152166
Two wires of equal diameters of resistivities $\rho_{1}$ and $\rho_{2}$ and lengths $x_{1}$ and $x_{2}$ respectively are joined in series. The equivalent resistivity of the combination is
A Given, First wire resistivity $=\rho_{1}$ Second wire resistivity $=\rho_{2}$ Length of first wire $=x_{1}$ Length of second wire $=\mathrm{x}_{2}$ Diameter of both wire are equal So, $\quad A_{1}=A_{2}=A$ We know that, $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ For first wire - $R_{1}=\frac{\rho_{1} x_{1}}{A}$ For second wire - $R_{2}=\frac{\rho_{2} x_{2}}{A}$ Since, both resistance are joined in series, Hence, $R_{\text {eq }}=R_{1}+R_{2}$ $R_{\text {eq }}=\frac{1}{A}\left(\rho_{1} x_{1}+\rho_{2} x_{2}\right)$ $\rho_{\text {eq }} \cdot \frac{\left(x_{1}+x_{2}\right)}{A}=\frac{1}{A}\left(\rho_{1} x_{1}+\rho_{2} x_{2}\right)$ $\rho_{\text {eq }}=\frac{\rho_{1} x_{1}+\rho_{2} x_{2}}{x_{1}+x_{2}}$
EAMCET-2000
Current Electricity
152167
A nichrome wire $50 \mathrm{~cm}$ long and $1 \mathrm{~mm}^{2}$ crosssection carries a current of 4 A. When connected to $2 \mathrm{~V}$ battery. The resistivity of nichrome wire in $\Omega-m$ is
1 $1 \times 10^{-6}$
2 $4 \times 10^{-7}$
3 $3 \times 10^{-7}$
4 $2 \times 10^{-7}$
Explanation:
A Given, Length of wire $(l)=50 \mathrm{~cm}=0.5 \mathrm{~m}$ Area of wire $(\mathrm{A})=1 \mathrm{~mm}^{2}=1 \times 10^{-6} \mathrm{~m}^{2}$ Voltage $(\mathrm{V})=2 \mathrm{~V}$ Current $(\mathrm{I})=4 \mathrm{~A}$ We know that, $\mathrm{V}=\mathrm{IR}$ $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{2}{4}$ $\mathrm{R}=0.5 \Omega$ $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ $\rho=\frac{\mathrm{AR}}{l}$ $\rho=\frac{1 \times 10^{-6} \times 0.5}{0.5}$ $\rho=1 \times 10^{-6} \Omega \mathrm{m} .$ $\because \quad \mathrm{R}=\rho \frac{l}{\mathrm{~A}}$
EAMCET-2001
Current Electricity
152168
n conducting wires of same dimensions but having resistivities $1,2,3$, connected in series. The equivalent resistivity of the combination is
1 $\frac{\mathrm{n}(\mathrm{n}+1)}{2}$
2 $\frac{\mathrm{n}+1}{2}$
3 $\frac{\mathrm{n}+1}{2 \mathrm{n}}$
4 $\frac{2 \mathrm{n}}{\mathrm{n}+1}$
Explanation:
A Given, Number of wire $=\mathrm{n}$ Resistivity are given as, $\rho_{1}=1, \rho_{2}=2, \rho_{3}=3, \rho_{4}=4 \ldots \ldots \ldots . \rho_{n}=n$ We know that, $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}$ $\therefore \mathrm{R}_{1}=\frac{\rho_{\mathrm{l}} l}{\mathrm{~A}}$ $\mathrm{R}_{2}=\frac{\rho_{1} l}{\mathrm{~A}}, \mathrm{R}_{3}=\frac{\rho_{3} l}{\mathrm{~A}}, \mathrm{R}_{\mathrm{n}}=\frac{\rho_{\mathrm{n}} l}{\mathrm{~A}}$ Since, all resistance are joined in series. $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}+\ldots \ldots \ldots \ldots+\mathrm{R}_{\mathrm{n}}$ $\rho_{\mathrm{eq}} \frac{l}{\mathrm{~A}}=\frac{l}{\mathrm{~A}}\left(\rho_{1}+\rho_{2}+\rho_{3}+\ldots \ldots \ldots \rho_{\mathrm{n}}\right)$ $\rho_{\text {eq }}=(1+2+3+4+\ldots \ldots \ldots \ldots+\mathrm{n})$ $\rho_{\text {eq }}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}$
EAMCET -2000
Current Electricity
152169
An aluminum (resistivity $\rho=2.2 \times 10^{-8} \Omega-\mathrm{m}$ ) wire of a diameter $1.4 \mathrm{~mm}$ is used to make a $4 \Omega$ resistor. The length of the wire is
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Current Electricity
152166
Two wires of equal diameters of resistivities $\rho_{1}$ and $\rho_{2}$ and lengths $x_{1}$ and $x_{2}$ respectively are joined in series. The equivalent resistivity of the combination is
A Given, First wire resistivity $=\rho_{1}$ Second wire resistivity $=\rho_{2}$ Length of first wire $=x_{1}$ Length of second wire $=\mathrm{x}_{2}$ Diameter of both wire are equal So, $\quad A_{1}=A_{2}=A$ We know that, $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ For first wire - $R_{1}=\frac{\rho_{1} x_{1}}{A}$ For second wire - $R_{2}=\frac{\rho_{2} x_{2}}{A}$ Since, both resistance are joined in series, Hence, $R_{\text {eq }}=R_{1}+R_{2}$ $R_{\text {eq }}=\frac{1}{A}\left(\rho_{1} x_{1}+\rho_{2} x_{2}\right)$ $\rho_{\text {eq }} \cdot \frac{\left(x_{1}+x_{2}\right)}{A}=\frac{1}{A}\left(\rho_{1} x_{1}+\rho_{2} x_{2}\right)$ $\rho_{\text {eq }}=\frac{\rho_{1} x_{1}+\rho_{2} x_{2}}{x_{1}+x_{2}}$
EAMCET-2000
Current Electricity
152167
A nichrome wire $50 \mathrm{~cm}$ long and $1 \mathrm{~mm}^{2}$ crosssection carries a current of 4 A. When connected to $2 \mathrm{~V}$ battery. The resistivity of nichrome wire in $\Omega-m$ is
1 $1 \times 10^{-6}$
2 $4 \times 10^{-7}$
3 $3 \times 10^{-7}$
4 $2 \times 10^{-7}$
Explanation:
A Given, Length of wire $(l)=50 \mathrm{~cm}=0.5 \mathrm{~m}$ Area of wire $(\mathrm{A})=1 \mathrm{~mm}^{2}=1 \times 10^{-6} \mathrm{~m}^{2}$ Voltage $(\mathrm{V})=2 \mathrm{~V}$ Current $(\mathrm{I})=4 \mathrm{~A}$ We know that, $\mathrm{V}=\mathrm{IR}$ $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{2}{4}$ $\mathrm{R}=0.5 \Omega$ $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ $\rho=\frac{\mathrm{AR}}{l}$ $\rho=\frac{1 \times 10^{-6} \times 0.5}{0.5}$ $\rho=1 \times 10^{-6} \Omega \mathrm{m} .$ $\because \quad \mathrm{R}=\rho \frac{l}{\mathrm{~A}}$
EAMCET-2001
Current Electricity
152168
n conducting wires of same dimensions but having resistivities $1,2,3$, connected in series. The equivalent resistivity of the combination is
1 $\frac{\mathrm{n}(\mathrm{n}+1)}{2}$
2 $\frac{\mathrm{n}+1}{2}$
3 $\frac{\mathrm{n}+1}{2 \mathrm{n}}$
4 $\frac{2 \mathrm{n}}{\mathrm{n}+1}$
Explanation:
A Given, Number of wire $=\mathrm{n}$ Resistivity are given as, $\rho_{1}=1, \rho_{2}=2, \rho_{3}=3, \rho_{4}=4 \ldots \ldots \ldots . \rho_{n}=n$ We know that, $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}$ $\therefore \mathrm{R}_{1}=\frac{\rho_{\mathrm{l}} l}{\mathrm{~A}}$ $\mathrm{R}_{2}=\frac{\rho_{1} l}{\mathrm{~A}}, \mathrm{R}_{3}=\frac{\rho_{3} l}{\mathrm{~A}}, \mathrm{R}_{\mathrm{n}}=\frac{\rho_{\mathrm{n}} l}{\mathrm{~A}}$ Since, all resistance are joined in series. $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}+\ldots \ldots \ldots \ldots+\mathrm{R}_{\mathrm{n}}$ $\rho_{\mathrm{eq}} \frac{l}{\mathrm{~A}}=\frac{l}{\mathrm{~A}}\left(\rho_{1}+\rho_{2}+\rho_{3}+\ldots \ldots \ldots \rho_{\mathrm{n}}\right)$ $\rho_{\text {eq }}=(1+2+3+4+\ldots \ldots \ldots \ldots+\mathrm{n})$ $\rho_{\text {eq }}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}$
EAMCET -2000
Current Electricity
152169
An aluminum (resistivity $\rho=2.2 \times 10^{-8} \Omega-\mathrm{m}$ ) wire of a diameter $1.4 \mathrm{~mm}$ is used to make a $4 \Omega$ resistor. The length of the wire is
152166
Two wires of equal diameters of resistivities $\rho_{1}$ and $\rho_{2}$ and lengths $x_{1}$ and $x_{2}$ respectively are joined in series. The equivalent resistivity of the combination is
A Given, First wire resistivity $=\rho_{1}$ Second wire resistivity $=\rho_{2}$ Length of first wire $=x_{1}$ Length of second wire $=\mathrm{x}_{2}$ Diameter of both wire are equal So, $\quad A_{1}=A_{2}=A$ We know that, $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ For first wire - $R_{1}=\frac{\rho_{1} x_{1}}{A}$ For second wire - $R_{2}=\frac{\rho_{2} x_{2}}{A}$ Since, both resistance are joined in series, Hence, $R_{\text {eq }}=R_{1}+R_{2}$ $R_{\text {eq }}=\frac{1}{A}\left(\rho_{1} x_{1}+\rho_{2} x_{2}\right)$ $\rho_{\text {eq }} \cdot \frac{\left(x_{1}+x_{2}\right)}{A}=\frac{1}{A}\left(\rho_{1} x_{1}+\rho_{2} x_{2}\right)$ $\rho_{\text {eq }}=\frac{\rho_{1} x_{1}+\rho_{2} x_{2}}{x_{1}+x_{2}}$
EAMCET-2000
Current Electricity
152167
A nichrome wire $50 \mathrm{~cm}$ long and $1 \mathrm{~mm}^{2}$ crosssection carries a current of 4 A. When connected to $2 \mathrm{~V}$ battery. The resistivity of nichrome wire in $\Omega-m$ is
1 $1 \times 10^{-6}$
2 $4 \times 10^{-7}$
3 $3 \times 10^{-7}$
4 $2 \times 10^{-7}$
Explanation:
A Given, Length of wire $(l)=50 \mathrm{~cm}=0.5 \mathrm{~m}$ Area of wire $(\mathrm{A})=1 \mathrm{~mm}^{2}=1 \times 10^{-6} \mathrm{~m}^{2}$ Voltage $(\mathrm{V})=2 \mathrm{~V}$ Current $(\mathrm{I})=4 \mathrm{~A}$ We know that, $\mathrm{V}=\mathrm{IR}$ $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{2}{4}$ $\mathrm{R}=0.5 \Omega$ $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ $\rho=\frac{\mathrm{AR}}{l}$ $\rho=\frac{1 \times 10^{-6} \times 0.5}{0.5}$ $\rho=1 \times 10^{-6} \Omega \mathrm{m} .$ $\because \quad \mathrm{R}=\rho \frac{l}{\mathrm{~A}}$
EAMCET-2001
Current Electricity
152168
n conducting wires of same dimensions but having resistivities $1,2,3$, connected in series. The equivalent resistivity of the combination is
1 $\frac{\mathrm{n}(\mathrm{n}+1)}{2}$
2 $\frac{\mathrm{n}+1}{2}$
3 $\frac{\mathrm{n}+1}{2 \mathrm{n}}$
4 $\frac{2 \mathrm{n}}{\mathrm{n}+1}$
Explanation:
A Given, Number of wire $=\mathrm{n}$ Resistivity are given as, $\rho_{1}=1, \rho_{2}=2, \rho_{3}=3, \rho_{4}=4 \ldots \ldots \ldots . \rho_{n}=n$ We know that, $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}$ $\therefore \mathrm{R}_{1}=\frac{\rho_{\mathrm{l}} l}{\mathrm{~A}}$ $\mathrm{R}_{2}=\frac{\rho_{1} l}{\mathrm{~A}}, \mathrm{R}_{3}=\frac{\rho_{3} l}{\mathrm{~A}}, \mathrm{R}_{\mathrm{n}}=\frac{\rho_{\mathrm{n}} l}{\mathrm{~A}}$ Since, all resistance are joined in series. $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}+\ldots \ldots \ldots \ldots+\mathrm{R}_{\mathrm{n}}$ $\rho_{\mathrm{eq}} \frac{l}{\mathrm{~A}}=\frac{l}{\mathrm{~A}}\left(\rho_{1}+\rho_{2}+\rho_{3}+\ldots \ldots \ldots \rho_{\mathrm{n}}\right)$ $\rho_{\text {eq }}=(1+2+3+4+\ldots \ldots \ldots \ldots+\mathrm{n})$ $\rho_{\text {eq }}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}$
EAMCET -2000
Current Electricity
152169
An aluminum (resistivity $\rho=2.2 \times 10^{-8} \Omega-\mathrm{m}$ ) wire of a diameter $1.4 \mathrm{~mm}$ is used to make a $4 \Omega$ resistor. The length of the wire is
152166
Two wires of equal diameters of resistivities $\rho_{1}$ and $\rho_{2}$ and lengths $x_{1}$ and $x_{2}$ respectively are joined in series. The equivalent resistivity of the combination is
A Given, First wire resistivity $=\rho_{1}$ Second wire resistivity $=\rho_{2}$ Length of first wire $=x_{1}$ Length of second wire $=\mathrm{x}_{2}$ Diameter of both wire are equal So, $\quad A_{1}=A_{2}=A$ We know that, $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ For first wire - $R_{1}=\frac{\rho_{1} x_{1}}{A}$ For second wire - $R_{2}=\frac{\rho_{2} x_{2}}{A}$ Since, both resistance are joined in series, Hence, $R_{\text {eq }}=R_{1}+R_{2}$ $R_{\text {eq }}=\frac{1}{A}\left(\rho_{1} x_{1}+\rho_{2} x_{2}\right)$ $\rho_{\text {eq }} \cdot \frac{\left(x_{1}+x_{2}\right)}{A}=\frac{1}{A}\left(\rho_{1} x_{1}+\rho_{2} x_{2}\right)$ $\rho_{\text {eq }}=\frac{\rho_{1} x_{1}+\rho_{2} x_{2}}{x_{1}+x_{2}}$
EAMCET-2000
Current Electricity
152167
A nichrome wire $50 \mathrm{~cm}$ long and $1 \mathrm{~mm}^{2}$ crosssection carries a current of 4 A. When connected to $2 \mathrm{~V}$ battery. The resistivity of nichrome wire in $\Omega-m$ is
1 $1 \times 10^{-6}$
2 $4 \times 10^{-7}$
3 $3 \times 10^{-7}$
4 $2 \times 10^{-7}$
Explanation:
A Given, Length of wire $(l)=50 \mathrm{~cm}=0.5 \mathrm{~m}$ Area of wire $(\mathrm{A})=1 \mathrm{~mm}^{2}=1 \times 10^{-6} \mathrm{~m}^{2}$ Voltage $(\mathrm{V})=2 \mathrm{~V}$ Current $(\mathrm{I})=4 \mathrm{~A}$ We know that, $\mathrm{V}=\mathrm{IR}$ $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{2}{4}$ $\mathrm{R}=0.5 \Omega$ $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ $\rho=\frac{\mathrm{AR}}{l}$ $\rho=\frac{1 \times 10^{-6} \times 0.5}{0.5}$ $\rho=1 \times 10^{-6} \Omega \mathrm{m} .$ $\because \quad \mathrm{R}=\rho \frac{l}{\mathrm{~A}}$
EAMCET-2001
Current Electricity
152168
n conducting wires of same dimensions but having resistivities $1,2,3$, connected in series. The equivalent resistivity of the combination is
1 $\frac{\mathrm{n}(\mathrm{n}+1)}{2}$
2 $\frac{\mathrm{n}+1}{2}$
3 $\frac{\mathrm{n}+1}{2 \mathrm{n}}$
4 $\frac{2 \mathrm{n}}{\mathrm{n}+1}$
Explanation:
A Given, Number of wire $=\mathrm{n}$ Resistivity are given as, $\rho_{1}=1, \rho_{2}=2, \rho_{3}=3, \rho_{4}=4 \ldots \ldots \ldots . \rho_{n}=n$ We know that, $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}$ $\therefore \mathrm{R}_{1}=\frac{\rho_{\mathrm{l}} l}{\mathrm{~A}}$ $\mathrm{R}_{2}=\frac{\rho_{1} l}{\mathrm{~A}}, \mathrm{R}_{3}=\frac{\rho_{3} l}{\mathrm{~A}}, \mathrm{R}_{\mathrm{n}}=\frac{\rho_{\mathrm{n}} l}{\mathrm{~A}}$ Since, all resistance are joined in series. $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}+\ldots \ldots \ldots \ldots+\mathrm{R}_{\mathrm{n}}$ $\rho_{\mathrm{eq}} \frac{l}{\mathrm{~A}}=\frac{l}{\mathrm{~A}}\left(\rho_{1}+\rho_{2}+\rho_{3}+\ldots \ldots \ldots \rho_{\mathrm{n}}\right)$ $\rho_{\text {eq }}=(1+2+3+4+\ldots \ldots \ldots \ldots+\mathrm{n})$ $\rho_{\text {eq }}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}$
EAMCET -2000
Current Electricity
152169
An aluminum (resistivity $\rho=2.2 \times 10^{-8} \Omega-\mathrm{m}$ ) wire of a diameter $1.4 \mathrm{~mm}$ is used to make a $4 \Omega$ resistor. The length of the wire is