QBManager

Current Electricity

151925 In a Wheatstone's bridge, three resistances $P$ , $Q$ and $R$ are connected in the three arms and the fourth arm is formed by two resistances $S_{1}$ and $S_{2}$ connected in parallel. The condition for the bridge to be balanced will be

1

\frac{P}{Q} = \frac{2 R}{S_{1} + S_{2}}

2

\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{S_{1} S_{2}}

3

\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{2 S_{1} S_{2}}

4

\frac{P}{Q} = \frac{R}{S_{1} + S_{2}}

Explanation:

B The wheatstone bridge diagram, according to question-

$S_{eq} = \frac{S_{1} S_{2}}{S_{1} + S_{2}}$
Where, $S_{1}$ and $S_{2}$ are connected in parallel Now, wheat stone bridge is balanced,
$\frac{P}{R} = \frac{Q}{S_{eq}}$
$\frac{P}{Q} = \frac{R}{S_{eq}} = \frac{R}{(\frac{S_{1} S_{2}}{S_{1} + S_{2}})}$
$\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{S_{1} S_{2}}$

UPSEE - 2010

Current Electricity

151926 Two wires of equal length and equal diameter and having resistivities $ρ_{1}$ and $ρ_{2}$ are connected in series. The equivalent resistivity of the combination is

1

\sqrt{ρ_{1} ρ_{2}}

2

\frac{ρ_{1} + ρ_{2}}{2}

3

\frac{ρ_{1} ρ_{2}}{ρ_{1} + ρ_{2}}

4

(ρ_{1} + ρ_{2})

Explanation:

B We know that,
Resistance, $R = \frac{ρ L}{A}$
Then,
$R_{1} = \frac{ρ_{1} L}{A}, R_{2} = \frac{ρ_{2} L}{A}$
For series connection
$R_{eq} = R_{1} + R_{2}$
$R_{eq} = \frac{ρ_{1} L}{A} + \frac{ρ_{2} L}{A}$
$R_{eq} = (\frac{ρ_{1} + ρ_{2}}{A}) L$
$\frac{ρ_{eq} \times 2 L}{A} = (\frac{ρ_{1} + ρ_{2}}{A}) L$
$ρ_{eq} = \frac{ρ_{1} + ρ_{2}}{2}$

GUJCET 2016

Current Electricity

151927 A uniform wire of resistance $R$ , of the radius $r$ is uniformly drawn until its radius is reduced to $r / n$ . Its new resistance is

1

nR

2

n^{3} R

3

n^{2} R

4

n^{4} R

Explanation:

D :Given,
Uniform wire of resistance $= R$
$Radius = r$
Resistance, $R = \frac{ρ . l}{A} = \frac{ρ l}{π r^{2}} (∵ A = π r^{2})$
$∵ (r^{'} = \frac{r}{n})$
Volume remains constant-
Initial volume $=$ Final volume
$r^{2} l = {(\frac{r}{n})}^{2} l^{'}$
$r^{2} l = \frac{r^{2}}{n^{2}} l^{'}$
$\frac{l}{l^{'}} = \frac{1}{n^{2}}$
$l^{'} = n^{2} l$
New resistance, $R^{'} = \frac{ρ l^{'}}{A^{'}} = \frac{ρ l^{'}}{π r^{' 2}}$
$R^{'} = \frac{ρ}{π} \times \frac{n^{2} l}{(r / n)^{2}} = n^{4} (\frac{ρ l}{π r^{2}}) = n^{4} R$

COMEDK 2011

Current Electricity

151929 The maximum current that flow in the fuse wire before it blows out, varies with the radius $r$ as

1

r^{3 / 2}

2

r

3

r^{2 / 3}

4

r^{1 / 2}

Explanation:

A The maximum current that flow in the fuse wire before it blows out varies with the radius $r^{3 / 2}$
If $l =$ length of wire
$r =$ radius of wire
$I =$ Current
$Q = h =$ the rate of heat loss then,
Resistance, $R = \frac{ρ l}{A} = \frac{ρ l}{π r^{2}}$
At steady state-
$I_{max}^{2} R = hA$
$I_{max}^{2} \times (\frac{ρ l}{π r^{2}}) = h \times 2 π rl$
$I_{max}^{2} = \frac{2 π^{2} h}{ρ} r^{3}$
$I_{max} \propto r^{3 / 2}$

Manipal UGET-2013

Current Electricity

151930 The resistance of a wire at $300 K$ is found to be $0.3 Ω$ . If the temperature coefficient of resistance of wire is $1.5 \times 10^{- 3} K^{- 1}$ the temperature at which the resistance becomes $0.6 Ω$ is

1

720 K

2

345 K

3

993 K

4

690 K

Explanation:

C Given,
$R_{300} = 0.3 Ω, R_{t} = 0.6 Ω$
$T = 300 K = 27^{\circ} C$
$∵$ Temperature coefficient of resistance
$α = 1.5 \times 10^{- 3} K^{- 1}$
$R_{300} = R_{0} (1 + α \times 27)$
$0.3 = R_{0} (1 + 1.5 \times 10^{- 3} \times 27)$
Again, $R_{t} = R_{0} (1 + α t)$
$0.6 = R_{0} (1 + 1.5 \times 10^{- 3} \times t)$
Dividing equation (ii) and (i), we get-
$\frac{0.6}{0.3} = \frac{1 + 1.5 \times 10^{- 3} t}{1 + 1.5 \times 10^{- 3} \times 27}$
$2 (1 + 1.5 \times 10^{- 3} \times 27) = 1 + 1.5 \times 10^{- 3} t$
$2 + 81 \times 10^{- 3} = 1 + 1.5 \times 10^{- 3} t$
$t = \frac{1.081}{1.5 \times 10^{- 3}} = 720^{\circ} C$
$t = 993 K$

Karnataka CET-2009

Current Electricity

151925 In a Wheatstone's bridge, three resistances $P$ , $Q$ and $R$ are connected in the three arms and the fourth arm is formed by two resistances $S_{1}$ and $S_{2}$ connected in parallel. The condition for the bridge to be balanced will be

1

\frac{P}{Q} = \frac{2 R}{S_{1} + S_{2}}

2

\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{S_{1} S_{2}}

3

\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{2 S_{1} S_{2}}

4

\frac{P}{Q} = \frac{R}{S_{1} + S_{2}}

Explanation:

B The wheatstone bridge diagram, according to question-

$S_{eq} = \frac{S_{1} S_{2}}{S_{1} + S_{2}}$
Where, $S_{1}$ and $S_{2}$ are connected in parallel Now, wheat stone bridge is balanced,
$\frac{P}{R} = \frac{Q}{S_{eq}}$
$\frac{P}{Q} = \frac{R}{S_{eq}} = \frac{R}{(\frac{S_{1} S_{2}}{S_{1} + S_{2}})}$
$\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{S_{1} S_{2}}$

UPSEE - 2010

Current Electricity

151926 Two wires of equal length and equal diameter and having resistivities $ρ_{1}$ and $ρ_{2}$ are connected in series. The equivalent resistivity of the combination is

1

\sqrt{ρ_{1} ρ_{2}}

2

\frac{ρ_{1} + ρ_{2}}{2}

3

\frac{ρ_{1} ρ_{2}}{ρ_{1} + ρ_{2}}

4

(ρ_{1} + ρ_{2})

Explanation:

B We know that,
Resistance, $R = \frac{ρ L}{A}$
Then,
$R_{1} = \frac{ρ_{1} L}{A}, R_{2} = \frac{ρ_{2} L}{A}$
For series connection
$R_{eq} = R_{1} + R_{2}$
$R_{eq} = \frac{ρ_{1} L}{A} + \frac{ρ_{2} L}{A}$
$R_{eq} = (\frac{ρ_{1} + ρ_{2}}{A}) L$
$\frac{ρ_{eq} \times 2 L}{A} = (\frac{ρ_{1} + ρ_{2}}{A}) L$
$ρ_{eq} = \frac{ρ_{1} + ρ_{2}}{2}$

GUJCET 2016

Current Electricity

151927 A uniform wire of resistance $R$ , of the radius $r$ is uniformly drawn until its radius is reduced to $r / n$ . Its new resistance is

1

nR

2

n^{3} R

3

n^{2} R

4

n^{4} R

Explanation:

D :Given,
Uniform wire of resistance $= R$
$Radius = r$
Resistance, $R = \frac{ρ . l}{A} = \frac{ρ l}{π r^{2}} (∵ A = π r^{2})$
$∵ (r^{'} = \frac{r}{n})$
Volume remains constant-
Initial volume $=$ Final volume
$r^{2} l = {(\frac{r}{n})}^{2} l^{'}$
$r^{2} l = \frac{r^{2}}{n^{2}} l^{'}$
$\frac{l}{l^{'}} = \frac{1}{n^{2}}$
$l^{'} = n^{2} l$
New resistance, $R^{'} = \frac{ρ l^{'}}{A^{'}} = \frac{ρ l^{'}}{π r^{' 2}}$
$R^{'} = \frac{ρ}{π} \times \frac{n^{2} l}{(r / n)^{2}} = n^{4} (\frac{ρ l}{π r^{2}}) = n^{4} R$

COMEDK 2011

Current Electricity

151929 The maximum current that flow in the fuse wire before it blows out, varies with the radius $r$ as

1

r^{3 / 2}

2

r

3

r^{2 / 3}

4

r^{1 / 2}

Explanation:

A The maximum current that flow in the fuse wire before it blows out varies with the radius $r^{3 / 2}$
If $l =$ length of wire
$r =$ radius of wire
$I =$ Current
$Q = h =$ the rate of heat loss then,
Resistance, $R = \frac{ρ l}{A} = \frac{ρ l}{π r^{2}}$
At steady state-
$I_{max}^{2} R = hA$
$I_{max}^{2} \times (\frac{ρ l}{π r^{2}}) = h \times 2 π rl$
$I_{max}^{2} = \frac{2 π^{2} h}{ρ} r^{3}$
$I_{max} \propto r^{3 / 2}$

Manipal UGET-2013

Current Electricity

151930 The resistance of a wire at $300 K$ is found to be $0.3 Ω$ . If the temperature coefficient of resistance of wire is $1.5 \times 10^{- 3} K^{- 1}$ the temperature at which the resistance becomes $0.6 Ω$ is

1

720 K

2

345 K

3

993 K

4

690 K

Explanation:

C Given,
$R_{300} = 0.3 Ω, R_{t} = 0.6 Ω$
$T = 300 K = 27^{\circ} C$
$∵$ Temperature coefficient of resistance
$α = 1.5 \times 10^{- 3} K^{- 1}$
$R_{300} = R_{0} (1 + α \times 27)$
$0.3 = R_{0} (1 + 1.5 \times 10^{- 3} \times 27)$
Again, $R_{t} = R_{0} (1 + α t)$
$0.6 = R_{0} (1 + 1.5 \times 10^{- 3} \times t)$
Dividing equation (ii) and (i), we get-
$\frac{0.6}{0.3} = \frac{1 + 1.5 \times 10^{- 3} t}{1 + 1.5 \times 10^{- 3} \times 27}$
$2 (1 + 1.5 \times 10^{- 3} \times 27) = 1 + 1.5 \times 10^{- 3} t$
$2 + 81 \times 10^{- 3} = 1 + 1.5 \times 10^{- 3} t$
$t = \frac{1.081}{1.5 \times 10^{- 3}} = 720^{\circ} C$
$t = 993 K$

Karnataka CET-2009

Current Electricity

151925 In a Wheatstone's bridge, three resistances $P$ , $Q$ and $R$ are connected in the three arms and the fourth arm is formed by two resistances $S_{1}$ and $S_{2}$ connected in parallel. The condition for the bridge to be balanced will be

1

\frac{P}{Q} = \frac{2 R}{S_{1} + S_{2}}

2

\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{S_{1} S_{2}}

3

\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{2 S_{1} S_{2}}

4

\frac{P}{Q} = \frac{R}{S_{1} + S_{2}}

Explanation:

B The wheatstone bridge diagram, according to question-

$S_{eq} = \frac{S_{1} S_{2}}{S_{1} + S_{2}}$
Where, $S_{1}$ and $S_{2}$ are connected in parallel Now, wheat stone bridge is balanced,
$\frac{P}{R} = \frac{Q}{S_{eq}}$
$\frac{P}{Q} = \frac{R}{S_{eq}} = \frac{R}{(\frac{S_{1} S_{2}}{S_{1} + S_{2}})}$
$\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{S_{1} S_{2}}$

UPSEE - 2010

Current Electricity

151926 Two wires of equal length and equal diameter and having resistivities $ρ_{1}$ and $ρ_{2}$ are connected in series. The equivalent resistivity of the combination is

1

\sqrt{ρ_{1} ρ_{2}}

2

\frac{ρ_{1} + ρ_{2}}{2}

3

\frac{ρ_{1} ρ_{2}}{ρ_{1} + ρ_{2}}

4

(ρ_{1} + ρ_{2})

Explanation:

B We know that,
Resistance, $R = \frac{ρ L}{A}$
Then,
$R_{1} = \frac{ρ_{1} L}{A}, R_{2} = \frac{ρ_{2} L}{A}$
For series connection
$R_{eq} = R_{1} + R_{2}$
$R_{eq} = \frac{ρ_{1} L}{A} + \frac{ρ_{2} L}{A}$
$R_{eq} = (\frac{ρ_{1} + ρ_{2}}{A}) L$
$\frac{ρ_{eq} \times 2 L}{A} = (\frac{ρ_{1} + ρ_{2}}{A}) L$
$ρ_{eq} = \frac{ρ_{1} + ρ_{2}}{2}$

GUJCET 2016

Current Electricity

151927 A uniform wire of resistance $R$ , of the radius $r$ is uniformly drawn until its radius is reduced to $r / n$ . Its new resistance is

1

nR

2

n^{3} R

3

n^{2} R

4

n^{4} R

Explanation:

D :Given,
Uniform wire of resistance $= R$
$Radius = r$
Resistance, $R = \frac{ρ . l}{A} = \frac{ρ l}{π r^{2}} (∵ A = π r^{2})$
$∵ (r^{'} = \frac{r}{n})$
Volume remains constant-
Initial volume $=$ Final volume
$r^{2} l = {(\frac{r}{n})}^{2} l^{'}$
$r^{2} l = \frac{r^{2}}{n^{2}} l^{'}$
$\frac{l}{l^{'}} = \frac{1}{n^{2}}$
$l^{'} = n^{2} l$
New resistance, $R^{'} = \frac{ρ l^{'}}{A^{'}} = \frac{ρ l^{'}}{π r^{' 2}}$
$R^{'} = \frac{ρ}{π} \times \frac{n^{2} l}{(r / n)^{2}} = n^{4} (\frac{ρ l}{π r^{2}}) = n^{4} R$

COMEDK 2011

Current Electricity

151929 The maximum current that flow in the fuse wire before it blows out, varies with the radius $r$ as

1

r^{3 / 2}

2

r

3

r^{2 / 3}

4

r^{1 / 2}

Explanation:

A The maximum current that flow in the fuse wire before it blows out varies with the radius $r^{3 / 2}$
If $l =$ length of wire
$r =$ radius of wire
$I =$ Current
$Q = h =$ the rate of heat loss then,
Resistance, $R = \frac{ρ l}{A} = \frac{ρ l}{π r^{2}}$
At steady state-
$I_{max}^{2} R = hA$
$I_{max}^{2} \times (\frac{ρ l}{π r^{2}}) = h \times 2 π rl$
$I_{max}^{2} = \frac{2 π^{2} h}{ρ} r^{3}$
$I_{max} \propto r^{3 / 2}$

Manipal UGET-2013

Current Electricity

151930 The resistance of a wire at $300 K$ is found to be $0.3 Ω$ . If the temperature coefficient of resistance of wire is $1.5 \times 10^{- 3} K^{- 1}$ the temperature at which the resistance becomes $0.6 Ω$ is

1

720 K

2

345 K

3

993 K

4

690 K

Explanation:

C Given,
$R_{300} = 0.3 Ω, R_{t} = 0.6 Ω$
$T = 300 K = 27^{\circ} C$
$∵$ Temperature coefficient of resistance
$α = 1.5 \times 10^{- 3} K^{- 1}$
$R_{300} = R_{0} (1 + α \times 27)$
$0.3 = R_{0} (1 + 1.5 \times 10^{- 3} \times 27)$
Again, $R_{t} = R_{0} (1 + α t)$
$0.6 = R_{0} (1 + 1.5 \times 10^{- 3} \times t)$
Dividing equation (ii) and (i), we get-
$\frac{0.6}{0.3} = \frac{1 + 1.5 \times 10^{- 3} t}{1 + 1.5 \times 10^{- 3} \times 27}$
$2 (1 + 1.5 \times 10^{- 3} \times 27) = 1 + 1.5 \times 10^{- 3} t$
$2 + 81 \times 10^{- 3} = 1 + 1.5 \times 10^{- 3} t$
$t = \frac{1.081}{1.5 \times 10^{- 3}} = 720^{\circ} C$
$t = 993 K$

Karnataka CET-2009

Current Electricity

151925 In a Wheatstone's bridge, three resistances $P$ , $Q$ and $R$ are connected in the three arms and the fourth arm is formed by two resistances $S_{1}$ and $S_{2}$ connected in parallel. The condition for the bridge to be balanced will be

1

\frac{P}{Q} = \frac{2 R}{S_{1} + S_{2}}

2

\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{S_{1} S_{2}}

3

\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{2 S_{1} S_{2}}

4

\frac{P}{Q} = \frac{R}{S_{1} + S_{2}}

Explanation:

B The wheatstone bridge diagram, according to question-

$S_{eq} = \frac{S_{1} S_{2}}{S_{1} + S_{2}}$
Where, $S_{1}$ and $S_{2}$ are connected in parallel Now, wheat stone bridge is balanced,
$\frac{P}{R} = \frac{Q}{S_{eq}}$
$\frac{P}{Q} = \frac{R}{S_{eq}} = \frac{R}{(\frac{S_{1} S_{2}}{S_{1} + S_{2}})}$
$\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{S_{1} S_{2}}$

UPSEE - 2010

Current Electricity

151926 Two wires of equal length and equal diameter and having resistivities $ρ_{1}$ and $ρ_{2}$ are connected in series. The equivalent resistivity of the combination is

1

\sqrt{ρ_{1} ρ_{2}}

2

\frac{ρ_{1} + ρ_{2}}{2}

3

\frac{ρ_{1} ρ_{2}}{ρ_{1} + ρ_{2}}

4

(ρ_{1} + ρ_{2})

Explanation:

B We know that,
Resistance, $R = \frac{ρ L}{A}$
Then,
$R_{1} = \frac{ρ_{1} L}{A}, R_{2} = \frac{ρ_{2} L}{A}$
For series connection
$R_{eq} = R_{1} + R_{2}$
$R_{eq} = \frac{ρ_{1} L}{A} + \frac{ρ_{2} L}{A}$
$R_{eq} = (\frac{ρ_{1} + ρ_{2}}{A}) L$
$\frac{ρ_{eq} \times 2 L}{A} = (\frac{ρ_{1} + ρ_{2}}{A}) L$
$ρ_{eq} = \frac{ρ_{1} + ρ_{2}}{2}$

GUJCET 2016

Current Electricity

151927 A uniform wire of resistance $R$ , of the radius $r$ is uniformly drawn until its radius is reduced to $r / n$ . Its new resistance is

1

nR

2

n^{3} R

3

n^{2} R

4

n^{4} R

Explanation:

D :Given,
Uniform wire of resistance $= R$
$Radius = r$
Resistance, $R = \frac{ρ . l}{A} = \frac{ρ l}{π r^{2}} (∵ A = π r^{2})$
$∵ (r^{'} = \frac{r}{n})$
Volume remains constant-
Initial volume $=$ Final volume
$r^{2} l = {(\frac{r}{n})}^{2} l^{'}$
$r^{2} l = \frac{r^{2}}{n^{2}} l^{'}$
$\frac{l}{l^{'}} = \frac{1}{n^{2}}$
$l^{'} = n^{2} l$
New resistance, $R^{'} = \frac{ρ l^{'}}{A^{'}} = \frac{ρ l^{'}}{π r^{' 2}}$
$R^{'} = \frac{ρ}{π} \times \frac{n^{2} l}{(r / n)^{2}} = n^{4} (\frac{ρ l}{π r^{2}}) = n^{4} R$

COMEDK 2011

Current Electricity

151929 The maximum current that flow in the fuse wire before it blows out, varies with the radius $r$ as

1

r^{3 / 2}

2

r

3

r^{2 / 3}

4

r^{1 / 2}

Explanation:

A The maximum current that flow in the fuse wire before it blows out varies with the radius $r^{3 / 2}$
If $l =$ length of wire
$r =$ radius of wire
$I =$ Current
$Q = h =$ the rate of heat loss then,
Resistance, $R = \frac{ρ l}{A} = \frac{ρ l}{π r^{2}}$
At steady state-
$I_{max}^{2} R = hA$
$I_{max}^{2} \times (\frac{ρ l}{π r^{2}}) = h \times 2 π rl$
$I_{max}^{2} = \frac{2 π^{2} h}{ρ} r^{3}$
$I_{max} \propto r^{3 / 2}$

Manipal UGET-2013

Current Electricity

151930 The resistance of a wire at $300 K$ is found to be $0.3 Ω$ . If the temperature coefficient of resistance of wire is $1.5 \times 10^{- 3} K^{- 1}$ the temperature at which the resistance becomes $0.6 Ω$ is

1

720 K

2

345 K

3

993 K

4

690 K

Explanation:

C Given,
$R_{300} = 0.3 Ω, R_{t} = 0.6 Ω$
$T = 300 K = 27^{\circ} C$
$∵$ Temperature coefficient of resistance
$α = 1.5 \times 10^{- 3} K^{- 1}$
$R_{300} = R_{0} (1 + α \times 27)$
$0.3 = R_{0} (1 + 1.5 \times 10^{- 3} \times 27)$
Again, $R_{t} = R_{0} (1 + α t)$
$0.6 = R_{0} (1 + 1.5 \times 10^{- 3} \times t)$
Dividing equation (ii) and (i), we get-
$\frac{0.6}{0.3} = \frac{1 + 1.5 \times 10^{- 3} t}{1 + 1.5 \times 10^{- 3} \times 27}$
$2 (1 + 1.5 \times 10^{- 3} \times 27) = 1 + 1.5 \times 10^{- 3} t$
$2 + 81 \times 10^{- 3} = 1 + 1.5 \times 10^{- 3} t$
$t = \frac{1.081}{1.5 \times 10^{- 3}} = 720^{\circ} C$
$t = 993 K$

Karnataka CET-2009

Current Electricity

151925 In a Wheatstone's bridge, three resistances $P$ , $Q$ and $R$ are connected in the three arms and the fourth arm is formed by two resistances $S_{1}$ and $S_{2}$ connected in parallel. The condition for the bridge to be balanced will be

1

\frac{P}{Q} = \frac{2 R}{S_{1} + S_{2}}

2

\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{S_{1} S_{2}}

3

\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{2 S_{1} S_{2}}

4

\frac{P}{Q} = \frac{R}{S_{1} + S_{2}}

Explanation:

B The wheatstone bridge diagram, according to question-

$S_{eq} = \frac{S_{1} S_{2}}{S_{1} + S_{2}}$
Where, $S_{1}$ and $S_{2}$ are connected in parallel Now, wheat stone bridge is balanced,
$\frac{P}{R} = \frac{Q}{S_{eq}}$
$\frac{P}{Q} = \frac{R}{S_{eq}} = \frac{R}{(\frac{S_{1} S_{2}}{S_{1} + S_{2}})}$
$\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{S_{1} S_{2}}$

UPSEE - 2010

Current Electricity

151926 Two wires of equal length and equal diameter and having resistivities $ρ_{1}$ and $ρ_{2}$ are connected in series. The equivalent resistivity of the combination is

1

\sqrt{ρ_{1} ρ_{2}}

2

\frac{ρ_{1} + ρ_{2}}{2}

3

\frac{ρ_{1} ρ_{2}}{ρ_{1} + ρ_{2}}

4

(ρ_{1} + ρ_{2})

Explanation:

B We know that,
Resistance, $R = \frac{ρ L}{A}$
Then,
$R_{1} = \frac{ρ_{1} L}{A}, R_{2} = \frac{ρ_{2} L}{A}$
For series connection
$R_{eq} = R_{1} + R_{2}$
$R_{eq} = \frac{ρ_{1} L}{A} + \frac{ρ_{2} L}{A}$
$R_{eq} = (\frac{ρ_{1} + ρ_{2}}{A}) L$
$\frac{ρ_{eq} \times 2 L}{A} = (\frac{ρ_{1} + ρ_{2}}{A}) L$
$ρ_{eq} = \frac{ρ_{1} + ρ_{2}}{2}$

GUJCET 2016

Current Electricity

151927 A uniform wire of resistance $R$ , of the radius $r$ is uniformly drawn until its radius is reduced to $r / n$ . Its new resistance is

1

nR

2

n^{3} R

3

n^{2} R

4

n^{4} R

Explanation:

D :Given,
Uniform wire of resistance $= R$
$Radius = r$
Resistance, $R = \frac{ρ . l}{A} = \frac{ρ l}{π r^{2}} (∵ A = π r^{2})$
$∵ (r^{'} = \frac{r}{n})$
Volume remains constant-
Initial volume $=$ Final volume
$r^{2} l = {(\frac{r}{n})}^{2} l^{'}$
$r^{2} l = \frac{r^{2}}{n^{2}} l^{'}$
$\frac{l}{l^{'}} = \frac{1}{n^{2}}$
$l^{'} = n^{2} l$
New resistance, $R^{'} = \frac{ρ l^{'}}{A^{'}} = \frac{ρ l^{'}}{π r^{' 2}}$
$R^{'} = \frac{ρ}{π} \times \frac{n^{2} l}{(r / n)^{2}} = n^{4} (\frac{ρ l}{π r^{2}}) = n^{4} R$

COMEDK 2011

Current Electricity

151929 The maximum current that flow in the fuse wire before it blows out, varies with the radius $r$ as

1

r^{3 / 2}

2

r

3

r^{2 / 3}

4

r^{1 / 2}

Explanation:

A The maximum current that flow in the fuse wire before it blows out varies with the radius $r^{3 / 2}$
If $l =$ length of wire
$r =$ radius of wire
$I =$ Current
$Q = h =$ the rate of heat loss then,
Resistance, $R = \frac{ρ l}{A} = \frac{ρ l}{π r^{2}}$
At steady state-
$I_{max}^{2} R = hA$
$I_{max}^{2} \times (\frac{ρ l}{π r^{2}}) = h \times 2 π rl$
$I_{max}^{2} = \frac{2 π^{2} h}{ρ} r^{3}$
$I_{max} \propto r^{3 / 2}$

Manipal UGET-2013

Current Electricity

151930 The resistance of a wire at $300 K$ is found to be $0.3 Ω$ . If the temperature coefficient of resistance of wire is $1.5 \times 10^{- 3} K^{- 1}$ the temperature at which the resistance becomes $0.6 Ω$ is

1

720 K

2

345 K

3

993 K

4

690 K

Explanation:

C Given,
$R_{300} = 0.3 Ω, R_{t} = 0.6 Ω$
$T = 300 K = 27^{\circ} C$
$∵$ Temperature coefficient of resistance
$α = 1.5 \times 10^{- 3} K^{- 1}$
$R_{300} = R_{0} (1 + α \times 27)$
$0.3 = R_{0} (1 + 1.5 \times 10^{- 3} \times 27)$
Again, $R_{t} = R_{0} (1 + α t)$
$0.6 = R_{0} (1 + 1.5 \times 10^{- 3} \times t)$
Dividing equation (ii) and (i), we get-
$\frac{0.6}{0.3} = \frac{1 + 1.5 \times 10^{- 3} t}{1 + 1.5 \times 10^{- 3} \times 27}$
$2 (1 + 1.5 \times 10^{- 3} \times 27) = 1 + 1.5 \times 10^{- 3} t$
$2 + 81 \times 10^{- 3} = 1 + 1.5 \times 10^{- 3} t$
$t = \frac{1.081}{1.5 \times 10^{- 3}} = 720^{\circ} C$
$t = 993 K$

Karnataka CET-2009

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