QBManager

Current Electricity

151965 A wire of length $L$ is drawn such that its diameter is reduced to half of its original diameter. If the initial resistance of the wire were $10 \Omega$, its new resistance would be

1 $40 \Omega$

2 $80 \Omega$

3 $120 \Omega$

4 $160 \Omega$

Explanation:

D Given,
$V_1 \longrightarrow \frac{1}{2}, D_1 r=10 \Omega$
$V_1 \longrightarrow R^{\prime}$
Resistance, $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$
$\mathrm{R}^{\prime} =\rho \frac{l^{\prime}}{\mathrm{A}^{\prime}}$
$\mathrm{V}_1 =\mathrm{A} l=\frac{\pi \mathrm{D}^2}{4} l$
$\mathrm{~V}_2 =\mathrm{A}^{\prime} \times l^{\prime}$
$=\frac{\pi \mathrm{D}^2}{16} \times l^{\prime}$
$\mathrm{V}_1 =\mathrm{V}_2$
$\frac{\pi \mathrm{D}^2}{4} l=\frac{\pi \mathrm{D}^2}{16} l^{\prime}$
$l =\frac{l^{\prime}}{4}$
$\frac{l}{l^{\prime}} =\frac{1}{4}$
$\frac{\mathrm{R}^{\prime}}{\mathrm{R}} =\frac{l^{\prime}}{\mathrm{A}^{\prime}} \times \frac{\mathrm{A}}{l}$
$\frac{\mathrm{R}^{\prime}}{10} =\frac{l^{\prime}}{l} \times \frac{\mathrm{A}}{\mathrm{A}^{\prime}}$
$\frac{\mathrm{R}^{\prime}}{10} \left.=4 \times\left(\frac{\pi \mathrm{D}^2}{4}\right)=4 \times \frac{\pi \mathrm{D}^2}{16}\right)=16$
$\frac{\mathrm{R}^{\prime}}{10} =16, \quad \frac{\pi \mathrm{D}^2}{16}=160 \Omega$
Resistance, $\mathrm{R}=\rho \frac{\mathrm{l}}{\mathrm{A}}$
$\mathrm{R}^{\prime}=\rho \frac{l^{\prime}}{\mathrm{A}^{\prime}}$
$\mathrm{V}_{1}=\mathrm{A} l=\frac{\pi \mathrm{D}^{2}}{4} l$
$\mathrm{~V}_{2}=\mathrm{A}^{\prime} \times l^{\prime}$
$=\frac{\pi \mathrm{D}^{2}}{16} \times l^{\prime}$
$\mathrm{V}_{1}=\mathrm{V}_{2}$
$\frac{\pi \mathrm{D}^{2}}{4} l=\frac{\pi \mathrm{D}^{2}}{16} l^{\prime}$
$l=\frac{l^{\prime}}{4}$
$\frac{l^{\prime}}{l^{\prime}}=\frac{1}{4}$
$\frac{\mathrm{R}^{\prime}}{\mathrm{R}}=\frac{l^{\prime}}{\mathrm{A}^{\prime}} \times \frac{\mathrm{A}}{l}$
$\frac{\mathrm{R}^{\prime}}{10}=\frac{l^{\prime}}{l} \times \frac{\mathrm{A}}{\mathrm{A}^{\prime}}$
$\frac{\mathrm{R}^{\prime}}{10}=4 \times\left(\frac{\frac{\pi \mathrm{D}^{2}}{4}}{\frac{\pi \mathrm{D}^{2}}{16}}\right)=4 \times \frac{\pi \mathrm{D}^{2} \times 16}{4 \times \pi \mathrm{D}^{2}}$
$\frac{\mathrm{R}^{\prime}}{10}=16, \quad \mathrm{R}^{\prime}=160 \Omega$
$\frac{\mathrm{A}}{10}$

AIIMS-2003

Current Electricity

151966 A wire of radius $r$ has resistance $R$. If it is stretched to a radius $\frac{r}{2}$, its resistance will be:

1 $16 \mathrm{R}$

2 $2 \mathrm{R}$

3 $4 \mathrm{R}$

4 zero

J and K CET-2003

Current Electricity

151967 The ratio of the coefficient of thermal conductivity of two different materials is $5: 3$. If the thermal resistance of the rods of same thickness of these materials is same, then the ratio of the length of these rods will be-

1 $3: 5$

2 $5: 3$

3 $3: 4$

4 $3: 2$

BCECE-2007

Current Electricity

151969 The resistance of a bulb filament is $100 \Omega$ at a temperature of $100^{\circ} \mathrm{C}$. If temperature coefficient of resistance be $0.005 /{ }^{0} \mathrm{C}$, its resistance will become $200 \Omega$ at a temperature of:

1 $300^{\circ} \mathrm{C}$

2 $400^{\circ} \mathrm{C}$

3 $500^{\circ} \mathrm{C}$

4 $200^{\circ} \mathrm{C}$

BCECE-2004

Current Electricity

151970 Steel and aluminium wires have equal resistances and masses. Which of the wires is longer and how many times? (Given, densities of steel and aluminium are $7.8 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ and $2.7 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ and their resistivities are 0.15 $\mu \Omega-m$ and $0.028 \mu \Omega-m$ respectively)

1 The aluminium wire is 3.9 times longer

2 The aluminium wire is 1.3 times longer

3 The aluminium wire is 2.6 times longer

4 The steel wire is 3.9 times longer

Explanation:

A Let length \& Area of steel $=\mathrm{L}_{\mathrm{s}} \& \mathrm{~A}_{\mathrm{s}}$ Length \& Area of Aluminium $=L_{a} \& A_{a}$
$\because \quad$ Mass of steel & Aluminium is equal
$\therefore \quad \rho_{\mathrm{s}} \mathrm{L}_{\mathrm{s}} \mathrm{A}_{\mathrm{s}}=\rho_{\mathrm{a}} \mathrm{L}_{\mathrm{a}} \mathrm{A}_{\mathrm{a}}$
$\rho_{\mathrm{s}} \& \rho_{\mathrm{a}}$ are density of steel & aluminium
Given,
$\rho_{\mathrm{s}}=7.8 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$
$\rho_{\mathrm{a}}=2.7 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$
$7.8 \times 10^{3} \mathrm{~L}_{\mathrm{s}} \mathrm{A}_{\mathrm{s}}=2.7 \times 10^{3} \mathrm{~L}_{\mathrm{a}} \mathrm{A}_{\mathrm{a}}$
$\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}} \times \frac{\mathrm{A}_{\mathrm{s}}}{\mathrm{A}_{\mathrm{a}}}=\frac{2.7}{7.8}$
Resistivity of steel $\left(\rho_{\mathrm{s}}\right)=0.15 \mu \Omega-\mathrm{m}$
Resistivity of aluminium $\left(\rho_{\mathrm{Al}}\right)=0.028 \mu \Omega-\mathrm{m}$
$\therefore \quad \rho_{\mathrm{s}} \frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{A}_{\mathrm{s}}}=\rho_{\mathrm{a}} \frac{\mathrm{L}_{\mathrm{a}}}{\mathrm{A}_{\mathrm{a}}}$
$0.15 \frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{A}_{\mathrm{s}}}=0.028 \frac{\mathrm{L}_{\mathrm{a}}}{\mathrm{A}_{\mathrm{a}}}$
$\therefore \quad \frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}} \times \frac{\mathrm{A}_{\mathrm{a}}}{\mathrm{A}_{\mathrm{s}}}=\frac{0.028}{0.15}$
From equation (i)
$\frac{\mathrm{A}_{\mathrm{a}}}{\mathrm{A}_{\mathrm{s}}}=\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}} \times \frac{7.8}{2.7}$
Putting this value in equation (ii)
$\frac{\mathrm{L}_{s}}{\mathrm{~L}_{a}} \times \frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}} \times \frac{7.8}{2.7}=\frac{0.028}{0.15}$
$\left(\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}}\right)^{2}=\frac{0.028}{0.15} \times \frac{2.7}{7.8}$
$\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}}=\sqrt{\frac{0.028 \times 2.7}{0.15 \times 7.8}}$
$\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}}=0.25419$
$\mathrm{~L}_{\mathrm{a}}=\frac{\mathrm{L}_{\mathrm{s}}}{0.25419}$
$\mathrm{~L}_{\mathrm{a}}=3.93 \mathrm{~L}_{\mathrm{s}}$

BCECE-2010

Current Electricity

151965 A wire of length $L$ is drawn such that its diameter is reduced to half of its original diameter. If the initial resistance of the wire were $10 \Omega$, its new resistance would be

1 $40 \Omega$

2 $80 \Omega$

3 $120 \Omega$

4 $160 \Omega$

Explanation:

D Given,
$V_1 \longrightarrow \frac{1}{2}, D_1 r=10 \Omega$
$V_1 \longrightarrow R^{\prime}$
Resistance, $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$
$\mathrm{R}^{\prime} =\rho \frac{l^{\prime}}{\mathrm{A}^{\prime}}$
$\mathrm{V}_1 =\mathrm{A} l=\frac{\pi \mathrm{D}^2}{4} l$
$\mathrm{~V}_2 =\mathrm{A}^{\prime} \times l^{\prime}$
$=\frac{\pi \mathrm{D}^2}{16} \times l^{\prime}$
$\mathrm{V}_1 =\mathrm{V}_2$
$\frac{\pi \mathrm{D}^2}{4} l=\frac{\pi \mathrm{D}^2}{16} l^{\prime}$
$l =\frac{l^{\prime}}{4}$
$\frac{l}{l^{\prime}} =\frac{1}{4}$
$\frac{\mathrm{R}^{\prime}}{\mathrm{R}} =\frac{l^{\prime}}{\mathrm{A}^{\prime}} \times \frac{\mathrm{A}}{l}$
$\frac{\mathrm{R}^{\prime}}{10} =\frac{l^{\prime}}{l} \times \frac{\mathrm{A}}{\mathrm{A}^{\prime}}$
$\frac{\mathrm{R}^{\prime}}{10} \left.=4 \times\left(\frac{\pi \mathrm{D}^2}{4}\right)=4 \times \frac{\pi \mathrm{D}^2}{16}\right)=16$
$\frac{\mathrm{R}^{\prime}}{10} =16, \quad \frac{\pi \mathrm{D}^2}{16}=160 \Omega$
Resistance, $\mathrm{R}=\rho \frac{\mathrm{l}}{\mathrm{A}}$
$\mathrm{R}^{\prime}=\rho \frac{l^{\prime}}{\mathrm{A}^{\prime}}$
$\mathrm{V}_{1}=\mathrm{A} l=\frac{\pi \mathrm{D}^{2}}{4} l$
$\mathrm{~V}_{2}=\mathrm{A}^{\prime} \times l^{\prime}$
$=\frac{\pi \mathrm{D}^{2}}{16} \times l^{\prime}$
$\mathrm{V}_{1}=\mathrm{V}_{2}$
$\frac{\pi \mathrm{D}^{2}}{4} l=\frac{\pi \mathrm{D}^{2}}{16} l^{\prime}$
$l=\frac{l^{\prime}}{4}$
$\frac{l^{\prime}}{l^{\prime}}=\frac{1}{4}$
$\frac{\mathrm{R}^{\prime}}{\mathrm{R}}=\frac{l^{\prime}}{\mathrm{A}^{\prime}} \times \frac{\mathrm{A}}{l}$
$\frac{\mathrm{R}^{\prime}}{10}=\frac{l^{\prime}}{l} \times \frac{\mathrm{A}}{\mathrm{A}^{\prime}}$
$\frac{\mathrm{R}^{\prime}}{10}=4 \times\left(\frac{\frac{\pi \mathrm{D}^{2}}{4}}{\frac{\pi \mathrm{D}^{2}}{16}}\right)=4 \times \frac{\pi \mathrm{D}^{2} \times 16}{4 \times \pi \mathrm{D}^{2}}$
$\frac{\mathrm{R}^{\prime}}{10}=16, \quad \mathrm{R}^{\prime}=160 \Omega$
$\frac{\mathrm{A}}{10}$

AIIMS-2003

Current Electricity

151966 A wire of radius $r$ has resistance $R$. If it is stretched to a radius $\frac{r}{2}$, its resistance will be:

1 $16 \mathrm{R}$

2 $2 \mathrm{R}$

3 $4 \mathrm{R}$

4 zero

J and K CET-2003

Current Electricity

151967 The ratio of the coefficient of thermal conductivity of two different materials is $5: 3$. If the thermal resistance of the rods of same thickness of these materials is same, then the ratio of the length of these rods will be-

1 $3: 5$

2 $5: 3$

3 $3: 4$

4 $3: 2$

BCECE-2007

Current Electricity

151969 The resistance of a bulb filament is $100 \Omega$ at a temperature of $100^{\circ} \mathrm{C}$. If temperature coefficient of resistance be $0.005 /{ }^{0} \mathrm{C}$, its resistance will become $200 \Omega$ at a temperature of:

1 $300^{\circ} \mathrm{C}$

2 $400^{\circ} \mathrm{C}$

3 $500^{\circ} \mathrm{C}$

4 $200^{\circ} \mathrm{C}$

BCECE-2004

Current Electricity

151970 Steel and aluminium wires have equal resistances and masses. Which of the wires is longer and how many times? (Given, densities of steel and aluminium are $7.8 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ and $2.7 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ and their resistivities are 0.15 $\mu \Omega-m$ and $0.028 \mu \Omega-m$ respectively)

1 The aluminium wire is 3.9 times longer

2 The aluminium wire is 1.3 times longer

3 The aluminium wire is 2.6 times longer

4 The steel wire is 3.9 times longer

Explanation:

A Let length \& Area of steel $=\mathrm{L}_{\mathrm{s}} \& \mathrm{~A}_{\mathrm{s}}$ Length \& Area of Aluminium $=L_{a} \& A_{a}$
$\because \quad$ Mass of steel & Aluminium is equal
$\therefore \quad \rho_{\mathrm{s}} \mathrm{L}_{\mathrm{s}} \mathrm{A}_{\mathrm{s}}=\rho_{\mathrm{a}} \mathrm{L}_{\mathrm{a}} \mathrm{A}_{\mathrm{a}}$
$\rho_{\mathrm{s}} \& \rho_{\mathrm{a}}$ are density of steel & aluminium
Given,
$\rho_{\mathrm{s}}=7.8 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$
$\rho_{\mathrm{a}}=2.7 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$
$7.8 \times 10^{3} \mathrm{~L}_{\mathrm{s}} \mathrm{A}_{\mathrm{s}}=2.7 \times 10^{3} \mathrm{~L}_{\mathrm{a}} \mathrm{A}_{\mathrm{a}}$
$\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}} \times \frac{\mathrm{A}_{\mathrm{s}}}{\mathrm{A}_{\mathrm{a}}}=\frac{2.7}{7.8}$
Resistivity of steel $\left(\rho_{\mathrm{s}}\right)=0.15 \mu \Omega-\mathrm{m}$
Resistivity of aluminium $\left(\rho_{\mathrm{Al}}\right)=0.028 \mu \Omega-\mathrm{m}$
$\therefore \quad \rho_{\mathrm{s}} \frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{A}_{\mathrm{s}}}=\rho_{\mathrm{a}} \frac{\mathrm{L}_{\mathrm{a}}}{\mathrm{A}_{\mathrm{a}}}$
$0.15 \frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{A}_{\mathrm{s}}}=0.028 \frac{\mathrm{L}_{\mathrm{a}}}{\mathrm{A}_{\mathrm{a}}}$
$\therefore \quad \frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}} \times \frac{\mathrm{A}_{\mathrm{a}}}{\mathrm{A}_{\mathrm{s}}}=\frac{0.028}{0.15}$
From equation (i)
$\frac{\mathrm{A}_{\mathrm{a}}}{\mathrm{A}_{\mathrm{s}}}=\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}} \times \frac{7.8}{2.7}$
Putting this value in equation (ii)
$\frac{\mathrm{L}_{s}}{\mathrm{~L}_{a}} \times \frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}} \times \frac{7.8}{2.7}=\frac{0.028}{0.15}$
$\left(\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}}\right)^{2}=\frac{0.028}{0.15} \times \frac{2.7}{7.8}$
$\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}}=\sqrt{\frac{0.028 \times 2.7}{0.15 \times 7.8}}$
$\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}}=0.25419$
$\mathrm{~L}_{\mathrm{a}}=\frac{\mathrm{L}_{\mathrm{s}}}{0.25419}$
$\mathrm{~L}_{\mathrm{a}}=3.93 \mathrm{~L}_{\mathrm{s}}$

BCECE-2010

Current Electricity

151965 A wire of length $L$ is drawn such that its diameter is reduced to half of its original diameter. If the initial resistance of the wire were $10 \Omega$, its new resistance would be

1 $40 \Omega$

2 $80 \Omega$

3 $120 \Omega$

4 $160 \Omega$

Explanation:

D Given,
$V_1 \longrightarrow \frac{1}{2}, D_1 r=10 \Omega$
$V_1 \longrightarrow R^{\prime}$
Resistance, $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$
$\mathrm{R}^{\prime} =\rho \frac{l^{\prime}}{\mathrm{A}^{\prime}}$
$\mathrm{V}_1 =\mathrm{A} l=\frac{\pi \mathrm{D}^2}{4} l$
$\mathrm{~V}_2 =\mathrm{A}^{\prime} \times l^{\prime}$
$=\frac{\pi \mathrm{D}^2}{16} \times l^{\prime}$
$\mathrm{V}_1 =\mathrm{V}_2$
$\frac{\pi \mathrm{D}^2}{4} l=\frac{\pi \mathrm{D}^2}{16} l^{\prime}$
$l =\frac{l^{\prime}}{4}$
$\frac{l}{l^{\prime}} =\frac{1}{4}$
$\frac{\mathrm{R}^{\prime}}{\mathrm{R}} =\frac{l^{\prime}}{\mathrm{A}^{\prime}} \times \frac{\mathrm{A}}{l}$
$\frac{\mathrm{R}^{\prime}}{10} =\frac{l^{\prime}}{l} \times \frac{\mathrm{A}}{\mathrm{A}^{\prime}}$
$\frac{\mathrm{R}^{\prime}}{10} \left.=4 \times\left(\frac{\pi \mathrm{D}^2}{4}\right)=4 \times \frac{\pi \mathrm{D}^2}{16}\right)=16$
$\frac{\mathrm{R}^{\prime}}{10} =16, \quad \frac{\pi \mathrm{D}^2}{16}=160 \Omega$
Resistance, $\mathrm{R}=\rho \frac{\mathrm{l}}{\mathrm{A}}$
$\mathrm{R}^{\prime}=\rho \frac{l^{\prime}}{\mathrm{A}^{\prime}}$
$\mathrm{V}_{1}=\mathrm{A} l=\frac{\pi \mathrm{D}^{2}}{4} l$
$\mathrm{~V}_{2}=\mathrm{A}^{\prime} \times l^{\prime}$
$=\frac{\pi \mathrm{D}^{2}}{16} \times l^{\prime}$
$\mathrm{V}_{1}=\mathrm{V}_{2}$
$\frac{\pi \mathrm{D}^{2}}{4} l=\frac{\pi \mathrm{D}^{2}}{16} l^{\prime}$
$l=\frac{l^{\prime}}{4}$
$\frac{l^{\prime}}{l^{\prime}}=\frac{1}{4}$
$\frac{\mathrm{R}^{\prime}}{\mathrm{R}}=\frac{l^{\prime}}{\mathrm{A}^{\prime}} \times \frac{\mathrm{A}}{l}$
$\frac{\mathrm{R}^{\prime}}{10}=\frac{l^{\prime}}{l} \times \frac{\mathrm{A}}{\mathrm{A}^{\prime}}$
$\frac{\mathrm{R}^{\prime}}{10}=4 \times\left(\frac{\frac{\pi \mathrm{D}^{2}}{4}}{\frac{\pi \mathrm{D}^{2}}{16}}\right)=4 \times \frac{\pi \mathrm{D}^{2} \times 16}{4 \times \pi \mathrm{D}^{2}}$
$\frac{\mathrm{R}^{\prime}}{10}=16, \quad \mathrm{R}^{\prime}=160 \Omega$
$\frac{\mathrm{A}}{10}$

AIIMS-2003

Current Electricity

151966 A wire of radius $r$ has resistance $R$. If it is stretched to a radius $\frac{r}{2}$, its resistance will be:

1 $16 \mathrm{R}$

2 $2 \mathrm{R}$

3 $4 \mathrm{R}$

4 zero

J and K CET-2003

Current Electricity

151967 The ratio of the coefficient of thermal conductivity of two different materials is $5: 3$. If the thermal resistance of the rods of same thickness of these materials is same, then the ratio of the length of these rods will be-

1 $3: 5$

2 $5: 3$

3 $3: 4$

4 $3: 2$

BCECE-2007

Current Electricity

151969 The resistance of a bulb filament is $100 \Omega$ at a temperature of $100^{\circ} \mathrm{C}$. If temperature coefficient of resistance be $0.005 /{ }^{0} \mathrm{C}$, its resistance will become $200 \Omega$ at a temperature of:

1 $300^{\circ} \mathrm{C}$

2 $400^{\circ} \mathrm{C}$

3 $500^{\circ} \mathrm{C}$

4 $200^{\circ} \mathrm{C}$

BCECE-2004

Current Electricity

151970 Steel and aluminium wires have equal resistances and masses. Which of the wires is longer and how many times? (Given, densities of steel and aluminium are $7.8 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ and $2.7 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ and their resistivities are 0.15 $\mu \Omega-m$ and $0.028 \mu \Omega-m$ respectively)

1 The aluminium wire is 3.9 times longer

2 The aluminium wire is 1.3 times longer

3 The aluminium wire is 2.6 times longer

4 The steel wire is 3.9 times longer

Explanation:

A Let length \& Area of steel $=\mathrm{L}_{\mathrm{s}} \& \mathrm{~A}_{\mathrm{s}}$ Length \& Area of Aluminium $=L_{a} \& A_{a}$
$\because \quad$ Mass of steel & Aluminium is equal
$\therefore \quad \rho_{\mathrm{s}} \mathrm{L}_{\mathrm{s}} \mathrm{A}_{\mathrm{s}}=\rho_{\mathrm{a}} \mathrm{L}_{\mathrm{a}} \mathrm{A}_{\mathrm{a}}$
$\rho_{\mathrm{s}} \& \rho_{\mathrm{a}}$ are density of steel & aluminium
Given,
$\rho_{\mathrm{s}}=7.8 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$
$\rho_{\mathrm{a}}=2.7 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$
$7.8 \times 10^{3} \mathrm{~L}_{\mathrm{s}} \mathrm{A}_{\mathrm{s}}=2.7 \times 10^{3} \mathrm{~L}_{\mathrm{a}} \mathrm{A}_{\mathrm{a}}$
$\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}} \times \frac{\mathrm{A}_{\mathrm{s}}}{\mathrm{A}_{\mathrm{a}}}=\frac{2.7}{7.8}$
Resistivity of steel $\left(\rho_{\mathrm{s}}\right)=0.15 \mu \Omega-\mathrm{m}$
Resistivity of aluminium $\left(\rho_{\mathrm{Al}}\right)=0.028 \mu \Omega-\mathrm{m}$
$\therefore \quad \rho_{\mathrm{s}} \frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{A}_{\mathrm{s}}}=\rho_{\mathrm{a}} \frac{\mathrm{L}_{\mathrm{a}}}{\mathrm{A}_{\mathrm{a}}}$
$0.15 \frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{A}_{\mathrm{s}}}=0.028 \frac{\mathrm{L}_{\mathrm{a}}}{\mathrm{A}_{\mathrm{a}}}$
$\therefore \quad \frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}} \times \frac{\mathrm{A}_{\mathrm{a}}}{\mathrm{A}_{\mathrm{s}}}=\frac{0.028}{0.15}$
From equation (i)
$\frac{\mathrm{A}_{\mathrm{a}}}{\mathrm{A}_{\mathrm{s}}}=\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}} \times \frac{7.8}{2.7}$
Putting this value in equation (ii)
$\frac{\mathrm{L}_{s}}{\mathrm{~L}_{a}} \times \frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}} \times \frac{7.8}{2.7}=\frac{0.028}{0.15}$
$\left(\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}}\right)^{2}=\frac{0.028}{0.15} \times \frac{2.7}{7.8}$
$\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}}=\sqrt{\frac{0.028 \times 2.7}{0.15 \times 7.8}}$
$\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}}=0.25419$
$\mathrm{~L}_{\mathrm{a}}=\frac{\mathrm{L}_{\mathrm{s}}}{0.25419}$
$\mathrm{~L}_{\mathrm{a}}=3.93 \mathrm{~L}_{\mathrm{s}}$

BCECE-2010

Current Electricity

151965 A wire of length $L$ is drawn such that its diameter is reduced to half of its original diameter. If the initial resistance of the wire were $10 \Omega$, its new resistance would be

1 $40 \Omega$

2 $80 \Omega$

3 $120 \Omega$

4 $160 \Omega$

Explanation:

D Given,
$V_1 \longrightarrow \frac{1}{2}, D_1 r=10 \Omega$
$V_1 \longrightarrow R^{\prime}$
Resistance, $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$
$\mathrm{R}^{\prime} =\rho \frac{l^{\prime}}{\mathrm{A}^{\prime}}$
$\mathrm{V}_1 =\mathrm{A} l=\frac{\pi \mathrm{D}^2}{4} l$
$\mathrm{~V}_2 =\mathrm{A}^{\prime} \times l^{\prime}$
$=\frac{\pi \mathrm{D}^2}{16} \times l^{\prime}$
$\mathrm{V}_1 =\mathrm{V}_2$
$\frac{\pi \mathrm{D}^2}{4} l=\frac{\pi \mathrm{D}^2}{16} l^{\prime}$
$l =\frac{l^{\prime}}{4}$
$\frac{l}{l^{\prime}} =\frac{1}{4}$
$\frac{\mathrm{R}^{\prime}}{\mathrm{R}} =\frac{l^{\prime}}{\mathrm{A}^{\prime}} \times \frac{\mathrm{A}}{l}$
$\frac{\mathrm{R}^{\prime}}{10} =\frac{l^{\prime}}{l} \times \frac{\mathrm{A}}{\mathrm{A}^{\prime}}$
$\frac{\mathrm{R}^{\prime}}{10} \left.=4 \times\left(\frac{\pi \mathrm{D}^2}{4}\right)=4 \times \frac{\pi \mathrm{D}^2}{16}\right)=16$
$\frac{\mathrm{R}^{\prime}}{10} =16, \quad \frac{\pi \mathrm{D}^2}{16}=160 \Omega$
Resistance, $\mathrm{R}=\rho \frac{\mathrm{l}}{\mathrm{A}}$
$\mathrm{R}^{\prime}=\rho \frac{l^{\prime}}{\mathrm{A}^{\prime}}$
$\mathrm{V}_{1}=\mathrm{A} l=\frac{\pi \mathrm{D}^{2}}{4} l$
$\mathrm{~V}_{2}=\mathrm{A}^{\prime} \times l^{\prime}$
$=\frac{\pi \mathrm{D}^{2}}{16} \times l^{\prime}$
$\mathrm{V}_{1}=\mathrm{V}_{2}$
$\frac{\pi \mathrm{D}^{2}}{4} l=\frac{\pi \mathrm{D}^{2}}{16} l^{\prime}$
$l=\frac{l^{\prime}}{4}$
$\frac{l^{\prime}}{l^{\prime}}=\frac{1}{4}$
$\frac{\mathrm{R}^{\prime}}{\mathrm{R}}=\frac{l^{\prime}}{\mathrm{A}^{\prime}} \times \frac{\mathrm{A}}{l}$
$\frac{\mathrm{R}^{\prime}}{10}=\frac{l^{\prime}}{l} \times \frac{\mathrm{A}}{\mathrm{A}^{\prime}}$
$\frac{\mathrm{R}^{\prime}}{10}=4 \times\left(\frac{\frac{\pi \mathrm{D}^{2}}{4}}{\frac{\pi \mathrm{D}^{2}}{16}}\right)=4 \times \frac{\pi \mathrm{D}^{2} \times 16}{4 \times \pi \mathrm{D}^{2}}$
$\frac{\mathrm{R}^{\prime}}{10}=16, \quad \mathrm{R}^{\prime}=160 \Omega$
$\frac{\mathrm{A}}{10}$

AIIMS-2003

Current Electricity

151966 A wire of radius $r$ has resistance $R$. If it is stretched to a radius $\frac{r}{2}$, its resistance will be:

1 $16 \mathrm{R}$

2 $2 \mathrm{R}$

3 $4 \mathrm{R}$

4 zero

J and K CET-2003

Current Electricity

151967 The ratio of the coefficient of thermal conductivity of two different materials is $5: 3$. If the thermal resistance of the rods of same thickness of these materials is same, then the ratio of the length of these rods will be-

1 $3: 5$

2 $5: 3$

3 $3: 4$

4 $3: 2$

BCECE-2007

Current Electricity

151969 The resistance of a bulb filament is $100 \Omega$ at a temperature of $100^{\circ} \mathrm{C}$. If temperature coefficient of resistance be $0.005 /{ }^{0} \mathrm{C}$, its resistance will become $200 \Omega$ at a temperature of:

1 $300^{\circ} \mathrm{C}$

2 $400^{\circ} \mathrm{C}$

3 $500^{\circ} \mathrm{C}$

4 $200^{\circ} \mathrm{C}$

BCECE-2004

Current Electricity

151970 Steel and aluminium wires have equal resistances and masses. Which of the wires is longer and how many times? (Given, densities of steel and aluminium are $7.8 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ and $2.7 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ and their resistivities are 0.15 $\mu \Omega-m$ and $0.028 \mu \Omega-m$ respectively)

1 The aluminium wire is 3.9 times longer

2 The aluminium wire is 1.3 times longer

3 The aluminium wire is 2.6 times longer

4 The steel wire is 3.9 times longer

Explanation:

A Let length \& Area of steel $=\mathrm{L}_{\mathrm{s}} \& \mathrm{~A}_{\mathrm{s}}$ Length \& Area of Aluminium $=L_{a} \& A_{a}$
$\because \quad$ Mass of steel & Aluminium is equal
$\therefore \quad \rho_{\mathrm{s}} \mathrm{L}_{\mathrm{s}} \mathrm{A}_{\mathrm{s}}=\rho_{\mathrm{a}} \mathrm{L}_{\mathrm{a}} \mathrm{A}_{\mathrm{a}}$
$\rho_{\mathrm{s}} \& \rho_{\mathrm{a}}$ are density of steel & aluminium
Given,
$\rho_{\mathrm{s}}=7.8 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$
$\rho_{\mathrm{a}}=2.7 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$
$7.8 \times 10^{3} \mathrm{~L}_{\mathrm{s}} \mathrm{A}_{\mathrm{s}}=2.7 \times 10^{3} \mathrm{~L}_{\mathrm{a}} \mathrm{A}_{\mathrm{a}}$
$\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}} \times \frac{\mathrm{A}_{\mathrm{s}}}{\mathrm{A}_{\mathrm{a}}}=\frac{2.7}{7.8}$
Resistivity of steel $\left(\rho_{\mathrm{s}}\right)=0.15 \mu \Omega-\mathrm{m}$
Resistivity of aluminium $\left(\rho_{\mathrm{Al}}\right)=0.028 \mu \Omega-\mathrm{m}$
$\therefore \quad \rho_{\mathrm{s}} \frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{A}_{\mathrm{s}}}=\rho_{\mathrm{a}} \frac{\mathrm{L}_{\mathrm{a}}}{\mathrm{A}_{\mathrm{a}}}$
$0.15 \frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{A}_{\mathrm{s}}}=0.028 \frac{\mathrm{L}_{\mathrm{a}}}{\mathrm{A}_{\mathrm{a}}}$
$\therefore \quad \frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}} \times \frac{\mathrm{A}_{\mathrm{a}}}{\mathrm{A}_{\mathrm{s}}}=\frac{0.028}{0.15}$
From equation (i)
$\frac{\mathrm{A}_{\mathrm{a}}}{\mathrm{A}_{\mathrm{s}}}=\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}} \times \frac{7.8}{2.7}$
Putting this value in equation (ii)
$\frac{\mathrm{L}_{s}}{\mathrm{~L}_{a}} \times \frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}} \times \frac{7.8}{2.7}=\frac{0.028}{0.15}$
$\left(\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}}\right)^{2}=\frac{0.028}{0.15} \times \frac{2.7}{7.8}$
$\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}}=\sqrt{\frac{0.028 \times 2.7}{0.15 \times 7.8}}$
$\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}}=0.25419$
$\mathrm{~L}_{\mathrm{a}}=\frac{\mathrm{L}_{\mathrm{s}}}{0.25419}$
$\mathrm{~L}_{\mathrm{a}}=3.93 \mathrm{~L}_{\mathrm{s}}$

BCECE-2010

Current Electricity

151965 A wire of length $L$ is drawn such that its diameter is reduced to half of its original diameter. If the initial resistance of the wire were $10 \Omega$, its new resistance would be

1 $40 \Omega$

2 $80 \Omega$

3 $120 \Omega$

4 $160 \Omega$

Explanation:

D Given,
$V_1 \longrightarrow \frac{1}{2}, D_1 r=10 \Omega$
$V_1 \longrightarrow R^{\prime}$
Resistance, $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$
$\mathrm{R}^{\prime} =\rho \frac{l^{\prime}}{\mathrm{A}^{\prime}}$
$\mathrm{V}_1 =\mathrm{A} l=\frac{\pi \mathrm{D}^2}{4} l$
$\mathrm{~V}_2 =\mathrm{A}^{\prime} \times l^{\prime}$
$=\frac{\pi \mathrm{D}^2}{16} \times l^{\prime}$
$\mathrm{V}_1 =\mathrm{V}_2$
$\frac{\pi \mathrm{D}^2}{4} l=\frac{\pi \mathrm{D}^2}{16} l^{\prime}$
$l =\frac{l^{\prime}}{4}$
$\frac{l}{l^{\prime}} =\frac{1}{4}$
$\frac{\mathrm{R}^{\prime}}{\mathrm{R}} =\frac{l^{\prime}}{\mathrm{A}^{\prime}} \times \frac{\mathrm{A}}{l}$
$\frac{\mathrm{R}^{\prime}}{10} =\frac{l^{\prime}}{l} \times \frac{\mathrm{A}}{\mathrm{A}^{\prime}}$
$\frac{\mathrm{R}^{\prime}}{10} \left.=4 \times\left(\frac{\pi \mathrm{D}^2}{4}\right)=4 \times \frac{\pi \mathrm{D}^2}{16}\right)=16$
$\frac{\mathrm{R}^{\prime}}{10} =16, \quad \frac{\pi \mathrm{D}^2}{16}=160 \Omega$
Resistance, $\mathrm{R}=\rho \frac{\mathrm{l}}{\mathrm{A}}$
$\mathrm{R}^{\prime}=\rho \frac{l^{\prime}}{\mathrm{A}^{\prime}}$
$\mathrm{V}_{1}=\mathrm{A} l=\frac{\pi \mathrm{D}^{2}}{4} l$
$\mathrm{~V}_{2}=\mathrm{A}^{\prime} \times l^{\prime}$
$=\frac{\pi \mathrm{D}^{2}}{16} \times l^{\prime}$
$\mathrm{V}_{1}=\mathrm{V}_{2}$
$\frac{\pi \mathrm{D}^{2}}{4} l=\frac{\pi \mathrm{D}^{2}}{16} l^{\prime}$
$l=\frac{l^{\prime}}{4}$
$\frac{l^{\prime}}{l^{\prime}}=\frac{1}{4}$
$\frac{\mathrm{R}^{\prime}}{\mathrm{R}}=\frac{l^{\prime}}{\mathrm{A}^{\prime}} \times \frac{\mathrm{A}}{l}$
$\frac{\mathrm{R}^{\prime}}{10}=\frac{l^{\prime}}{l} \times \frac{\mathrm{A}}{\mathrm{A}^{\prime}}$
$\frac{\mathrm{R}^{\prime}}{10}=4 \times\left(\frac{\frac{\pi \mathrm{D}^{2}}{4}}{\frac{\pi \mathrm{D}^{2}}{16}}\right)=4 \times \frac{\pi \mathrm{D}^{2} \times 16}{4 \times \pi \mathrm{D}^{2}}$
$\frac{\mathrm{R}^{\prime}}{10}=16, \quad \mathrm{R}^{\prime}=160 \Omega$
$\frac{\mathrm{A}}{10}$

AIIMS-2003

Current Electricity

151966 A wire of radius $r$ has resistance $R$. If it is stretched to a radius $\frac{r}{2}$, its resistance will be:

1 $16 \mathrm{R}$

2 $2 \mathrm{R}$

3 $4 \mathrm{R}$

4 zero

J and K CET-2003

Current Electricity

151967 The ratio of the coefficient of thermal conductivity of two different materials is $5: 3$. If the thermal resistance of the rods of same thickness of these materials is same, then the ratio of the length of these rods will be-

1 $3: 5$

2 $5: 3$

3 $3: 4$

4 $3: 2$

BCECE-2007

Current Electricity

151969 The resistance of a bulb filament is $100 \Omega$ at a temperature of $100^{\circ} \mathrm{C}$. If temperature coefficient of resistance be $0.005 /{ }^{0} \mathrm{C}$, its resistance will become $200 \Omega$ at a temperature of:

1 $300^{\circ} \mathrm{C}$

2 $400^{\circ} \mathrm{C}$

3 $500^{\circ} \mathrm{C}$

4 $200^{\circ} \mathrm{C}$

BCECE-2004

Current Electricity

151970 Steel and aluminium wires have equal resistances and masses. Which of the wires is longer and how many times? (Given, densities of steel and aluminium are $7.8 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ and $2.7 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ and their resistivities are 0.15 $\mu \Omega-m$ and $0.028 \mu \Omega-m$ respectively)

1 The aluminium wire is 3.9 times longer

2 The aluminium wire is 1.3 times longer

3 The aluminium wire is 2.6 times longer

4 The steel wire is 3.9 times longer

Explanation:

A Let length \& Area of steel $=\mathrm{L}_{\mathrm{s}} \& \mathrm{~A}_{\mathrm{s}}$ Length \& Area of Aluminium $=L_{a} \& A_{a}$
$\because \quad$ Mass of steel & Aluminium is equal
$\therefore \quad \rho_{\mathrm{s}} \mathrm{L}_{\mathrm{s}} \mathrm{A}_{\mathrm{s}}=\rho_{\mathrm{a}} \mathrm{L}_{\mathrm{a}} \mathrm{A}_{\mathrm{a}}$
$\rho_{\mathrm{s}} \& \rho_{\mathrm{a}}$ are density of steel & aluminium
Given,
$\rho_{\mathrm{s}}=7.8 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$
$\rho_{\mathrm{a}}=2.7 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$
$7.8 \times 10^{3} \mathrm{~L}_{\mathrm{s}} \mathrm{A}_{\mathrm{s}}=2.7 \times 10^{3} \mathrm{~L}_{\mathrm{a}} \mathrm{A}_{\mathrm{a}}$
$\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}} \times \frac{\mathrm{A}_{\mathrm{s}}}{\mathrm{A}_{\mathrm{a}}}=\frac{2.7}{7.8}$
Resistivity of steel $\left(\rho_{\mathrm{s}}\right)=0.15 \mu \Omega-\mathrm{m}$
Resistivity of aluminium $\left(\rho_{\mathrm{Al}}\right)=0.028 \mu \Omega-\mathrm{m}$
$\therefore \quad \rho_{\mathrm{s}} \frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{A}_{\mathrm{s}}}=\rho_{\mathrm{a}} \frac{\mathrm{L}_{\mathrm{a}}}{\mathrm{A}_{\mathrm{a}}}$
$0.15 \frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{A}_{\mathrm{s}}}=0.028 \frac{\mathrm{L}_{\mathrm{a}}}{\mathrm{A}_{\mathrm{a}}}$
$\therefore \quad \frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}} \times \frac{\mathrm{A}_{\mathrm{a}}}{\mathrm{A}_{\mathrm{s}}}=\frac{0.028}{0.15}$
From equation (i)
$\frac{\mathrm{A}_{\mathrm{a}}}{\mathrm{A}_{\mathrm{s}}}=\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}} \times \frac{7.8}{2.7}$
Putting this value in equation (ii)
$\frac{\mathrm{L}_{s}}{\mathrm{~L}_{a}} \times \frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}} \times \frac{7.8}{2.7}=\frac{0.028}{0.15}$
$\left(\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}}\right)^{2}=\frac{0.028}{0.15} \times \frac{2.7}{7.8}$
$\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}}=\sqrt{\frac{0.028 \times 2.7}{0.15 \times 7.8}}$
$\frac{\mathrm{L}_{\mathrm{s}}}{\mathrm{L}_{\mathrm{a}}}=0.25419$
$\mathrm{~L}_{\mathrm{a}}=\frac{\mathrm{L}_{\mathrm{s}}}{0.25419}$
$\mathrm{~L}_{\mathrm{a}}=3.93 \mathrm{~L}_{\mathrm{s}}$

BCECE-2010

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