QBManager

Current Electricity

151961 Assertion: Bending a wire does not effect electrical resistance.
Reason: Resistance of wire is proportional of resistivity of material.

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

A Resistance wire
$R = \frac{ρ l}{A}$
Which does not depend on the geometry of wire since when wire is bended, resistivity, length and area of cross section do not change, therefore resistance of of wire also remains same.

AIIMS-2016

Current Electricity

151962 Two wires of same metal have the same length but their cross sections are in the ratio $3 : 1$ . They are joined in series.
The resistance of the thicker wire is $10 Ω$ . The total resistance of the combination is .

1

5 / 2 Ω

2

40 / 3 Ω

3

40 Ω

4

100 Ω

Explanation:

C Given, $R_{1} = 10 Ω$
$A_{1} = 3 A$
$A_{2} = A$
We know that
Resistance, $R = \frac{ρ l}{A}$
For same material and same length
$\frac{R_{2}}{R_{1}} = \frac{A_{1}}{A_{2}} = \frac{3 A}{A} = \frac{3}{1}$
$R_{2} = 3 R_{1}$
$R_{2} = 3 \times 10 = 30 Ω$
Total resistance of series combination
$R_{eq} = R_{1} + R_{2} = 10 + 30$
$R_{eq} = 40 Ω$

AIIMS-2012

Current Electricity

151963 When the power delivered by a 100 volt battery is 40 watts the equivalent resistance of a circuit is :

1

100 ohms

2

250 ohms

3

300 ohms

4

350 ohms

Explanation:

B Given,
$V = 100 volt$
$P = 40 watt$
We know that
$R = \frac{V^{2}}{P} = \frac{100 \times 100}{40}$
$R = 250 ohms$

AIIMS-2011

Current Electricity

151964 Four wires of the same diameter are connected in turn between two points, maintained at the constant potential difference. Their resistivities are; $ρ$ and $L$ (wire1)., $1.2 ρ$ and 1.2L (wire2), $0.9 ρ$ and $0.9 L$ (wire 3) and $ρ$ and $1.5 L$ (wire4). Rank the wires according to the rates at which energy is dissipated as heat, greatest first

1

4 > 3 > 1 > 2

2

4 > 2 > 1 > 3

3

1 > 2 > 3 > 4

4

3 > 1 > 2 > 4

Explanation:

D We know that,
$R = \frac{ρ L}{A}$
$R_{1} = \frac{ρ L}{A}$
$R_{2} = \frac{(1.2 ρ) (1.2 L)}{A} = \frac{1.44 ρ L}{A}$
$R_{3} = \frac{(0.9 ρ) (0.9 L)}{A} = \frac{0.81 ρ L}{A}$
$R_{4} = \frac{ρ (1.5 L)}{A} = \frac{1.5 ρ . L}{A}$
$R_{3} < R_{1} < R_{2} < R_{4}$
For constant potential
$P \propto \frac{1}{R}$
Then, $3 > 1 > 2 > 4$

Current Electricity

151961 Assertion: Bending a wire does not effect electrical resistance.
Reason: Resistance of wire is proportional of resistivity of material.

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

A Resistance wire
$R = \frac{ρ l}{A}$
Which does not depend on the geometry of wire since when wire is bended, resistivity, length and area of cross section do not change, therefore resistance of of wire also remains same.

AIIMS-2016

Current Electricity

151962 Two wires of same metal have the same length but their cross sections are in the ratio $3 : 1$ . They are joined in series.
The resistance of the thicker wire is $10 Ω$ . The total resistance of the combination is .

1

5 / 2 Ω

2

40 / 3 Ω

3

40 Ω

4

100 Ω

Explanation:

C Given, $R_{1} = 10 Ω$
$A_{1} = 3 A$
$A_{2} = A$
We know that
Resistance, $R = \frac{ρ l}{A}$
For same material and same length
$\frac{R_{2}}{R_{1}} = \frac{A_{1}}{A_{2}} = \frac{3 A}{A} = \frac{3}{1}$
$R_{2} = 3 R_{1}$
$R_{2} = 3 \times 10 = 30 Ω$
Total resistance of series combination
$R_{eq} = R_{1} + R_{2} = 10 + 30$
$R_{eq} = 40 Ω$

AIIMS-2012

Current Electricity

151963 When the power delivered by a 100 volt battery is 40 watts the equivalent resistance of a circuit is :

1

100 ohms

2

250 ohms

3

300 ohms

4

350 ohms

Explanation:

B Given,
$V = 100 volt$
$P = 40 watt$
We know that
$R = \frac{V^{2}}{P} = \frac{100 \times 100}{40}$
$R = 250 ohms$

AIIMS-2011

Current Electricity

151964 Four wires of the same diameter are connected in turn between two points, maintained at the constant potential difference. Their resistivities are; $ρ$ and $L$ (wire1)., $1.2 ρ$ and 1.2L (wire2), $0.9 ρ$ and $0.9 L$ (wire 3) and $ρ$ and $1.5 L$ (wire4). Rank the wires according to the rates at which energy is dissipated as heat, greatest first

1

4 > 3 > 1 > 2

2

4 > 2 > 1 > 3

3

1 > 2 > 3 > 4

4

3 > 1 > 2 > 4

Explanation:

D We know that,
$R = \frac{ρ L}{A}$
$R_{1} = \frac{ρ L}{A}$
$R_{2} = \frac{(1.2 ρ) (1.2 L)}{A} = \frac{1.44 ρ L}{A}$
$R_{3} = \frac{(0.9 ρ) (0.9 L)}{A} = \frac{0.81 ρ L}{A}$
$R_{4} = \frac{ρ (1.5 L)}{A} = \frac{1.5 ρ . L}{A}$
$R_{3} < R_{1} < R_{2} < R_{4}$
For constant potential
$P \propto \frac{1}{R}$
Then, $3 > 1 > 2 > 4$

Current Electricity

151961 Assertion: Bending a wire does not effect electrical resistance.
Reason: Resistance of wire is proportional of resistivity of material.

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

A Resistance wire
$R = \frac{ρ l}{A}$
Which does not depend on the geometry of wire since when wire is bended, resistivity, length and area of cross section do not change, therefore resistance of of wire also remains same.

AIIMS-2016

Current Electricity

151962 Two wires of same metal have the same length but their cross sections are in the ratio $3 : 1$ . They are joined in series.
The resistance of the thicker wire is $10 Ω$ . The total resistance of the combination is .

1

5 / 2 Ω

2

40 / 3 Ω

3

40 Ω

4

100 Ω

Explanation:

C Given, $R_{1} = 10 Ω$
$A_{1} = 3 A$
$A_{2} = A$
We know that
Resistance, $R = \frac{ρ l}{A}$
For same material and same length
$\frac{R_{2}}{R_{1}} = \frac{A_{1}}{A_{2}} = \frac{3 A}{A} = \frac{3}{1}$
$R_{2} = 3 R_{1}$
$R_{2} = 3 \times 10 = 30 Ω$
Total resistance of series combination
$R_{eq} = R_{1} + R_{2} = 10 + 30$
$R_{eq} = 40 Ω$

AIIMS-2012

Current Electricity

151963 When the power delivered by a 100 volt battery is 40 watts the equivalent resistance of a circuit is :

1

100 ohms

2

250 ohms

3

300 ohms

4

350 ohms

Explanation:

B Given,
$V = 100 volt$
$P = 40 watt$
We know that
$R = \frac{V^{2}}{P} = \frac{100 \times 100}{40}$
$R = 250 ohms$

AIIMS-2011

Current Electricity

151964 Four wires of the same diameter are connected in turn between two points, maintained at the constant potential difference. Their resistivities are; $ρ$ and $L$ (wire1)., $1.2 ρ$ and 1.2L (wire2), $0.9 ρ$ and $0.9 L$ (wire 3) and $ρ$ and $1.5 L$ (wire4). Rank the wires according to the rates at which energy is dissipated as heat, greatest first

1

4 > 3 > 1 > 2

2

4 > 2 > 1 > 3

3

1 > 2 > 3 > 4

4

3 > 1 > 2 > 4

Explanation:

D We know that,
$R = \frac{ρ L}{A}$
$R_{1} = \frac{ρ L}{A}$
$R_{2} = \frac{(1.2 ρ) (1.2 L)}{A} = \frac{1.44 ρ L}{A}$
$R_{3} = \frac{(0.9 ρ) (0.9 L)}{A} = \frac{0.81 ρ L}{A}$
$R_{4} = \frac{ρ (1.5 L)}{A} = \frac{1.5 ρ . L}{A}$
$R_{3} < R_{1} < R_{2} < R_{4}$
For constant potential
$P \propto \frac{1}{R}$
Then, $3 > 1 > 2 > 4$

Current Electricity

151961 Assertion: Bending a wire does not effect electrical resistance.
Reason: Resistance of wire is proportional of resistivity of material.

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

A Resistance wire
$R = \frac{ρ l}{A}$
Which does not depend on the geometry of wire since when wire is bended, resistivity, length and area of cross section do not change, therefore resistance of of wire also remains same.

AIIMS-2016

Current Electricity

151962 Two wires of same metal have the same length but their cross sections are in the ratio $3 : 1$ . They are joined in series.
The resistance of the thicker wire is $10 Ω$ . The total resistance of the combination is .

1

5 / 2 Ω

2

40 / 3 Ω

3

40 Ω

4

100 Ω

Explanation:

C Given, $R_{1} = 10 Ω$
$A_{1} = 3 A$
$A_{2} = A$
We know that
Resistance, $R = \frac{ρ l}{A}$
For same material and same length
$\frac{R_{2}}{R_{1}} = \frac{A_{1}}{A_{2}} = \frac{3 A}{A} = \frac{3}{1}$
$R_{2} = 3 R_{1}$
$R_{2} = 3 \times 10 = 30 Ω$
Total resistance of series combination
$R_{eq} = R_{1} + R_{2} = 10 + 30$
$R_{eq} = 40 Ω$

AIIMS-2012

Current Electricity

151963 When the power delivered by a 100 volt battery is 40 watts the equivalent resistance of a circuit is :

1

100 ohms

2

250 ohms

3

300 ohms

4

350 ohms

Explanation:

B Given,
$V = 100 volt$
$P = 40 watt$
We know that
$R = \frac{V^{2}}{P} = \frac{100 \times 100}{40}$
$R = 250 ohms$

AIIMS-2011

Current Electricity

151964 Four wires of the same diameter are connected in turn between two points, maintained at the constant potential difference. Their resistivities are; $ρ$ and $L$ (wire1)., $1.2 ρ$ and 1.2L (wire2), $0.9 ρ$ and $0.9 L$ (wire 3) and $ρ$ and $1.5 L$ (wire4). Rank the wires according to the rates at which energy is dissipated as heat, greatest first

1

4 > 3 > 1 > 2

2

4 > 2 > 1 > 3

3

1 > 2 > 3 > 4

4

3 > 1 > 2 > 4

Explanation:

D We know that,
$R = \frac{ρ L}{A}$
$R_{1} = \frac{ρ L}{A}$
$R_{2} = \frac{(1.2 ρ) (1.2 L)}{A} = \frac{1.44 ρ L}{A}$
$R_{3} = \frac{(0.9 ρ) (0.9 L)}{A} = \frac{0.81 ρ L}{A}$
$R_{4} = \frac{ρ (1.5 L)}{A} = \frac{1.5 ρ . L}{A}$
$R_{3} < R_{1} < R_{2} < R_{4}$
For constant potential
$P \propto \frac{1}{R}$
Then, $3 > 1 > 2 > 4$