152133 A wire of resistance $160 \Omega$ is melted and drawn in wire of one-fourth of its length. The new resistance of the wire will be

152134 A uniform metallic wire carries a current $2 \mathrm{~A}$, when $3.4 \mathrm{~V}$ battery is connected across it. The mass of uniform metallic wire is $8.92 \times 10^{-3} \mathrm{~kg}$, density is $8.92 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ and resistivity is 1.7 $\times 10^{-8} \Omega-\mathbf{m}$.
The length of wire is :

152135 The INCORRECT statement is

152137 The length of germanium rod is $0.928 \mathrm{~cm}$ and its area of cross section is $1 \mathbf{m m}^{2}$. If for germanium $\mathrm{n}_{\mathrm{i}}=2.5 \times 10^{19} \mathrm{~m}^{-3}, \mu_{\mathrm{n}}=0.15 \mathrm{~m}^{2} \mathrm{~V}^{-}$ $\mathrm{s}^{-1}, \mu_{\mathrm{e}}=0.35 \mathrm{~m}^{2} \mathrm{~V}^{-1} \mathrm{~s}^{-1}$ then resistivity is

152138 Consider a wire of length $L$ with a resistance of $5 \Omega$. Applying an external force, the wire is elongated such that its length becomes $3 \mathrm{~L}$. Assuming the resistivity and density of the material is unchanged, the resistance of the elongated wire is,

152133 A wire of resistance $160 \Omega$ is melted and drawn in wire of one-fourth of its length. The new resistance of the wire will be

152134 A uniform metallic wire carries a current $2 \mathrm{~A}$, when $3.4 \mathrm{~V}$ battery is connected across it. The mass of uniform metallic wire is $8.92 \times 10^{-3} \mathrm{~kg}$, density is $8.92 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ and resistivity is 1.7 $\times 10^{-8} \Omega-\mathbf{m}$.
The length of wire is :

152135 The INCORRECT statement is

152137 The length of germanium rod is $0.928 \mathrm{~cm}$ and its area of cross section is $1 \mathbf{m m}^{2}$. If for germanium $\mathrm{n}_{\mathrm{i}}=2.5 \times 10^{19} \mathrm{~m}^{-3}, \mu_{\mathrm{n}}=0.15 \mathrm{~m}^{2} \mathrm{~V}^{-}$ $\mathrm{s}^{-1}, \mu_{\mathrm{e}}=0.35 \mathrm{~m}^{2} \mathrm{~V}^{-1} \mathrm{~s}^{-1}$ then resistivity is

152138 Consider a wire of length $L$ with a resistance of $5 \Omega$. Applying an external force, the wire is elongated such that its length becomes $3 \mathrm{~L}$. Assuming the resistivity and density of the material is unchanged, the resistance of the elongated wire is,

152133 A wire of resistance $160 \Omega$ is melted and drawn in wire of one-fourth of its length. The new resistance of the wire will be

152134 A uniform metallic wire carries a current $2 \mathrm{~A}$, when $3.4 \mathrm{~V}$ battery is connected across it. The mass of uniform metallic wire is $8.92 \times 10^{-3} \mathrm{~kg}$, density is $8.92 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ and resistivity is 1.7 $\times 10^{-8} \Omega-\mathbf{m}$.
The length of wire is :

152135 The INCORRECT statement is

152137 The length of germanium rod is $0.928 \mathrm{~cm}$ and its area of cross section is $1 \mathbf{m m}^{2}$. If for germanium $\mathrm{n}_{\mathrm{i}}=2.5 \times 10^{19} \mathrm{~m}^{-3}, \mu_{\mathrm{n}}=0.15 \mathrm{~m}^{2} \mathrm{~V}^{-}$ $\mathrm{s}^{-1}, \mu_{\mathrm{e}}=0.35 \mathrm{~m}^{2} \mathrm{~V}^{-1} \mathrm{~s}^{-1}$ then resistivity is

152138 Consider a wire of length $L$ with a resistance of $5 \Omega$. Applying an external force, the wire is elongated such that its length becomes $3 \mathrm{~L}$. Assuming the resistivity and density of the material is unchanged, the resistance of the elongated wire is,

152133 A wire of resistance $160 \Omega$ is melted and drawn in wire of one-fourth of its length. The new resistance of the wire will be

152134 A uniform metallic wire carries a current $2 \mathrm{~A}$, when $3.4 \mathrm{~V}$ battery is connected across it. The mass of uniform metallic wire is $8.92 \times 10^{-3} \mathrm{~kg}$, density is $8.92 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ and resistivity is 1.7 $\times 10^{-8} \Omega-\mathbf{m}$.
The length of wire is :

152135 The INCORRECT statement is

152137 The length of germanium rod is $0.928 \mathrm{~cm}$ and its area of cross section is $1 \mathbf{m m}^{2}$. If for germanium $\mathrm{n}_{\mathrm{i}}=2.5 \times 10^{19} \mathrm{~m}^{-3}, \mu_{\mathrm{n}}=0.15 \mathrm{~m}^{2} \mathrm{~V}^{-}$ $\mathrm{s}^{-1}, \mu_{\mathrm{e}}=0.35 \mathrm{~m}^{2} \mathrm{~V}^{-1} \mathrm{~s}^{-1}$ then resistivity is

152138 Consider a wire of length $L$ with a resistance of $5 \Omega$. Applying an external force, the wire is elongated such that its length becomes $3 \mathrm{~L}$. Assuming the resistivity and density of the material is unchanged, the resistance of the elongated wire is,

152133 A wire of resistance $160 \Omega$ is melted and drawn in wire of one-fourth of its length. The new resistance of the wire will be

152134 A uniform metallic wire carries a current $2 \mathrm{~A}$, when $3.4 \mathrm{~V}$ battery is connected across it. The mass of uniform metallic wire is $8.92 \times 10^{-3} \mathrm{~kg}$, density is $8.92 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ and resistivity is 1.7 $\times 10^{-8} \Omega-\mathbf{m}$.
The length of wire is :

152135 The INCORRECT statement is

152137 The length of germanium rod is $0.928 \mathrm{~cm}$ and its area of cross section is $1 \mathbf{m m}^{2}$. If for germanium $\mathrm{n}_{\mathrm{i}}=2.5 \times 10^{19} \mathrm{~m}^{-3}, \mu_{\mathrm{n}}=0.15 \mathrm{~m}^{2} \mathrm{~V}^{-}$ $\mathrm{s}^{-1}, \mu_{\mathrm{e}}=0.35 \mathrm{~m}^{2} \mathrm{~V}^{-1} \mathrm{~s}^{-1}$ then resistivity is

152138 Consider a wire of length $L$ with a resistance of $5 \Omega$. Applying an external force, the wire is elongated such that its length becomes $3 \mathrm{~L}$. Assuming the resistivity and density of the material is unchanged, the resistance of the elongated wire is,