QBManager

Current Electricity

151936 A wire $X$ is half the diameter and half the length of a wire $Y$ of similar material. The ratio of resistance of $X$ to that of $Y$ is

1

8 : 1

2

4 : 1

3

2 : 1

4

1 : 1

Explanation:

C Resistance (R) $= ρ \frac{l}{A} = ρ \frac{4 l}{π D^{2}}$
$R \propto \frac{l}{D^{2}}$
$\frac{R_{X}}{R_{Y}} = \frac{l_{X} D_{X}^{2}}{I_{Y} D_{X}^{2}}$
$\frac{R_{X}}{R_{Y}} = \frac{\frac{l}{2} D^{2}}{{(\frac{D}{2})}^{2}}$
$\frac{R_{X}}{R_{Y}} = \frac{2}{1}$
$R_{X} : R_{Y} = 2 : 1$

BITSAT-2014

Current Electricity

151937 Two resistance at $0^{\circ} C$ with temperature coefficient of resistance $α_{1}$ and $α_{2}$ joined in series act as a single resistance in a circuit. The temperature coefficient of their single resistance will be

1

α_{1} + α_{2}

2

\frac{α_{1} α_{2}}{α_{1} + α_{2}}

3

\frac{α_{1} - α_{2}}{2}

4

\frac{α_{1} + α_{2}}{2}

Explanation:

D Given,
$R_{1} = R_{0} (1 + α_{1} t)$
$R_{2} = R_{0} (1 + α_{2} t)$
In series joint,
$R_{total} = R_{1} + R_{2}$
$R_{total} = R_{o} (1 + α_{1} t) + R_{o} (1 + α_{2} t)$
$R_{total} = R_{o} + R_{o} α_{1} \cdot t + R_{o} + R_{o} α_{2} \cdot t$
$= 2 R_{o} + R_{o} t (α_{1} + α_{2})$
$R_{t} = 2 R_{o} (1 + \frac{α_{1} + α_{2}}{2} \cdot t)$
Compare the standard equation of the wire resistance
$R_{t} = R_{o} (1 + α t)$
$α = \frac{α_{1} + α_{2}}{2}$

BITSAT-2016

Current Electricity

151939 A rod of a certain metal is $1.0 m$ long and 0.6 $cm$ in diameter. Its resistance is $3.0 \times 10^{- 3} Ω$ . Another disc made of the same metal is $2.0 cm$ in diameter and $1.0 mm$ thick. What is the resistance between the round faces of the disc?

1

1.35 \times 10^{- 8} Ω

2

2.70 \times 10^{- 7} Ω

3

4.05 \times 10^{- 6} Ω

4

8.10 \times 10^{- 5} Ω

Explanation:

B Given,
Length, $l = 1 m$
Diameter, $d = 0.6 cm$
Resistance, $R = 3.0 \times 10^{- 3} Ω$

Resistance, $R = ρ \frac{l}{A}$
$ρ = \frac{R \times A}{l} = \frac{3 \times 10^{- 3} \times π \times {(0.3 \times 10^{- 2})}^{2}}{1}$
$= 27 \times 10^{- 9} π Ω m$
$R_{disc} = ρ \times \frac{t}{A} = 27 \times 10^{- 9} \times π \times \frac{1 \times 10^{- 3}}{π {(1 \times 10^{- 2})}^{2}}$
$= 2.7 \times 10^{- 7} Ω$

UPSEE - 2012

Current Electricity

151941 An electric current passes through a circuit containing two wires of the same material connected in parallel. If the lengths of the wires are in the ratio of $4 / 3$ and radius of the wires are in the ratio of $2 / 3$ , then the ratio of the currents passing through the wires will be

1 3

2

1 / 3

3

8 / 9

4 None of these

Explanation:

B Given,
$\frac{l_{1}}{l_{2}} = \frac{4}{3}, \frac{r_{1}}{r_{2}} = \frac{2}{3}$
Resistance $(R) = ρ \frac{l}{A}$
$(∵ A = π r^{2})$
$\frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} \times \frac{A_{2}}{A_{1}} = \frac{l_{1}}{l_{2}} \times {(\frac{r_{2}}{r_{1}})}^{2}$
$\frac{R_{1}}{R_{2}} = \frac{4}{3} {(\frac{3}{2})}^{2} = \frac{3}{1}$
Applying Ohm's law-
$V = I_{1} R_{1} = I_{2} R_{2}$
$I_{1} = \frac{V}{R_{1}}, I_{2} = \frac{V}{R_{2}}$
$\frac{I_{1}}{I_{2}} = \frac{R_{2}}{R_{1}} = \frac{1}{3}$

UPSEE - 2008

Current Electricity

151936 A wire $X$ is half the diameter and half the length of a wire $Y$ of similar material. The ratio of resistance of $X$ to that of $Y$ is

1

8 : 1

2

4 : 1

3

2 : 1

4

1 : 1

Explanation:

C Resistance (R) $= ρ \frac{l}{A} = ρ \frac{4 l}{π D^{2}}$
$R \propto \frac{l}{D^{2}}$
$\frac{R_{X}}{R_{Y}} = \frac{l_{X} D_{X}^{2}}{I_{Y} D_{X}^{2}}$
$\frac{R_{X}}{R_{Y}} = \frac{\frac{l}{2} D^{2}}{{(\frac{D}{2})}^{2}}$
$\frac{R_{X}}{R_{Y}} = \frac{2}{1}$
$R_{X} : R_{Y} = 2 : 1$

BITSAT-2014

Current Electricity

151937 Two resistance at $0^{\circ} C$ with temperature coefficient of resistance $α_{1}$ and $α_{2}$ joined in series act as a single resistance in a circuit. The temperature coefficient of their single resistance will be

1

α_{1} + α_{2}

2

\frac{α_{1} α_{2}}{α_{1} + α_{2}}

3

\frac{α_{1} - α_{2}}{2}

4

\frac{α_{1} + α_{2}}{2}

Explanation:

D Given,
$R_{1} = R_{0} (1 + α_{1} t)$
$R_{2} = R_{0} (1 + α_{2} t)$
In series joint,
$R_{total} = R_{1} + R_{2}$
$R_{total} = R_{o} (1 + α_{1} t) + R_{o} (1 + α_{2} t)$
$R_{total} = R_{o} + R_{o} α_{1} \cdot t + R_{o} + R_{o} α_{2} \cdot t$
$= 2 R_{o} + R_{o} t (α_{1} + α_{2})$
$R_{t} = 2 R_{o} (1 + \frac{α_{1} + α_{2}}{2} \cdot t)$
Compare the standard equation of the wire resistance
$R_{t} = R_{o} (1 + α t)$
$α = \frac{α_{1} + α_{2}}{2}$

BITSAT-2016

Current Electricity

151939 A rod of a certain metal is $1.0 m$ long and 0.6 $cm$ in diameter. Its resistance is $3.0 \times 10^{- 3} Ω$ . Another disc made of the same metal is $2.0 cm$ in diameter and $1.0 mm$ thick. What is the resistance between the round faces of the disc?

1

1.35 \times 10^{- 8} Ω

2

2.70 \times 10^{- 7} Ω

3

4.05 \times 10^{- 6} Ω

4

8.10 \times 10^{- 5} Ω

Explanation:

B Given,
Length, $l = 1 m$
Diameter, $d = 0.6 cm$
Resistance, $R = 3.0 \times 10^{- 3} Ω$

Resistance, $R = ρ \frac{l}{A}$
$ρ = \frac{R \times A}{l} = \frac{3 \times 10^{- 3} \times π \times {(0.3 \times 10^{- 2})}^{2}}{1}$
$= 27 \times 10^{- 9} π Ω m$
$R_{disc} = ρ \times \frac{t}{A} = 27 \times 10^{- 9} \times π \times \frac{1 \times 10^{- 3}}{π {(1 \times 10^{- 2})}^{2}}$
$= 2.7 \times 10^{- 7} Ω$

UPSEE - 2012

Current Electricity

151941 An electric current passes through a circuit containing two wires of the same material connected in parallel. If the lengths of the wires are in the ratio of $4 / 3$ and radius of the wires are in the ratio of $2 / 3$ , then the ratio of the currents passing through the wires will be

1 3

2

1 / 3

3

8 / 9

4 None of these

Explanation:

B Given,
$\frac{l_{1}}{l_{2}} = \frac{4}{3}, \frac{r_{1}}{r_{2}} = \frac{2}{3}$
Resistance $(R) = ρ \frac{l}{A}$
$(∵ A = π r^{2})$
$\frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} \times \frac{A_{2}}{A_{1}} = \frac{l_{1}}{l_{2}} \times {(\frac{r_{2}}{r_{1}})}^{2}$
$\frac{R_{1}}{R_{2}} = \frac{4}{3} {(\frac{3}{2})}^{2} = \frac{3}{1}$
Applying Ohm's law-
$V = I_{1} R_{1} = I_{2} R_{2}$
$I_{1} = \frac{V}{R_{1}}, I_{2} = \frac{V}{R_{2}}$
$\frac{I_{1}}{I_{2}} = \frac{R_{2}}{R_{1}} = \frac{1}{3}$

UPSEE - 2008

Current Electricity

151936 A wire $X$ is half the diameter and half the length of a wire $Y$ of similar material. The ratio of resistance of $X$ to that of $Y$ is

1

8 : 1

2

4 : 1

3

2 : 1

4

1 : 1

Explanation:

C Resistance (R) $= ρ \frac{l}{A} = ρ \frac{4 l}{π D^{2}}$
$R \propto \frac{l}{D^{2}}$
$\frac{R_{X}}{R_{Y}} = \frac{l_{X} D_{X}^{2}}{I_{Y} D_{X}^{2}}$
$\frac{R_{X}}{R_{Y}} = \frac{\frac{l}{2} D^{2}}{{(\frac{D}{2})}^{2}}$
$\frac{R_{X}}{R_{Y}} = \frac{2}{1}$
$R_{X} : R_{Y} = 2 : 1$

BITSAT-2014

Current Electricity

151937 Two resistance at $0^{\circ} C$ with temperature coefficient of resistance $α_{1}$ and $α_{2}$ joined in series act as a single resistance in a circuit. The temperature coefficient of their single resistance will be

1

α_{1} + α_{2}

2

\frac{α_{1} α_{2}}{α_{1} + α_{2}}

3

\frac{α_{1} - α_{2}}{2}

4

\frac{α_{1} + α_{2}}{2}

Explanation:

D Given,
$R_{1} = R_{0} (1 + α_{1} t)$
$R_{2} = R_{0} (1 + α_{2} t)$
In series joint,
$R_{total} = R_{1} + R_{2}$
$R_{total} = R_{o} (1 + α_{1} t) + R_{o} (1 + α_{2} t)$
$R_{total} = R_{o} + R_{o} α_{1} \cdot t + R_{o} + R_{o} α_{2} \cdot t$
$= 2 R_{o} + R_{o} t (α_{1} + α_{2})$
$R_{t} = 2 R_{o} (1 + \frac{α_{1} + α_{2}}{2} \cdot t)$
Compare the standard equation of the wire resistance
$R_{t} = R_{o} (1 + α t)$
$α = \frac{α_{1} + α_{2}}{2}$

BITSAT-2016

Current Electricity

151939 A rod of a certain metal is $1.0 m$ long and 0.6 $cm$ in diameter. Its resistance is $3.0 \times 10^{- 3} Ω$ . Another disc made of the same metal is $2.0 cm$ in diameter and $1.0 mm$ thick. What is the resistance between the round faces of the disc?

1

1.35 \times 10^{- 8} Ω

2

2.70 \times 10^{- 7} Ω

3

4.05 \times 10^{- 6} Ω

4

8.10 \times 10^{- 5} Ω

Explanation:

B Given,
Length, $l = 1 m$
Diameter, $d = 0.6 cm$
Resistance, $R = 3.0 \times 10^{- 3} Ω$

Resistance, $R = ρ \frac{l}{A}$
$ρ = \frac{R \times A}{l} = \frac{3 \times 10^{- 3} \times π \times {(0.3 \times 10^{- 2})}^{2}}{1}$
$= 27 \times 10^{- 9} π Ω m$
$R_{disc} = ρ \times \frac{t}{A} = 27 \times 10^{- 9} \times π \times \frac{1 \times 10^{- 3}}{π {(1 \times 10^{- 2})}^{2}}$
$= 2.7 \times 10^{- 7} Ω$

UPSEE - 2012

Current Electricity

151941 An electric current passes through a circuit containing two wires of the same material connected in parallel. If the lengths of the wires are in the ratio of $4 / 3$ and radius of the wires are in the ratio of $2 / 3$ , then the ratio of the currents passing through the wires will be

1 3

2

1 / 3

3

8 / 9

4 None of these

Explanation:

B Given,
$\frac{l_{1}}{l_{2}} = \frac{4}{3}, \frac{r_{1}}{r_{2}} = \frac{2}{3}$
Resistance $(R) = ρ \frac{l}{A}$
$(∵ A = π r^{2})$
$\frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} \times \frac{A_{2}}{A_{1}} = \frac{l_{1}}{l_{2}} \times {(\frac{r_{2}}{r_{1}})}^{2}$
$\frac{R_{1}}{R_{2}} = \frac{4}{3} {(\frac{3}{2})}^{2} = \frac{3}{1}$
Applying Ohm's law-
$V = I_{1} R_{1} = I_{2} R_{2}$
$I_{1} = \frac{V}{R_{1}}, I_{2} = \frac{V}{R_{2}}$
$\frac{I_{1}}{I_{2}} = \frac{R_{2}}{R_{1}} = \frac{1}{3}$

UPSEE - 2008

Current Electricity

151936 A wire $X$ is half the diameter and half the length of a wire $Y$ of similar material. The ratio of resistance of $X$ to that of $Y$ is

1

8 : 1

2

4 : 1

3

2 : 1

4

1 : 1

Explanation:

C Resistance (R) $= ρ \frac{l}{A} = ρ \frac{4 l}{π D^{2}}$
$R \propto \frac{l}{D^{2}}$
$\frac{R_{X}}{R_{Y}} = \frac{l_{X} D_{X}^{2}}{I_{Y} D_{X}^{2}}$
$\frac{R_{X}}{R_{Y}} = \frac{\frac{l}{2} D^{2}}{{(\frac{D}{2})}^{2}}$
$\frac{R_{X}}{R_{Y}} = \frac{2}{1}$
$R_{X} : R_{Y} = 2 : 1$

BITSAT-2014

Current Electricity

151937 Two resistance at $0^{\circ} C$ with temperature coefficient of resistance $α_{1}$ and $α_{2}$ joined in series act as a single resistance in a circuit. The temperature coefficient of their single resistance will be

1

α_{1} + α_{2}

2

\frac{α_{1} α_{2}}{α_{1} + α_{2}}

3

\frac{α_{1} - α_{2}}{2}

4

\frac{α_{1} + α_{2}}{2}

Explanation:

D Given,
$R_{1} = R_{0} (1 + α_{1} t)$
$R_{2} = R_{0} (1 + α_{2} t)$
In series joint,
$R_{total} = R_{1} + R_{2}$
$R_{total} = R_{o} (1 + α_{1} t) + R_{o} (1 + α_{2} t)$
$R_{total} = R_{o} + R_{o} α_{1} \cdot t + R_{o} + R_{o} α_{2} \cdot t$
$= 2 R_{o} + R_{o} t (α_{1} + α_{2})$
$R_{t} = 2 R_{o} (1 + \frac{α_{1} + α_{2}}{2} \cdot t)$
Compare the standard equation of the wire resistance
$R_{t} = R_{o} (1 + α t)$
$α = \frac{α_{1} + α_{2}}{2}$

BITSAT-2016

Current Electricity

151939 A rod of a certain metal is $1.0 m$ long and 0.6 $cm$ in diameter. Its resistance is $3.0 \times 10^{- 3} Ω$ . Another disc made of the same metal is $2.0 cm$ in diameter and $1.0 mm$ thick. What is the resistance between the round faces of the disc?

1

1.35 \times 10^{- 8} Ω

2

2.70 \times 10^{- 7} Ω

3

4.05 \times 10^{- 6} Ω

4

8.10 \times 10^{- 5} Ω

Explanation:

B Given,
Length, $l = 1 m$
Diameter, $d = 0.6 cm$
Resistance, $R = 3.0 \times 10^{- 3} Ω$

Resistance, $R = ρ \frac{l}{A}$
$ρ = \frac{R \times A}{l} = \frac{3 \times 10^{- 3} \times π \times {(0.3 \times 10^{- 2})}^{2}}{1}$
$= 27 \times 10^{- 9} π Ω m$
$R_{disc} = ρ \times \frac{t}{A} = 27 \times 10^{- 9} \times π \times \frac{1 \times 10^{- 3}}{π {(1 \times 10^{- 2})}^{2}}$
$= 2.7 \times 10^{- 7} Ω$

UPSEE - 2012

Current Electricity

151941 An electric current passes through a circuit containing two wires of the same material connected in parallel. If the lengths of the wires are in the ratio of $4 / 3$ and radius of the wires are in the ratio of $2 / 3$ , then the ratio of the currents passing through the wires will be

1 3

2

1 / 3

3

8 / 9

4 None of these

Explanation:

B Given,
$\frac{l_{1}}{l_{2}} = \frac{4}{3}, \frac{r_{1}}{r_{2}} = \frac{2}{3}$
Resistance $(R) = ρ \frac{l}{A}$
$(∵ A = π r^{2})$
$\frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} \times \frac{A_{2}}{A_{1}} = \frac{l_{1}}{l_{2}} \times {(\frac{r_{2}}{r_{1}})}^{2}$
$\frac{R_{1}}{R_{2}} = \frac{4}{3} {(\frac{3}{2})}^{2} = \frac{3}{1}$
Applying Ohm's law-
$V = I_{1} R_{1} = I_{2} R_{2}$
$I_{1} = \frac{V}{R_{1}}, I_{2} = \frac{V}{R_{2}}$
$\frac{I_{1}}{I_{2}} = \frac{R_{2}}{R_{1}} = \frac{1}{3}$

UPSEE - 2008

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