357060
Assertion : Bending of a wire does not affect resistivity of a material. Reason : Resistance of wire is proportional to resistivity of material.
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:
The formula for resistance is \(R=\rho \dfrac{l}{A}\), where \(R\) is resistance, \(\rho\) is the resistivity of the material, \(l\) is the length of the wire, and \(A\) is the cross-sectional area of the wire. Importantly, resistivity is an inherent property of the material and is independent of the wire's geometry.Therefore, when a wire is bent or shaped, its resistivity, remain unchanged. So correct option is (2).
PHXII03:CURRENT ELECTRICITY
357061
The electric intensity \(E\), current density \(j\) and specific resistance \(k\) are related to each other by the relation
1 \(k = jE\)
2 \(E = jk\)
3 \(E = k/j\)
4 \(E = j/k\)
Explanation:
Conceptual Question
PHXII03:CURRENT ELECTRICITY
357062
A wire of resistance \(160 \Omega\) is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be
1 \(40 \Omega\)
2 \(16 \Omega\)
3 \(10 \Omega\)
4 \(640 \Omega\)
Explanation:
Volume of the wire before being melted and after will remain same therefore. Volume of wire before melting = Volume of wire after melting \(A^{\prime} l^{\prime}=A l\) \(\pi r^{\prime 2} l^{\prime}=\pi r^{2} l\) As \(l=\dfrac{l^{\prime}}{4} ; \pi r^{\prime 2} l^{\prime}=\pi r^{2} \dfrac{l^{\prime}}{4} r^{2}\) \(r = 2r'\frac{R}{{R'}} = \frac{{\rho l}}{A} \times \frac{{A'}}{{\rho l'}} = \frac{l}{{l'}} \times \frac{{\pi {{r'}^2}}}{{\pi {r^2}}} = \frac{l}{{l'}}\frac{{{{r'}^2}}}{{{r^2}}}\) \(R=R^{\prime}\left(\dfrac{l^{\prime}}{4 l^{\prime}} \times \dfrac{r^{\prime 2}}{\left(2 r^{\prime}\right)^{2}}\right)=\dfrac{160}{16}=10 \Omega\)
JEE - 2023
PHXII03:CURRENT ELECTRICITY
357063
Masses of three wires of copper are in the ratio 1 : 3 : 5 and their lengths are in the ratio 5:3:1. The ratio of their electrical resistance are
1 \(1\,{\rm{:}}\,15\,{\rm{:}}\,25\)
2 \(1\,{\rm{:}}\,3\,{\rm{:}}\,5\)
3 \(125\,{\rm{:}}\,15\,{\rm{:}}\,1\)
4 \(5\,{\rm{:}}\,3\,{\rm{:}}\,1\)
Explanation:
As all three wires have same material (copper) in them, the ratio of their masses equals the ratio of their volumes. \(\therefore \quad {V_1}:{V_2}:{V_3} = 1:3:5\) Also \({l_1}:{l_2}:{l_3} = 5:3:1\) (given) As resistance \(R = (\rho l){\rm{/}}A = (\rho {l^2}){\rm{/}}Al = (\rho {l^2}){\rm{/}}V\) \(\therefore \;\;{R_1}:{R_2}:{R_3} = (l_1^2/{V_1}):(l_2^2/{V_2}):(l_3^2/{V_3})\) \( = ({5^2}{\rm{/}}1\,)\,:\,\,({3^2}{\rm{/3)}}:({1^2}{\rm{/}}5) = 125:15:1\)
KCET - 2019
PHXII03:CURRENT ELECTRICITY
357064
The length of the wire is doubled. Its conductance will be
1 remain unchanged
2 halved
3 doubled
4 quadrupled
Explanation:
Conductance is given by \(\sigma=\dfrac{1}{R}\)\(\text { Substituting } R=\rho \dfrac{l}{A}, \text { we get } \sigma=\dfrac{A}{\rho l}\) If \(A\) is constant, then we have \(\sigma \propto \dfrac{1}{l}\) \(\Rightarrow \quad \dfrac{\sigma_{2}}{\sigma_{1}}=\dfrac{l_{1}}{l_{2}}\) Substituting, \(l_{2}=2 l_{1}\), we get \(\sigma_{2}=\dfrac{\sigma_{1}}{2}\) We see that, conductance will be halved.
357060
Assertion : Bending of a wire does not affect resistivity of a material. Reason : Resistance of wire is proportional to resistivity of material.
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:
The formula for resistance is \(R=\rho \dfrac{l}{A}\), where \(R\) is resistance, \(\rho\) is the resistivity of the material, \(l\) is the length of the wire, and \(A\) is the cross-sectional area of the wire. Importantly, resistivity is an inherent property of the material and is independent of the wire's geometry.Therefore, when a wire is bent or shaped, its resistivity, remain unchanged. So correct option is (2).
PHXII03:CURRENT ELECTRICITY
357061
The electric intensity \(E\), current density \(j\) and specific resistance \(k\) are related to each other by the relation
1 \(k = jE\)
2 \(E = jk\)
3 \(E = k/j\)
4 \(E = j/k\)
Explanation:
Conceptual Question
PHXII03:CURRENT ELECTRICITY
357062
A wire of resistance \(160 \Omega\) is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be
1 \(40 \Omega\)
2 \(16 \Omega\)
3 \(10 \Omega\)
4 \(640 \Omega\)
Explanation:
Volume of the wire before being melted and after will remain same therefore. Volume of wire before melting = Volume of wire after melting \(A^{\prime} l^{\prime}=A l\) \(\pi r^{\prime 2} l^{\prime}=\pi r^{2} l\) As \(l=\dfrac{l^{\prime}}{4} ; \pi r^{\prime 2} l^{\prime}=\pi r^{2} \dfrac{l^{\prime}}{4} r^{2}\) \(r = 2r'\frac{R}{{R'}} = \frac{{\rho l}}{A} \times \frac{{A'}}{{\rho l'}} = \frac{l}{{l'}} \times \frac{{\pi {{r'}^2}}}{{\pi {r^2}}} = \frac{l}{{l'}}\frac{{{{r'}^2}}}{{{r^2}}}\) \(R=R^{\prime}\left(\dfrac{l^{\prime}}{4 l^{\prime}} \times \dfrac{r^{\prime 2}}{\left(2 r^{\prime}\right)^{2}}\right)=\dfrac{160}{16}=10 \Omega\)
JEE - 2023
PHXII03:CURRENT ELECTRICITY
357063
Masses of three wires of copper are in the ratio 1 : 3 : 5 and their lengths are in the ratio 5:3:1. The ratio of their electrical resistance are
1 \(1\,{\rm{:}}\,15\,{\rm{:}}\,25\)
2 \(1\,{\rm{:}}\,3\,{\rm{:}}\,5\)
3 \(125\,{\rm{:}}\,15\,{\rm{:}}\,1\)
4 \(5\,{\rm{:}}\,3\,{\rm{:}}\,1\)
Explanation:
As all three wires have same material (copper) in them, the ratio of their masses equals the ratio of their volumes. \(\therefore \quad {V_1}:{V_2}:{V_3} = 1:3:5\) Also \({l_1}:{l_2}:{l_3} = 5:3:1\) (given) As resistance \(R = (\rho l){\rm{/}}A = (\rho {l^2}){\rm{/}}Al = (\rho {l^2}){\rm{/}}V\) \(\therefore \;\;{R_1}:{R_2}:{R_3} = (l_1^2/{V_1}):(l_2^2/{V_2}):(l_3^2/{V_3})\) \( = ({5^2}{\rm{/}}1\,)\,:\,\,({3^2}{\rm{/3)}}:({1^2}{\rm{/}}5) = 125:15:1\)
KCET - 2019
PHXII03:CURRENT ELECTRICITY
357064
The length of the wire is doubled. Its conductance will be
1 remain unchanged
2 halved
3 doubled
4 quadrupled
Explanation:
Conductance is given by \(\sigma=\dfrac{1}{R}\)\(\text { Substituting } R=\rho \dfrac{l}{A}, \text { we get } \sigma=\dfrac{A}{\rho l}\) If \(A\) is constant, then we have \(\sigma \propto \dfrac{1}{l}\) \(\Rightarrow \quad \dfrac{\sigma_{2}}{\sigma_{1}}=\dfrac{l_{1}}{l_{2}}\) Substituting, \(l_{2}=2 l_{1}\), we get \(\sigma_{2}=\dfrac{\sigma_{1}}{2}\) We see that, conductance will be halved.
357060
Assertion : Bending of a wire does not affect resistivity of a material. Reason : Resistance of wire is proportional to resistivity of material.
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:
The formula for resistance is \(R=\rho \dfrac{l}{A}\), where \(R\) is resistance, \(\rho\) is the resistivity of the material, \(l\) is the length of the wire, and \(A\) is the cross-sectional area of the wire. Importantly, resistivity is an inherent property of the material and is independent of the wire's geometry.Therefore, when a wire is bent or shaped, its resistivity, remain unchanged. So correct option is (2).
PHXII03:CURRENT ELECTRICITY
357061
The electric intensity \(E\), current density \(j\) and specific resistance \(k\) are related to each other by the relation
1 \(k = jE\)
2 \(E = jk\)
3 \(E = k/j\)
4 \(E = j/k\)
Explanation:
Conceptual Question
PHXII03:CURRENT ELECTRICITY
357062
A wire of resistance \(160 \Omega\) is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be
1 \(40 \Omega\)
2 \(16 \Omega\)
3 \(10 \Omega\)
4 \(640 \Omega\)
Explanation:
Volume of the wire before being melted and after will remain same therefore. Volume of wire before melting = Volume of wire after melting \(A^{\prime} l^{\prime}=A l\) \(\pi r^{\prime 2} l^{\prime}=\pi r^{2} l\) As \(l=\dfrac{l^{\prime}}{4} ; \pi r^{\prime 2} l^{\prime}=\pi r^{2} \dfrac{l^{\prime}}{4} r^{2}\) \(r = 2r'\frac{R}{{R'}} = \frac{{\rho l}}{A} \times \frac{{A'}}{{\rho l'}} = \frac{l}{{l'}} \times \frac{{\pi {{r'}^2}}}{{\pi {r^2}}} = \frac{l}{{l'}}\frac{{{{r'}^2}}}{{{r^2}}}\) \(R=R^{\prime}\left(\dfrac{l^{\prime}}{4 l^{\prime}} \times \dfrac{r^{\prime 2}}{\left(2 r^{\prime}\right)^{2}}\right)=\dfrac{160}{16}=10 \Omega\)
JEE - 2023
PHXII03:CURRENT ELECTRICITY
357063
Masses of three wires of copper are in the ratio 1 : 3 : 5 and their lengths are in the ratio 5:3:1. The ratio of their electrical resistance are
1 \(1\,{\rm{:}}\,15\,{\rm{:}}\,25\)
2 \(1\,{\rm{:}}\,3\,{\rm{:}}\,5\)
3 \(125\,{\rm{:}}\,15\,{\rm{:}}\,1\)
4 \(5\,{\rm{:}}\,3\,{\rm{:}}\,1\)
Explanation:
As all three wires have same material (copper) in them, the ratio of their masses equals the ratio of their volumes. \(\therefore \quad {V_1}:{V_2}:{V_3} = 1:3:5\) Also \({l_1}:{l_2}:{l_3} = 5:3:1\) (given) As resistance \(R = (\rho l){\rm{/}}A = (\rho {l^2}){\rm{/}}Al = (\rho {l^2}){\rm{/}}V\) \(\therefore \;\;{R_1}:{R_2}:{R_3} = (l_1^2/{V_1}):(l_2^2/{V_2}):(l_3^2/{V_3})\) \( = ({5^2}{\rm{/}}1\,)\,:\,\,({3^2}{\rm{/3)}}:({1^2}{\rm{/}}5) = 125:15:1\)
KCET - 2019
PHXII03:CURRENT ELECTRICITY
357064
The length of the wire is doubled. Its conductance will be
1 remain unchanged
2 halved
3 doubled
4 quadrupled
Explanation:
Conductance is given by \(\sigma=\dfrac{1}{R}\)\(\text { Substituting } R=\rho \dfrac{l}{A}, \text { we get } \sigma=\dfrac{A}{\rho l}\) If \(A\) is constant, then we have \(\sigma \propto \dfrac{1}{l}\) \(\Rightarrow \quad \dfrac{\sigma_{2}}{\sigma_{1}}=\dfrac{l_{1}}{l_{2}}\) Substituting, \(l_{2}=2 l_{1}\), we get \(\sigma_{2}=\dfrac{\sigma_{1}}{2}\) We see that, conductance will be halved.
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII03:CURRENT ELECTRICITY
357060
Assertion : Bending of a wire does not affect resistivity of a material. Reason : Resistance of wire is proportional to resistivity of material.
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:
The formula for resistance is \(R=\rho \dfrac{l}{A}\), where \(R\) is resistance, \(\rho\) is the resistivity of the material, \(l\) is the length of the wire, and \(A\) is the cross-sectional area of the wire. Importantly, resistivity is an inherent property of the material and is independent of the wire's geometry.Therefore, when a wire is bent or shaped, its resistivity, remain unchanged. So correct option is (2).
PHXII03:CURRENT ELECTRICITY
357061
The electric intensity \(E\), current density \(j\) and specific resistance \(k\) are related to each other by the relation
1 \(k = jE\)
2 \(E = jk\)
3 \(E = k/j\)
4 \(E = j/k\)
Explanation:
Conceptual Question
PHXII03:CURRENT ELECTRICITY
357062
A wire of resistance \(160 \Omega\) is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be
1 \(40 \Omega\)
2 \(16 \Omega\)
3 \(10 \Omega\)
4 \(640 \Omega\)
Explanation:
Volume of the wire before being melted and after will remain same therefore. Volume of wire before melting = Volume of wire after melting \(A^{\prime} l^{\prime}=A l\) \(\pi r^{\prime 2} l^{\prime}=\pi r^{2} l\) As \(l=\dfrac{l^{\prime}}{4} ; \pi r^{\prime 2} l^{\prime}=\pi r^{2} \dfrac{l^{\prime}}{4} r^{2}\) \(r = 2r'\frac{R}{{R'}} = \frac{{\rho l}}{A} \times \frac{{A'}}{{\rho l'}} = \frac{l}{{l'}} \times \frac{{\pi {{r'}^2}}}{{\pi {r^2}}} = \frac{l}{{l'}}\frac{{{{r'}^2}}}{{{r^2}}}\) \(R=R^{\prime}\left(\dfrac{l^{\prime}}{4 l^{\prime}} \times \dfrac{r^{\prime 2}}{\left(2 r^{\prime}\right)^{2}}\right)=\dfrac{160}{16}=10 \Omega\)
JEE - 2023
PHXII03:CURRENT ELECTRICITY
357063
Masses of three wires of copper are in the ratio 1 : 3 : 5 and their lengths are in the ratio 5:3:1. The ratio of their electrical resistance are
1 \(1\,{\rm{:}}\,15\,{\rm{:}}\,25\)
2 \(1\,{\rm{:}}\,3\,{\rm{:}}\,5\)
3 \(125\,{\rm{:}}\,15\,{\rm{:}}\,1\)
4 \(5\,{\rm{:}}\,3\,{\rm{:}}\,1\)
Explanation:
As all three wires have same material (copper) in them, the ratio of their masses equals the ratio of their volumes. \(\therefore \quad {V_1}:{V_2}:{V_3} = 1:3:5\) Also \({l_1}:{l_2}:{l_3} = 5:3:1\) (given) As resistance \(R = (\rho l){\rm{/}}A = (\rho {l^2}){\rm{/}}Al = (\rho {l^2}){\rm{/}}V\) \(\therefore \;\;{R_1}:{R_2}:{R_3} = (l_1^2/{V_1}):(l_2^2/{V_2}):(l_3^2/{V_3})\) \( = ({5^2}{\rm{/}}1\,)\,:\,\,({3^2}{\rm{/3)}}:({1^2}{\rm{/}}5) = 125:15:1\)
KCET - 2019
PHXII03:CURRENT ELECTRICITY
357064
The length of the wire is doubled. Its conductance will be
1 remain unchanged
2 halved
3 doubled
4 quadrupled
Explanation:
Conductance is given by \(\sigma=\dfrac{1}{R}\)\(\text { Substituting } R=\rho \dfrac{l}{A}, \text { we get } \sigma=\dfrac{A}{\rho l}\) If \(A\) is constant, then we have \(\sigma \propto \dfrac{1}{l}\) \(\Rightarrow \quad \dfrac{\sigma_{2}}{\sigma_{1}}=\dfrac{l_{1}}{l_{2}}\) Substituting, \(l_{2}=2 l_{1}\), we get \(\sigma_{2}=\dfrac{\sigma_{1}}{2}\) We see that, conductance will be halved.
357060
Assertion : Bending of a wire does not affect resistivity of a material. Reason : Resistance of wire is proportional to resistivity of material.
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:
The formula for resistance is \(R=\rho \dfrac{l}{A}\), where \(R\) is resistance, \(\rho\) is the resistivity of the material, \(l\) is the length of the wire, and \(A\) is the cross-sectional area of the wire. Importantly, resistivity is an inherent property of the material and is independent of the wire's geometry.Therefore, when a wire is bent or shaped, its resistivity, remain unchanged. So correct option is (2).
PHXII03:CURRENT ELECTRICITY
357061
The electric intensity \(E\), current density \(j\) and specific resistance \(k\) are related to each other by the relation
1 \(k = jE\)
2 \(E = jk\)
3 \(E = k/j\)
4 \(E = j/k\)
Explanation:
Conceptual Question
PHXII03:CURRENT ELECTRICITY
357062
A wire of resistance \(160 \Omega\) is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be
1 \(40 \Omega\)
2 \(16 \Omega\)
3 \(10 \Omega\)
4 \(640 \Omega\)
Explanation:
Volume of the wire before being melted and after will remain same therefore. Volume of wire before melting = Volume of wire after melting \(A^{\prime} l^{\prime}=A l\) \(\pi r^{\prime 2} l^{\prime}=\pi r^{2} l\) As \(l=\dfrac{l^{\prime}}{4} ; \pi r^{\prime 2} l^{\prime}=\pi r^{2} \dfrac{l^{\prime}}{4} r^{2}\) \(r = 2r'\frac{R}{{R'}} = \frac{{\rho l}}{A} \times \frac{{A'}}{{\rho l'}} = \frac{l}{{l'}} \times \frac{{\pi {{r'}^2}}}{{\pi {r^2}}} = \frac{l}{{l'}}\frac{{{{r'}^2}}}{{{r^2}}}\) \(R=R^{\prime}\left(\dfrac{l^{\prime}}{4 l^{\prime}} \times \dfrac{r^{\prime 2}}{\left(2 r^{\prime}\right)^{2}}\right)=\dfrac{160}{16}=10 \Omega\)
JEE - 2023
PHXII03:CURRENT ELECTRICITY
357063
Masses of three wires of copper are in the ratio 1 : 3 : 5 and their lengths are in the ratio 5:3:1. The ratio of their electrical resistance are
1 \(1\,{\rm{:}}\,15\,{\rm{:}}\,25\)
2 \(1\,{\rm{:}}\,3\,{\rm{:}}\,5\)
3 \(125\,{\rm{:}}\,15\,{\rm{:}}\,1\)
4 \(5\,{\rm{:}}\,3\,{\rm{:}}\,1\)
Explanation:
As all three wires have same material (copper) in them, the ratio of their masses equals the ratio of their volumes. \(\therefore \quad {V_1}:{V_2}:{V_3} = 1:3:5\) Also \({l_1}:{l_2}:{l_3} = 5:3:1\) (given) As resistance \(R = (\rho l){\rm{/}}A = (\rho {l^2}){\rm{/}}Al = (\rho {l^2}){\rm{/}}V\) \(\therefore \;\;{R_1}:{R_2}:{R_3} = (l_1^2/{V_1}):(l_2^2/{V_2}):(l_3^2/{V_3})\) \( = ({5^2}{\rm{/}}1\,)\,:\,\,({3^2}{\rm{/3)}}:({1^2}{\rm{/}}5) = 125:15:1\)
KCET - 2019
PHXII03:CURRENT ELECTRICITY
357064
The length of the wire is doubled. Its conductance will be
1 remain unchanged
2 halved
3 doubled
4 quadrupled
Explanation:
Conductance is given by \(\sigma=\dfrac{1}{R}\)\(\text { Substituting } R=\rho \dfrac{l}{A}, \text { we get } \sigma=\dfrac{A}{\rho l}\) If \(A\) is constant, then we have \(\sigma \propto \dfrac{1}{l}\) \(\Rightarrow \quad \dfrac{\sigma_{2}}{\sigma_{1}}=\dfrac{l_{1}}{l_{2}}\) Substituting, \(l_{2}=2 l_{1}\), we get \(\sigma_{2}=\dfrac{\sigma_{1}}{2}\) We see that, conductance will be halved.