357043
A wire of a certain material is stretched slowly by 10 percent. Its new resistance and specific resistance become respectively
1 1.21 times, same
2 1.1 times, 1.1 times
3 1.2 times, 1.1 times
4 Both remain the same
Explanation:
After stretching, specific resistance \(\left( \rho \right)\) will remain same. Original resistance of wire, \(R = \frac{{\rho I}}{A}\) Ratio of resistance before and after stretching, \(\frac{{{R_2}}}{{{R_1}}} = \frac{{{l_2}}}{{{l_1}}} \times \frac{{{A_1}}}{{{A_2}}} = {\left[ {\frac{{{l_2}}}{{{l_1}}}} \right]^2}\) \({l_2} = 1.1\;{l_1} \Rightarrow {R_2} = 1.21{R_1}\)
PHXII03:CURRENT ELECTRICITY
357044
Dimensions of a block are \(1\;cm \times 1\;cm \times \) \(100\;cm\). If specific resistance of its material is \(3 \times {10^{ - 7}}\Omega \,m,\) then the resistance between the opposite rectangular faces is
1 \(3 \times 10^{-7} \Omega\)
2 \(3 \times 10^{-9} \Omega\)
3 \(3 \times 10^{-5} \Omega\)
4 \(3 \times 10^{-3} \Omega\)
Explanation:
Resistance, \(R=\dfrac{\rho l}{A}\) Dimensions of the block, \(1\;cm \times 1\;cm \times 100\;cm,\) \(\rho = 3 \times {10^{ - 7}}\Omega\, m,l = 1\;cm = {10^{ - 2}}\;m\) Area of rectangular face of the block, \(A\) \( = 1cm \times 100{\mkern 1mu} cm = 100 \times {10^{ - 4}}{m^2} = 1 \times {10^{ - 2}}{m^2}\) \(\therefore R = 3 \times {10^{ - 7}} \times \frac{{{{10}^{ - 2}}}}{{1 \times {{10}^{ - 2}}}} = 3 \times {10^{ - 7}}\,\Omega \)
PHXII03:CURRENT ELECTRICITY
357045
If \(n,e,\tau ,m,\) are representing electron density, charge, relaxation time and mass of an electron respectively then the resistance of a wire of length \(l\) and cross sectional area \(A\) is given by
1 \(\frac{{ml}}{{n{e^2}\tau A}}\)
2 \(\frac{{2mA}}{{n{e^2}\tau }}\)
3 \(n{e^2}\tau A\)
4 \(\frac{{n{e^2}\tau A}}{{2m}}\)
Explanation:
Conceptual Question
PHXII03:CURRENT ELECTRICITY
357046
Find \({R_{\text {max }}: R_{\text {min }}}\) for the given slab.
357047
A wire of resistance \(3\Omega \) is stretched to twice its original length. The resistance of the new wire will be
1 \({\rm{1}}{\rm{.5\Omega }}\)
2 \({\rm{3}}\,{\rm{\Omega }}\)
3 \(6\,{\rm{\Omega }}\)
4 \(12\,{\rm{\Omega }}\)
Explanation:
For a wire of cross sectional area \(A\) and length \(l\), resistance\(R = \frac{{\rho l}}{A} = \frac{{\rho {l^2}}}{{Al}} = \frac{{\rho {l^2}}}{V}\), where \(\rho = \) resistivity of the material of the wire, and \(V = \) volume of the wire. When the wire is streached, the volume is constant. \(\therefore \quad R \propto {l^2} \Rightarrow \frac{{{R_2}}}{{{R_1}}} = {\left( {\frac{{{l_2}}}{{{l_1}}}} \right)^2}\) Given,\({R_1} = 3\Omega \) and \({l_2} = 2{l_1}\) \({R_2} = 3 \times {2^2} = 12\,\Omega \)
357043
A wire of a certain material is stretched slowly by 10 percent. Its new resistance and specific resistance become respectively
1 1.21 times, same
2 1.1 times, 1.1 times
3 1.2 times, 1.1 times
4 Both remain the same
Explanation:
After stretching, specific resistance \(\left( \rho \right)\) will remain same. Original resistance of wire, \(R = \frac{{\rho I}}{A}\) Ratio of resistance before and after stretching, \(\frac{{{R_2}}}{{{R_1}}} = \frac{{{l_2}}}{{{l_1}}} \times \frac{{{A_1}}}{{{A_2}}} = {\left[ {\frac{{{l_2}}}{{{l_1}}}} \right]^2}\) \({l_2} = 1.1\;{l_1} \Rightarrow {R_2} = 1.21{R_1}\)
PHXII03:CURRENT ELECTRICITY
357044
Dimensions of a block are \(1\;cm \times 1\;cm \times \) \(100\;cm\). If specific resistance of its material is \(3 \times {10^{ - 7}}\Omega \,m,\) then the resistance between the opposite rectangular faces is
1 \(3 \times 10^{-7} \Omega\)
2 \(3 \times 10^{-9} \Omega\)
3 \(3 \times 10^{-5} \Omega\)
4 \(3 \times 10^{-3} \Omega\)
Explanation:
Resistance, \(R=\dfrac{\rho l}{A}\) Dimensions of the block, \(1\;cm \times 1\;cm \times 100\;cm,\) \(\rho = 3 \times {10^{ - 7}}\Omega\, m,l = 1\;cm = {10^{ - 2}}\;m\) Area of rectangular face of the block, \(A\) \( = 1cm \times 100{\mkern 1mu} cm = 100 \times {10^{ - 4}}{m^2} = 1 \times {10^{ - 2}}{m^2}\) \(\therefore R = 3 \times {10^{ - 7}} \times \frac{{{{10}^{ - 2}}}}{{1 \times {{10}^{ - 2}}}} = 3 \times {10^{ - 7}}\,\Omega \)
PHXII03:CURRENT ELECTRICITY
357045
If \(n,e,\tau ,m,\) are representing electron density, charge, relaxation time and mass of an electron respectively then the resistance of a wire of length \(l\) and cross sectional area \(A\) is given by
1 \(\frac{{ml}}{{n{e^2}\tau A}}\)
2 \(\frac{{2mA}}{{n{e^2}\tau }}\)
3 \(n{e^2}\tau A\)
4 \(\frac{{n{e^2}\tau A}}{{2m}}\)
Explanation:
Conceptual Question
PHXII03:CURRENT ELECTRICITY
357046
Find \({R_{\text {max }}: R_{\text {min }}}\) for the given slab.
357047
A wire of resistance \(3\Omega \) is stretched to twice its original length. The resistance of the new wire will be
1 \({\rm{1}}{\rm{.5\Omega }}\)
2 \({\rm{3}}\,{\rm{\Omega }}\)
3 \(6\,{\rm{\Omega }}\)
4 \(12\,{\rm{\Omega }}\)
Explanation:
For a wire of cross sectional area \(A\) and length \(l\), resistance\(R = \frac{{\rho l}}{A} = \frac{{\rho {l^2}}}{{Al}} = \frac{{\rho {l^2}}}{V}\), where \(\rho = \) resistivity of the material of the wire, and \(V = \) volume of the wire. When the wire is streached, the volume is constant. \(\therefore \quad R \propto {l^2} \Rightarrow \frac{{{R_2}}}{{{R_1}}} = {\left( {\frac{{{l_2}}}{{{l_1}}}} \right)^2}\) Given,\({R_1} = 3\Omega \) and \({l_2} = 2{l_1}\) \({R_2} = 3 \times {2^2} = 12\,\Omega \)
357043
A wire of a certain material is stretched slowly by 10 percent. Its new resistance and specific resistance become respectively
1 1.21 times, same
2 1.1 times, 1.1 times
3 1.2 times, 1.1 times
4 Both remain the same
Explanation:
After stretching, specific resistance \(\left( \rho \right)\) will remain same. Original resistance of wire, \(R = \frac{{\rho I}}{A}\) Ratio of resistance before and after stretching, \(\frac{{{R_2}}}{{{R_1}}} = \frac{{{l_2}}}{{{l_1}}} \times \frac{{{A_1}}}{{{A_2}}} = {\left[ {\frac{{{l_2}}}{{{l_1}}}} \right]^2}\) \({l_2} = 1.1\;{l_1} \Rightarrow {R_2} = 1.21{R_1}\)
PHXII03:CURRENT ELECTRICITY
357044
Dimensions of a block are \(1\;cm \times 1\;cm \times \) \(100\;cm\). If specific resistance of its material is \(3 \times {10^{ - 7}}\Omega \,m,\) then the resistance between the opposite rectangular faces is
1 \(3 \times 10^{-7} \Omega\)
2 \(3 \times 10^{-9} \Omega\)
3 \(3 \times 10^{-5} \Omega\)
4 \(3 \times 10^{-3} \Omega\)
Explanation:
Resistance, \(R=\dfrac{\rho l}{A}\) Dimensions of the block, \(1\;cm \times 1\;cm \times 100\;cm,\) \(\rho = 3 \times {10^{ - 7}}\Omega\, m,l = 1\;cm = {10^{ - 2}}\;m\) Area of rectangular face of the block, \(A\) \( = 1cm \times 100{\mkern 1mu} cm = 100 \times {10^{ - 4}}{m^2} = 1 \times {10^{ - 2}}{m^2}\) \(\therefore R = 3 \times {10^{ - 7}} \times \frac{{{{10}^{ - 2}}}}{{1 \times {{10}^{ - 2}}}} = 3 \times {10^{ - 7}}\,\Omega \)
PHXII03:CURRENT ELECTRICITY
357045
If \(n,e,\tau ,m,\) are representing electron density, charge, relaxation time and mass of an electron respectively then the resistance of a wire of length \(l\) and cross sectional area \(A\) is given by
1 \(\frac{{ml}}{{n{e^2}\tau A}}\)
2 \(\frac{{2mA}}{{n{e^2}\tau }}\)
3 \(n{e^2}\tau A\)
4 \(\frac{{n{e^2}\tau A}}{{2m}}\)
Explanation:
Conceptual Question
PHXII03:CURRENT ELECTRICITY
357046
Find \({R_{\text {max }}: R_{\text {min }}}\) for the given slab.
357047
A wire of resistance \(3\Omega \) is stretched to twice its original length. The resistance of the new wire will be
1 \({\rm{1}}{\rm{.5\Omega }}\)
2 \({\rm{3}}\,{\rm{\Omega }}\)
3 \(6\,{\rm{\Omega }}\)
4 \(12\,{\rm{\Omega }}\)
Explanation:
For a wire of cross sectional area \(A\) and length \(l\), resistance\(R = \frac{{\rho l}}{A} = \frac{{\rho {l^2}}}{{Al}} = \frac{{\rho {l^2}}}{V}\), where \(\rho = \) resistivity of the material of the wire, and \(V = \) volume of the wire. When the wire is streached, the volume is constant. \(\therefore \quad R \propto {l^2} \Rightarrow \frac{{{R_2}}}{{{R_1}}} = {\left( {\frac{{{l_2}}}{{{l_1}}}} \right)^2}\) Given,\({R_1} = 3\Omega \) and \({l_2} = 2{l_1}\) \({R_2} = 3 \times {2^2} = 12\,\Omega \)
357043
A wire of a certain material is stretched slowly by 10 percent. Its new resistance and specific resistance become respectively
1 1.21 times, same
2 1.1 times, 1.1 times
3 1.2 times, 1.1 times
4 Both remain the same
Explanation:
After stretching, specific resistance \(\left( \rho \right)\) will remain same. Original resistance of wire, \(R = \frac{{\rho I}}{A}\) Ratio of resistance before and after stretching, \(\frac{{{R_2}}}{{{R_1}}} = \frac{{{l_2}}}{{{l_1}}} \times \frac{{{A_1}}}{{{A_2}}} = {\left[ {\frac{{{l_2}}}{{{l_1}}}} \right]^2}\) \({l_2} = 1.1\;{l_1} \Rightarrow {R_2} = 1.21{R_1}\)
PHXII03:CURRENT ELECTRICITY
357044
Dimensions of a block are \(1\;cm \times 1\;cm \times \) \(100\;cm\). If specific resistance of its material is \(3 \times {10^{ - 7}}\Omega \,m,\) then the resistance between the opposite rectangular faces is
1 \(3 \times 10^{-7} \Omega\)
2 \(3 \times 10^{-9} \Omega\)
3 \(3 \times 10^{-5} \Omega\)
4 \(3 \times 10^{-3} \Omega\)
Explanation:
Resistance, \(R=\dfrac{\rho l}{A}\) Dimensions of the block, \(1\;cm \times 1\;cm \times 100\;cm,\) \(\rho = 3 \times {10^{ - 7}}\Omega\, m,l = 1\;cm = {10^{ - 2}}\;m\) Area of rectangular face of the block, \(A\) \( = 1cm \times 100{\mkern 1mu} cm = 100 \times {10^{ - 4}}{m^2} = 1 \times {10^{ - 2}}{m^2}\) \(\therefore R = 3 \times {10^{ - 7}} \times \frac{{{{10}^{ - 2}}}}{{1 \times {{10}^{ - 2}}}} = 3 \times {10^{ - 7}}\,\Omega \)
PHXII03:CURRENT ELECTRICITY
357045
If \(n,e,\tau ,m,\) are representing electron density, charge, relaxation time and mass of an electron respectively then the resistance of a wire of length \(l\) and cross sectional area \(A\) is given by
1 \(\frac{{ml}}{{n{e^2}\tau A}}\)
2 \(\frac{{2mA}}{{n{e^2}\tau }}\)
3 \(n{e^2}\tau A\)
4 \(\frac{{n{e^2}\tau A}}{{2m}}\)
Explanation:
Conceptual Question
PHXII03:CURRENT ELECTRICITY
357046
Find \({R_{\text {max }}: R_{\text {min }}}\) for the given slab.
357047
A wire of resistance \(3\Omega \) is stretched to twice its original length. The resistance of the new wire will be
1 \({\rm{1}}{\rm{.5\Omega }}\)
2 \({\rm{3}}\,{\rm{\Omega }}\)
3 \(6\,{\rm{\Omega }}\)
4 \(12\,{\rm{\Omega }}\)
Explanation:
For a wire of cross sectional area \(A\) and length \(l\), resistance\(R = \frac{{\rho l}}{A} = \frac{{\rho {l^2}}}{{Al}} = \frac{{\rho {l^2}}}{V}\), where \(\rho = \) resistivity of the material of the wire, and \(V = \) volume of the wire. When the wire is streached, the volume is constant. \(\therefore \quad R \propto {l^2} \Rightarrow \frac{{{R_2}}}{{{R_1}}} = {\left( {\frac{{{l_2}}}{{{l_1}}}} \right)^2}\) Given,\({R_1} = 3\Omega \) and \({l_2} = 2{l_1}\) \({R_2} = 3 \times {2^2} = 12\,\Omega \)
357043
A wire of a certain material is stretched slowly by 10 percent. Its new resistance and specific resistance become respectively
1 1.21 times, same
2 1.1 times, 1.1 times
3 1.2 times, 1.1 times
4 Both remain the same
Explanation:
After stretching, specific resistance \(\left( \rho \right)\) will remain same. Original resistance of wire, \(R = \frac{{\rho I}}{A}\) Ratio of resistance before and after stretching, \(\frac{{{R_2}}}{{{R_1}}} = \frac{{{l_2}}}{{{l_1}}} \times \frac{{{A_1}}}{{{A_2}}} = {\left[ {\frac{{{l_2}}}{{{l_1}}}} \right]^2}\) \({l_2} = 1.1\;{l_1} \Rightarrow {R_2} = 1.21{R_1}\)
PHXII03:CURRENT ELECTRICITY
357044
Dimensions of a block are \(1\;cm \times 1\;cm \times \) \(100\;cm\). If specific resistance of its material is \(3 \times {10^{ - 7}}\Omega \,m,\) then the resistance between the opposite rectangular faces is
1 \(3 \times 10^{-7} \Omega\)
2 \(3 \times 10^{-9} \Omega\)
3 \(3 \times 10^{-5} \Omega\)
4 \(3 \times 10^{-3} \Omega\)
Explanation:
Resistance, \(R=\dfrac{\rho l}{A}\) Dimensions of the block, \(1\;cm \times 1\;cm \times 100\;cm,\) \(\rho = 3 \times {10^{ - 7}}\Omega\, m,l = 1\;cm = {10^{ - 2}}\;m\) Area of rectangular face of the block, \(A\) \( = 1cm \times 100{\mkern 1mu} cm = 100 \times {10^{ - 4}}{m^2} = 1 \times {10^{ - 2}}{m^2}\) \(\therefore R = 3 \times {10^{ - 7}} \times \frac{{{{10}^{ - 2}}}}{{1 \times {{10}^{ - 2}}}} = 3 \times {10^{ - 7}}\,\Omega \)
PHXII03:CURRENT ELECTRICITY
357045
If \(n,e,\tau ,m,\) are representing electron density, charge, relaxation time and mass of an electron respectively then the resistance of a wire of length \(l\) and cross sectional area \(A\) is given by
1 \(\frac{{ml}}{{n{e^2}\tau A}}\)
2 \(\frac{{2mA}}{{n{e^2}\tau }}\)
3 \(n{e^2}\tau A\)
4 \(\frac{{n{e^2}\tau A}}{{2m}}\)
Explanation:
Conceptual Question
PHXII03:CURRENT ELECTRICITY
357046
Find \({R_{\text {max }}: R_{\text {min }}}\) for the given slab.
357047
A wire of resistance \(3\Omega \) is stretched to twice its original length. The resistance of the new wire will be
1 \({\rm{1}}{\rm{.5\Omega }}\)
2 \({\rm{3}}\,{\rm{\Omega }}\)
3 \(6\,{\rm{\Omega }}\)
4 \(12\,{\rm{\Omega }}\)
Explanation:
For a wire of cross sectional area \(A\) and length \(l\), resistance\(R = \frac{{\rho l}}{A} = \frac{{\rho {l^2}}}{{Al}} = \frac{{\rho {l^2}}}{V}\), where \(\rho = \) resistivity of the material of the wire, and \(V = \) volume of the wire. When the wire is streached, the volume is constant. \(\therefore \quad R \propto {l^2} \Rightarrow \frac{{{R_2}}}{{{R_1}}} = {\left( {\frac{{{l_2}}}{{{l_1}}}} \right)^2}\) Given,\({R_1} = 3\Omega \) and \({l_2} = 2{l_1}\) \({R_2} = 3 \times {2^2} = 12\,\Omega \)
