {\left( {\frac{{2{l_1}}}{{{l_1}}}} \right)^2} = 4\) \( \Rightarrow {C_2} = \frac{{{C_1}}}{4}\)
PHXII03:CURRENT ELECTRICITY
357066
A steady current is passing through a linear conductor of nonuniform cross-section. The net quantity of charge crossing any cross-section per second is
1 Independent of area of cross-section
2 Directly proportional to the length of the conductor
3 Directly proportional to the area of cross section.
4 Inversely proportional to the area of the conductor
Explanation:
The net quantity of charge crossing any cross - section per second in a steady current is independent of the area of the cross - section. This is a fundamental principle of current flow in a conductor, as the same amount of charge passes through any cross - section in a given time, regardless of its area. The current remains constant throughout the conductor, so the charge crossing per second does not depend on the cross - sectional area. So option (1) is correct.
PHXII03:CURRENT ELECTRICITY
357067
A copper wire of cross-sectional area \(2.0\,m{m^2},\) resistivity \( = 1.7 \times {10^{ - 8}}\,\,\Omega m\), carries a current of 1 \(A\). The electric field in the copper wire is
357068
A wire of resistance \(R\) is elongated \(n\)-fold to make a new uniform wire. The resistance of new wire is:
1 \(n R\)
2 \(n^{2} R\)
3 \(2 n R\)
4 \(2 n^{2} R\)
Explanation:
As \(R \propto l\), so the length increases \(n\)-fold. Also, since the volume of the wire remains constant, so the area of the wire decreases \(n\) fold. \(\begin{aligned}& A_{1} l_{1}=A_{2}\left(n l_{1}\right) \Rightarrow A_{2}=\dfrac{A_{1}}{n} \\& R_{1}=\dfrac{\rho l_{1}}{A_{1}}, R_{2}=\dfrac{\rho l_{2}}{A_{2}}=n^{2}\left(\dfrac{\rho l_{1}}{A_{1}}\right)=n^{2} R_{1}=n^{2} R\end{aligned}\) Hence, the resistance increases to \(n^{2} R\).
357065
The length of a wire is doubled. Its conductance will be
1 Unchanged
2 Halved
3 Quadrupled
4 1/4 of the original value
Explanation:
Conductance \(C = \frac{1}{R} = \frac{A}{{\rho l}} = \frac{V}{{\rho {l^2}}}\) \(C \propto \frac{1}{{{l^2}}}\) ( \(V\) is the Volume of the conductor) \(\frac{{{C_1}}}{{{C_2}}} = {\left( {\frac{{{l_2}}}{{{l_1}}}} \right)^2} = {\left( {\frac{{2{l_1}}}{{{l_1}}}} \right)^2} = 4\) \( \Rightarrow {C_2} = \frac{{{C_1}}}{4}\)
PHXII03:CURRENT ELECTRICITY
357066
A steady current is passing through a linear conductor of nonuniform cross-section. The net quantity of charge crossing any cross-section per second is
1 Independent of area of cross-section
2 Directly proportional to the length of the conductor
3 Directly proportional to the area of cross section.
4 Inversely proportional to the area of the conductor
Explanation:
The net quantity of charge crossing any cross - section per second in a steady current is independent of the area of the cross - section. This is a fundamental principle of current flow in a conductor, as the same amount of charge passes through any cross - section in a given time, regardless of its area. The current remains constant throughout the conductor, so the charge crossing per second does not depend on the cross - sectional area. So option (1) is correct.
PHXII03:CURRENT ELECTRICITY
357067
A copper wire of cross-sectional area \(2.0\,m{m^2},\) resistivity \( = 1.7 \times {10^{ - 8}}\,\,\Omega m\), carries a current of 1 \(A\). The electric field in the copper wire is
357068
A wire of resistance \(R\) is elongated \(n\)-fold to make a new uniform wire. The resistance of new wire is:
1 \(n R\)
2 \(n^{2} R\)
3 \(2 n R\)
4 \(2 n^{2} R\)
Explanation:
As \(R \propto l\), so the length increases \(n\)-fold. Also, since the volume of the wire remains constant, so the area of the wire decreases \(n\) fold. \(\begin{aligned}& A_{1} l_{1}=A_{2}\left(n l_{1}\right) \Rightarrow A_{2}=\dfrac{A_{1}}{n} \\& R_{1}=\dfrac{\rho l_{1}}{A_{1}}, R_{2}=\dfrac{\rho l_{2}}{A_{2}}=n^{2}\left(\dfrac{\rho l_{1}}{A_{1}}\right)=n^{2} R_{1}=n^{2} R\end{aligned}\) Hence, the resistance increases to \(n^{2} R\).
357065
The length of a wire is doubled. Its conductance will be
1 Unchanged
2 Halved
3 Quadrupled
4 1/4 of the original value
Explanation:
Conductance \(C = \frac{1}{R} = \frac{A}{{\rho l}} = \frac{V}{{\rho {l^2}}}\) \(C \propto \frac{1}{{{l^2}}}\) ( \(V\) is the Volume of the conductor) \(\frac{{{C_1}}}{{{C_2}}} = {\left( {\frac{{{l_2}}}{{{l_1}}}} \right)^2} = {\left( {\frac{{2{l_1}}}{{{l_1}}}} \right)^2} = 4\) \( \Rightarrow {C_2} = \frac{{{C_1}}}{4}\)
PHXII03:CURRENT ELECTRICITY
357066
A steady current is passing through a linear conductor of nonuniform cross-section. The net quantity of charge crossing any cross-section per second is
1 Independent of area of cross-section
2 Directly proportional to the length of the conductor
3 Directly proportional to the area of cross section.
4 Inversely proportional to the area of the conductor
Explanation:
The net quantity of charge crossing any cross - section per second in a steady current is independent of the area of the cross - section. This is a fundamental principle of current flow in a conductor, as the same amount of charge passes through any cross - section in a given time, regardless of its area. The current remains constant throughout the conductor, so the charge crossing per second does not depend on the cross - sectional area. So option (1) is correct.
PHXII03:CURRENT ELECTRICITY
357067
A copper wire of cross-sectional area \(2.0\,m{m^2},\) resistivity \( = 1.7 \times {10^{ - 8}}\,\,\Omega m\), carries a current of 1 \(A\). The electric field in the copper wire is
357068
A wire of resistance \(R\) is elongated \(n\)-fold to make a new uniform wire. The resistance of new wire is:
1 \(n R\)
2 \(n^{2} R\)
3 \(2 n R\)
4 \(2 n^{2} R\)
Explanation:
As \(R \propto l\), so the length increases \(n\)-fold. Also, since the volume of the wire remains constant, so the area of the wire decreases \(n\) fold. \(\begin{aligned}& A_{1} l_{1}=A_{2}\left(n l_{1}\right) \Rightarrow A_{2}=\dfrac{A_{1}}{n} \\& R_{1}=\dfrac{\rho l_{1}}{A_{1}}, R_{2}=\dfrac{\rho l_{2}}{A_{2}}=n^{2}\left(\dfrac{\rho l_{1}}{A_{1}}\right)=n^{2} R_{1}=n^{2} R\end{aligned}\) Hence, the resistance increases to \(n^{2} R\).
