357052
Assertion : When the length of a conductor is doubled (area is constant) than resistance will also gets doubled. Reason : Resistance is directly proportional to the length of a conductor
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(R=\rho \dfrac{l}{A}\) So correct option is (1).
PHXII03:CURRENT ELECTRICITY
357053
We have two wires \(A\) and \(B\) of same mass and same material. The diameter of the wire \(A\) is half of that \(B\). If the resistance of wire \(A\) is 24 \(ohm\) then the resistance of wire \(B\) will be
1 \(3.0\,ohm\)
2 \(12\,ohm\)
3 \(1.5\,ohm\)
4 None of the above
Explanation:
Same mass same material, i.e., volume is constant \(R = \rho \frac{{\rm{l}}}{A} \Rightarrow \frac{{{R_1}}}{{{R_2}}} = \frac{{{{\rm{l}}_1}}}{{{{\rm{l}}_2}}} \times \frac{{{A_2}}}{{{A_1}}} = {\left( {\frac{{{A_2}}}{{{A_1}}}} \right)^2} = {\left( {\frac{{{d_2}}}{{{d_1}}}} \right)^4}\) \( \Rightarrow \frac{{24}}{{{R_2}}} = {\left( {\frac{d}{{d/2}}} \right)^4} = 16 \Rightarrow {R_2} = 1.5\Omega \)
PHXII03:CURRENT ELECTRICITY
357054
What is percentage change in its resistance if radius is increased by 1%?
1 \(1\% \)
2 \(2\% \)
3 \(3\% \)
4 \(4\% \)
Explanation:
\(\because R = \rho \frac{V}{{{A^2}}}\quad \Rightarrow \frac{{dR}}{R} = - 2\frac{{dA}}{A}\) As cross-sectional area \({\rm{A = \pi }}{{\rm{r}}^{\rm{2}}}\) So \(\frac{{dR}}{R} = - 2\frac{{dA}}{A} = - 4\frac{{dr}}{r}\) If radius is increased by 1%, the resistance is decreased by 4%.
PHXII03:CURRENT ELECTRICITY
357055
A copper wire of length 10 \(m\) and radius \(\left( {{{10}^{ - 2}}/\pi } \right)m\) has electrical resistance of \(10\,\Omega \). The current density in the wire for an electric field strength of 10 (\(V\)/\(m\)) is:
1 \({10^6}\,A/{m^2}\)
2 \({10^{ - 5}}\,A/{m^2}\)
3 \({10^5}\,A/{m^2}\)
4 \({10^4}\,A/{m^2}\)
Explanation:
Method I Radius of wire \( = \frac{{{{10}^{ - 2}}}}{{\sqrt \pi }}\,\,\) Cross sectional area \(A = \pi {r^2} = {10^{ - 4}}{m^2}\) Current density is given by \(j = \frac{i}{A} = \left( {\frac{V}{R}} \right) \cdot \frac{1}{A} = \frac{{E\ell }}{{RA}}\) \(j = \frac{{10 \times 10}}{{10 \times {{10}^{ - 4}}}} = {10^5}A/{m^2}\) Method II \(j = \sigma E \Rightarrow \frac{E}{\rho } = \frac{{E\ell }}{{RA}} = \frac{{10 \times 10 \times \pi }}{{10 \times {{10}^{ - 4}} \times \pi }}\) \(\left( {\because R = \frac{{\rho \ell }}{A}} \right)\) \(j = {10^5}\,\,A/{m^2}\)
357052
Assertion : When the length of a conductor is doubled (area is constant) than resistance will also gets doubled. Reason : Resistance is directly proportional to the length of a conductor
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(R=\rho \dfrac{l}{A}\) So correct option is (1).
PHXII03:CURRENT ELECTRICITY
357053
We have two wires \(A\) and \(B\) of same mass and same material. The diameter of the wire \(A\) is half of that \(B\). If the resistance of wire \(A\) is 24 \(ohm\) then the resistance of wire \(B\) will be
1 \(3.0\,ohm\)
2 \(12\,ohm\)
3 \(1.5\,ohm\)
4 None of the above
Explanation:
Same mass same material, i.e., volume is constant \(R = \rho \frac{{\rm{l}}}{A} \Rightarrow \frac{{{R_1}}}{{{R_2}}} = \frac{{{{\rm{l}}_1}}}{{{{\rm{l}}_2}}} \times \frac{{{A_2}}}{{{A_1}}} = {\left( {\frac{{{A_2}}}{{{A_1}}}} \right)^2} = {\left( {\frac{{{d_2}}}{{{d_1}}}} \right)^4}\) \( \Rightarrow \frac{{24}}{{{R_2}}} = {\left( {\frac{d}{{d/2}}} \right)^4} = 16 \Rightarrow {R_2} = 1.5\Omega \)
PHXII03:CURRENT ELECTRICITY
357054
What is percentage change in its resistance if radius is increased by 1%?
1 \(1\% \)
2 \(2\% \)
3 \(3\% \)
4 \(4\% \)
Explanation:
\(\because R = \rho \frac{V}{{{A^2}}}\quad \Rightarrow \frac{{dR}}{R} = - 2\frac{{dA}}{A}\) As cross-sectional area \({\rm{A = \pi }}{{\rm{r}}^{\rm{2}}}\) So \(\frac{{dR}}{R} = - 2\frac{{dA}}{A} = - 4\frac{{dr}}{r}\) If radius is increased by 1%, the resistance is decreased by 4%.
PHXII03:CURRENT ELECTRICITY
357055
A copper wire of length 10 \(m\) and radius \(\left( {{{10}^{ - 2}}/\pi } \right)m\) has electrical resistance of \(10\,\Omega \). The current density in the wire for an electric field strength of 10 (\(V\)/\(m\)) is:
1 \({10^6}\,A/{m^2}\)
2 \({10^{ - 5}}\,A/{m^2}\)
3 \({10^5}\,A/{m^2}\)
4 \({10^4}\,A/{m^2}\)
Explanation:
Method I Radius of wire \( = \frac{{{{10}^{ - 2}}}}{{\sqrt \pi }}\,\,\) Cross sectional area \(A = \pi {r^2} = {10^{ - 4}}{m^2}\) Current density is given by \(j = \frac{i}{A} = \left( {\frac{V}{R}} \right) \cdot \frac{1}{A} = \frac{{E\ell }}{{RA}}\) \(j = \frac{{10 \times 10}}{{10 \times {{10}^{ - 4}}}} = {10^5}A/{m^2}\) Method II \(j = \sigma E \Rightarrow \frac{E}{\rho } = \frac{{E\ell }}{{RA}} = \frac{{10 \times 10 \times \pi }}{{10 \times {{10}^{ - 4}} \times \pi }}\) \(\left( {\because R = \frac{{\rho \ell }}{A}} \right)\) \(j = {10^5}\,\,A/{m^2}\)
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PHXII03:CURRENT ELECTRICITY
357052
Assertion : When the length of a conductor is doubled (area is constant) than resistance will also gets doubled. Reason : Resistance is directly proportional to the length of a conductor
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(R=\rho \dfrac{l}{A}\) So correct option is (1).
PHXII03:CURRENT ELECTRICITY
357053
We have two wires \(A\) and \(B\) of same mass and same material. The diameter of the wire \(A\) is half of that \(B\). If the resistance of wire \(A\) is 24 \(ohm\) then the resistance of wire \(B\) will be
1 \(3.0\,ohm\)
2 \(12\,ohm\)
3 \(1.5\,ohm\)
4 None of the above
Explanation:
Same mass same material, i.e., volume is constant \(R = \rho \frac{{\rm{l}}}{A} \Rightarrow \frac{{{R_1}}}{{{R_2}}} = \frac{{{{\rm{l}}_1}}}{{{{\rm{l}}_2}}} \times \frac{{{A_2}}}{{{A_1}}} = {\left( {\frac{{{A_2}}}{{{A_1}}}} \right)^2} = {\left( {\frac{{{d_2}}}{{{d_1}}}} \right)^4}\) \( \Rightarrow \frac{{24}}{{{R_2}}} = {\left( {\frac{d}{{d/2}}} \right)^4} = 16 \Rightarrow {R_2} = 1.5\Omega \)
PHXII03:CURRENT ELECTRICITY
357054
What is percentage change in its resistance if radius is increased by 1%?
1 \(1\% \)
2 \(2\% \)
3 \(3\% \)
4 \(4\% \)
Explanation:
\(\because R = \rho \frac{V}{{{A^2}}}\quad \Rightarrow \frac{{dR}}{R} = - 2\frac{{dA}}{A}\) As cross-sectional area \({\rm{A = \pi }}{{\rm{r}}^{\rm{2}}}\) So \(\frac{{dR}}{R} = - 2\frac{{dA}}{A} = - 4\frac{{dr}}{r}\) If radius is increased by 1%, the resistance is decreased by 4%.
PHXII03:CURRENT ELECTRICITY
357055
A copper wire of length 10 \(m\) and radius \(\left( {{{10}^{ - 2}}/\pi } \right)m\) has electrical resistance of \(10\,\Omega \). The current density in the wire for an electric field strength of 10 (\(V\)/\(m\)) is:
1 \({10^6}\,A/{m^2}\)
2 \({10^{ - 5}}\,A/{m^2}\)
3 \({10^5}\,A/{m^2}\)
4 \({10^4}\,A/{m^2}\)
Explanation:
Method I Radius of wire \( = \frac{{{{10}^{ - 2}}}}{{\sqrt \pi }}\,\,\) Cross sectional area \(A = \pi {r^2} = {10^{ - 4}}{m^2}\) Current density is given by \(j = \frac{i}{A} = \left( {\frac{V}{R}} \right) \cdot \frac{1}{A} = \frac{{E\ell }}{{RA}}\) \(j = \frac{{10 \times 10}}{{10 \times {{10}^{ - 4}}}} = {10^5}A/{m^2}\) Method II \(j = \sigma E \Rightarrow \frac{E}{\rho } = \frac{{E\ell }}{{RA}} = \frac{{10 \times 10 \times \pi }}{{10 \times {{10}^{ - 4}} \times \pi }}\) \(\left( {\because R = \frac{{\rho \ell }}{A}} \right)\) \(j = {10^5}\,\,A/{m^2}\)
357052
Assertion : When the length of a conductor is doubled (area is constant) than resistance will also gets doubled. Reason : Resistance is directly proportional to the length of a conductor
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(R=\rho \dfrac{l}{A}\) So correct option is (1).
PHXII03:CURRENT ELECTRICITY
357053
We have two wires \(A\) and \(B\) of same mass and same material. The diameter of the wire \(A\) is half of that \(B\). If the resistance of wire \(A\) is 24 \(ohm\) then the resistance of wire \(B\) will be
1 \(3.0\,ohm\)
2 \(12\,ohm\)
3 \(1.5\,ohm\)
4 None of the above
Explanation:
Same mass same material, i.e., volume is constant \(R = \rho \frac{{\rm{l}}}{A} \Rightarrow \frac{{{R_1}}}{{{R_2}}} = \frac{{{{\rm{l}}_1}}}{{{{\rm{l}}_2}}} \times \frac{{{A_2}}}{{{A_1}}} = {\left( {\frac{{{A_2}}}{{{A_1}}}} \right)^2} = {\left( {\frac{{{d_2}}}{{{d_1}}}} \right)^4}\) \( \Rightarrow \frac{{24}}{{{R_2}}} = {\left( {\frac{d}{{d/2}}} \right)^4} = 16 \Rightarrow {R_2} = 1.5\Omega \)
PHXII03:CURRENT ELECTRICITY
357054
What is percentage change in its resistance if radius is increased by 1%?
1 \(1\% \)
2 \(2\% \)
3 \(3\% \)
4 \(4\% \)
Explanation:
\(\because R = \rho \frac{V}{{{A^2}}}\quad \Rightarrow \frac{{dR}}{R} = - 2\frac{{dA}}{A}\) As cross-sectional area \({\rm{A = \pi }}{{\rm{r}}^{\rm{2}}}\) So \(\frac{{dR}}{R} = - 2\frac{{dA}}{A} = - 4\frac{{dr}}{r}\) If radius is increased by 1%, the resistance is decreased by 4%.
PHXII03:CURRENT ELECTRICITY
357055
A copper wire of length 10 \(m\) and radius \(\left( {{{10}^{ - 2}}/\pi } \right)m\) has electrical resistance of \(10\,\Omega \). The current density in the wire for an electric field strength of 10 (\(V\)/\(m\)) is:
1 \({10^6}\,A/{m^2}\)
2 \({10^{ - 5}}\,A/{m^2}\)
3 \({10^5}\,A/{m^2}\)
4 \({10^4}\,A/{m^2}\)
Explanation:
Method I Radius of wire \( = \frac{{{{10}^{ - 2}}}}{{\sqrt \pi }}\,\,\) Cross sectional area \(A = \pi {r^2} = {10^{ - 4}}{m^2}\) Current density is given by \(j = \frac{i}{A} = \left( {\frac{V}{R}} \right) \cdot \frac{1}{A} = \frac{{E\ell }}{{RA}}\) \(j = \frac{{10 \times 10}}{{10 \times {{10}^{ - 4}}}} = {10^5}A/{m^2}\) Method II \(j = \sigma E \Rightarrow \frac{E}{\rho } = \frac{{E\ell }}{{RA}} = \frac{{10 \times 10 \times \pi }}{{10 \times {{10}^{ - 4}} \times \pi }}\) \(\left( {\because R = \frac{{\rho \ell }}{A}} \right)\) \(j = {10^5}\,\,A/{m^2}\)